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Assistance with uploading pictures

We are putting together a non-profit sit to help track missing children and part of the website asks for a face shot and a body shot picture of the child.

I need to be able to save the two files in a predetermined directory on the server.

I have the asp form where the officers can enter the information about the missing person and it allows them to select pics but I need help with an upload asp.

The missing person wil have an ID say 151 and there will be a headshot and body pic and we want to save them as 151_hs.jpg or 151_bdy.jpg no matter what the name of the input file is.

Any help would be greatly appreciated.
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pinkstonm
Asked:
pinkstonm
  • 15
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1 Solution
 
hc0904pcdCommented:
do you have an upload component available on the server to use, ie saupload or aspupload?
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hc0904pcdCommented:
if you do that will be easier & better
if not, this link might help http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=8525&lngWId=4
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hc0904pcdCommented:
could you put up the relevent code from your processing page, that handles the form submission?
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pinkstonmAuthor Commented:
i d not have an uploader but have been looking at several of the freebies out there
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pinkstonmAuthor Commented:
Hc,

Thanks I read the material from planet source and while it is a bit above my head I had a question before I get started.  One of the scenerios said that you can actually upload pictures into an Access DB which is what we are using.  I wanted to understand the pros and cons to doing this versus uploading the pics to a folder and having an entry in the DB pointing to them.

Thanks
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hc0904pcdCommented:
hi,
personally i prefer loading an image and having an entry in the db relating to it, it seems cleaner somehow than putting the actual image into the db, which I imagine would make the db a bit bloated.

Essentially you upload an image to a folder on your site, and then put the filename into the db record, along with any other info you want to track. To use the image you read the db record, get the filename, add it to the folder name, and that's where the image is.

I'd suggest using some sample code from the link above, and drop a copy of your code as you go, so I can chip in with advice.

It's fairly easy, and I'm sure you'll have no problem with it.


 
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pinkstonmAuthor Commented:
Okay keep your eyes out as I am no pro :)
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pinkstonmAuthor Commented:
Okay well I tried....

I followed the simple instructions and go an error.

***testForm.asp***
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>New Page 1</title>
</head>
<body>
<FORM method="post" encType="multipart/form-data" action="testUpload.asp">
<INPUT type="File" name="File1">
<INPUT type="Submit" value="Upload">
</FORM>
</body>
</html>


***testUpload.asp***
<!--#INCLUDE FILE="upload/clsUpload.asp"-->
<%
Dim Upload
Dim Folder
Set Upload = New clsUpload
Folder = Server.MapPath("submissions") & "\"
Upload("File1").SaveAs Folder & Upload("File1").FileName
Set Upload = Nothing
%>

The error I get is:
Microsoft VBScript runtime error '800a0046'
Permission denied
/upload/clsField.asp, line 227

 Folder submissions has write access set to general
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pinkstonmAuthor Commented:
Here is really what I am trying to accomplish with their code

***testForm.asp***
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>New Page 1</title>
</head>
<body>
<FORM method="post" encType="multipart/form-data" action="testUpload.asp">
<INPUT type="File" name="File1">
<INPUT type="File" name="File2">
<INPUT type="Submit" value="Upload">
</FORM>
</body>
</html>


***testUpload.asp***
<!--#INCLUDE FILE="upload/clsUpload.asp"-->
<%
Dim Upload
Dim Folder
clientid = 1500

Set Upload = New clsUpload
Folder = Server.MapPath("submissions") & "\"

' File 1 would be headshot
Upload.RenameFile Folder & FileName, clientid & "_head.jpg"
Upload("File1").SaveAs Folder & FileName

' File 1 would be body
Upload.RenameFile Folder & FileName, clientid & "_body.jpg"
Upload("File2").SaveAs Folder & FileName

Set Upload = Nothing
%>
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pinkstonmAuthor Commented:
When I tried that method I got

Microsoft VBScript runtime error '800a0035'

File not found

/upload/clsUpload.asp, line 820
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hc0904pcdCommented:
for the first error,

The error I get is:
Microsoft VBScript runtime error '800a0046'
Permission denied
/upload/clsField.asp, line 227

 Folder submissions has write access set to general

I should have mentioned this earlier ... the folder where you want to load images must be a folder which you can write to, as opposed to read-only access which most folders have by default.

This means that to set up the uploads script, you must first set permissions.
If you are using a hosting company, it's easy. You ask them to set permissions for /xxx/ folder so that it can be used for uploading files. They know what that means, it's a common request, and they know how to do it. When it's done, you won't see the "Permission denied" error again.

Here's a tip - if you think you might want flexibility to use subfolders, or to rename them, then create a generic folder and call it something like "uploads", or "file_library", or anything that makes sense to you. Then make a subfolder you're going to use for uploads. Now ask that the hosting company to set permissions on the first folder including any subdirectories. That way if you need to you can change the name of the subfolder, create new ones etc without having to get the permissions reset.

If you host your own server, then you probably know how to set permissions.

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pinkstonmAuthor Commented:
well the folder is set to read/write general and I am still getting errors.
0
 
pinkstonmAuthor Commented:
The error I am getting now is

Microsoft VBScript runtime error '800a0035'
File not found
/upload/clsUpload.asp, line 820

Based on the following code:

***testForm.asp***
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>New Page 1</title>
</head>
<body>
<FORM method="post" encType="multipart/form-data" action="testUpload.asp">
<INPUT type="File" name="File1">
<INPUT type="File" name="File2">
<INPUT type="Submit" value="Upload">
</FORM>
</body>
</html>


***testUpload.asp***
<!--#INCLUDE FILE="upload/clsUpload.asp"-->
<%
Dim Upload
Dim Folder
clientid = 1500

Set Upload = New clsUpload
Folder = Server.MapPath("submissions") & "\"

' File 1 would be headshot
Upload.RenameFile Folder & FileName, clientid & "_head.jpg"
Upload("File1").SaveAs Folder & FileName

' File 1 would be body
Upload.RenameFile Folder & FileName, clientid & "_body.jpg"
Upload("File2").SaveAs Folder & FileName

Set Upload = Nothing
%>
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pinkstonmAuthor Commented:
Does anyone have any better suggestions or a sample of how to do this.  The old version out on planet source apparently does not work.

Any help would be greatly appreciated!
0
 
pinkstonmAuthor Commented:
Got a base upload to work so the permissions error from the other sample code was not true.

Here is the Base Form:
<%@ Language=VBScript %>
<%
option explicit
Response.Expires = -1
Server.ScriptTimeout = 600
%>
<!-- #include file="Upload/freeaspupload.asp" -->
<%
  uploadsDirVar = "xxxxx"

function OutputForm()
%>
    <form name="frmSend" method="POST" enctype="multipart/form-data" action="uploadTester.asp" onSubmit="return onSubmitForm();">
      <B>File names:</B><br>
    File 1: <input name="attach1" type="file" size=35><br>
    File 2: <input name="attach2" type="file" size=35><br>
    <br>
    <input style="margin-top:4" type=submit value="Upload">
    </form>
<%
end function
function TestEnvironment()
    Dim fso, fileName, testFile, streamTest
    TestEnvironment = ""
    Set fso = Server.CreateObject("Scripting.FileSystemObject")
    if not fso.FolderExists(uploadsDirVar) then
        TestEnvironment = "<B>Folder " & uploadsDirVar & " does not exist.</B><br>The value of your uploadsDirVar is incorrect. Open uploadTester.asp in an editor and change the value of uploadsDirVar to the pathname of a directory with write permissions."
        exit function
    end if
    fileName = uploadsDirVar & "\test.txt"
    on error resume next
    Set testFile = fso.CreateTextFile(fileName, true)
    If Err.Number<>0 then
        TestEnvironment = "<B>Folder " & uploadsDirVar & " does not have write permissions.</B><br>The value of your uploadsDirVar is incorrect. Open uploadTester.asp in an editor and change the value of uploadsDirVar to the pathname of a directory with write permissions."
        exit function
    end if
    Err.Clear
    testFile.Close
    fso.DeleteFile(fileName)
    If Err.Number<>0 then
        TestEnvironment = "<B>Folder " & uploadsDirVar & " does not have delete permissions</B>, although it does have write permissions.<br>Change the permissions for IUSR_<I>computername</I> on this folder."
        exit function
    end if
    Err.Clear
    Set streamTest = Server.CreateObject("ADODB.Stream")
    If Err.Number<>0 then
        TestEnvironment = "<B>The ADODB object <I>Stream</I> is not available in your server.</B><br>Check the Requirements page for information about upgrading your ADODB libraries."
        exit function
    end if
    Set streamTest = Nothing
end function
function SaveFiles
    Dim Upload, fileName, fileSize, ks, i, fileKey

    Set Upload = New FreeASPUpload
    Upload.Save(uploadsDirVar)

      ' If something fails inside the script, but the exception is handled
      If Err.Number<>0 then Exit function

    SaveFiles = ""
    ks = Upload.UploadedFiles.keys
    if (UBound(ks) <> -1) then
        SaveFiles = "<B>Files uploaded:</B> "
        for each fileKey in Upload.UploadedFiles.keys
            SaveFiles = SaveFiles & Upload.UploadedFiles(fileKey).FileName & " (" & Upload.UploadedFiles(fileKey).Length & "B) "
        next
    else
        SaveFiles = "The file name specified in the upload form does not correspond to a valid file in the system."
    end if
      SaveFiles = SaveFiles & "<br>Enter a number = " & Upload.Form("enter_a_number") & "<br>"
      SaveFiles = SaveFiles & "Checkbox values = " & Upload.Form("checkbox_values") & "<br>"
end function
%>

<HTML>
<HEAD>
<TITLE>Test Free ASP Upload</TITLE>
<style>
BODY {background-color: white;font-family:arial; font-size:12}
</style>
<script>
function onSubmitForm() {
    var formDOMObj = document.frmSend;
    if (formDOMObj.attach1.value == "" && formDOMObj.attach2.value == "" && formDOMObj.attach3.value == "" && formDOMObj.attach4.value == "" )
        alert("Please press the browse button and pick a file.")
    else
        return true;
    return false;
}
</script>
</HEAD>
<BODY>
<br><br>
<div style="border-bottom: #A91905 2px solid;font-size:16">Upload files to your server</div>
<%
Dim diagnostics
if Request.ServerVariables("REQUEST_METHOD") <> "POST" then
    diagnostics = TestEnvironment()
    if diagnostics<>"" then
        response.write "<div style=""margin-left:20; margin-top:30; margin-right:30; margin-bottom:30;"">"
        response.write diagnostics
        response.write "<p>After you correct this problem, reload the page."
        response.write "</div>"
    else
        response.write "<div style=""margin-left:150"">"
        OutputForm()
        response.write "</div>"
    end if
else
    response.write "<div style=""margin-left:150"">"
    OutputForm()
    response.write SaveFiles()
    response.write "<br><br></div>"
end if
%>
</BODY>
</HTML>


Here is the called ASP that does the upload:
<%
'  For examples, documentation, and your own free copy, go to:
'  http://www.freeaspupload.net
'  Note: You can copy and use this script for free and you can make changes
'  to the code, but you cannot remove the above comment.

'Changes:
'Aug 2, 2005: Add support for checkboxes and other input elements with multiple values

Class FreeASPUpload
      Public UploadedFiles
      Public FormElements

      Private VarArrayBinRequest
      Private StreamRequest
      Private uploadedYet

      Private Sub Class_Initialize()
            Set UploadedFiles = Server.CreateObject("Scripting.Dictionary")
            Set FormElements = Server.CreateObject("Scripting.Dictionary")
            Set StreamRequest = Server.CreateObject("ADODB.Stream")
            StreamRequest.Type = 1 'adTypeBinary
            StreamRequest.Open
            uploadedYet = false
      End Sub
      
      Private Sub Class_Terminate()
            If IsObject(UploadedFiles) Then
                  UploadedFiles.RemoveAll()
                  Set UploadedFiles = Nothing
            End If
            If IsObject(FormElements) Then
                  FormElements.RemoveAll()
                  Set FormElements = Nothing
            End If
            StreamRequest.Close
            Set StreamRequest = Nothing
      End Sub

      Public Property Get Form(sIndex)
            Form = ""
            If FormElements.Exists(LCase(sIndex)) Then Form = FormElements.Item(LCase(sIndex))
      End Property

      Public Property Get Files()
            Files = UploadedFiles.Items
      End Property

      'Calls Upload to extract the data from the binary request and then saves the uploaded files
      Public Sub Save(path)
            Dim streamFile, fileItem

            if Right(path, 1) <> "\" then path = path & "\"

            if not uploadedYet then Upload

            For Each fileItem In UploadedFiles.Items
                  Set streamFile = Server.CreateObject("ADODB.Stream")
                  streamFile.Type = 1
                  streamFile.Open
                  StreamRequest.Position=fileItem.Start
                  StreamRequest.CopyTo streamFile, fileItem.Length
                  streamFile.SaveToFile path & fileItem.FileName, 2
                  streamFile.close
                  Set streamFile = Nothing
                  fileItem.Path = path & fileItem.FileName
             Next
      End Sub

      Public Function SaveBinRequest(path) ' For debugging purposes
            StreamRequest.SaveToFile path & "\debugStream.bin", 2
      End Function

      Public Sub DumpData() 'only works if files are plain text
            Dim i, aKeys, f
            response.write "Form Items:<br>"
            aKeys = FormElements.Keys
            For i = 0 To FormElements.Count -1 ' Iterate the array
                  response.write aKeys(i) & " = " & FormElements.Item(aKeys(i)) & "<BR>"
            Next
            response.write "Uploaded Files:<br>"
            For Each f In UploadedFiles.Items
                  response.write "Name: " & f.FileName & "<br>"
                  response.write "Type: " & f.ContentType & "<br>"
                  response.write "Start: " & f.Start & "<br>"
                  response.write "Size: " & f.Length & "<br>"
             Next
         End Sub

      Private Sub Upload()
            Dim nCurPos, nDataBoundPos, nLastSepPos
            Dim nPosFile, nPosBound
            Dim sFieldName, osPathSep, auxStr

            'RFC1867 Tokens
            Dim vDataSep
            Dim tNewLine, tDoubleQuotes, tTerm, tFilename, tName, tContentDisp, tContentType
            tNewLine = Byte2String(Chr(13))
            tDoubleQuotes = Byte2String(Chr(34))
            tTerm = Byte2String("--")
            tFilename = Byte2String("filename=""")
            tName = Byte2String("name=""")
            tContentDisp = Byte2String("Content-Disposition")
            tContentType = Byte2String("Content-Type:")

            uploadedYet = true

            on error resume next
            VarArrayBinRequest = Request.BinaryRead(Request.TotalBytes)
            if Err.Number <> 0 then
                  response.write "<br><br><B>System reported this error:</B><p>"
                  response.write Err.Description & "<p>"
                  response.write "The most likely cause for this error is the incorrect setup of AspMaxRequestEntityAllowed in IIS MetaBase. Please see instructions in the <A HREF='http://www.freeaspupload.net/freeaspupload/requirements.asp'>requirements page of freeaspupload.net</A>.<p>"
                  Exit Sub
            end if
            on error goto 0 'reset error handling

            nCurPos = FindToken(tNewLine,1) 'Note: nCurPos is 1-based (and so is InstrB, MidB, etc)

            If nCurPos <= 1  Then Exit Sub
             
            'vDataSep is a separator like -----------------------------21763138716045
            vDataSep = MidB(VarArrayBinRequest, 1, nCurPos-1)

            'Start of current separator
            nDataBoundPos = 1

            'Beginning of last line
            nLastSepPos = FindToken(vDataSep & tTerm, 1)

            Do Until nDataBoundPos = nLastSepPos
                  
                  nCurPos = SkipToken(tContentDisp, nDataBoundPos)
                  nCurPos = SkipToken(tName, nCurPos)
                  sFieldName = ExtractField(tDoubleQuotes, nCurPos)

                  nPosFile = FindToken(tFilename, nCurPos)
                  nPosBound = FindToken(vDataSep, nCurPos)
                  
                  If nPosFile <> 0 And  nPosFile < nPosBound Then
                        Dim oUploadFile
                        Set oUploadFile = New UploadedFile
                        
                        nCurPos = SkipToken(tFilename, nCurPos)
                        auxStr = ExtractField(tDoubleQuotes, nCurPos)
                ' We are interested only in the name of the file, not the whole path
                ' Path separator is \ in windows, / in UNIX
                ' While IE seems to put the whole pathname in the stream, Mozilla seem to
                ' only put the actual file name, so UNIX paths may be rare. But not impossible.
                osPathSep = "\"
                if InStr(auxStr, osPathSep) = 0 then osPathSep = "/"
                        oUploadFile.FileName = Right(auxStr, Len(auxStr)-InStrRev(auxStr, osPathSep))

                        if (Len(oUploadFile.FileName) > 0) then 'File field not left empty
                              nCurPos = SkipToken(tContentType, nCurPos)
                              
                    auxStr = ExtractField(tNewLine, nCurPos)
                    ' NN on UNIX puts things like this in the streaa:
                    '    ?? python py type=?? python application/x-python
                              oUploadFile.ContentType = Right(auxStr, Len(auxStr)-InStrRev(auxStr, " "))
                              nCurPos = FindToken(tNewLine, nCurPos) + 4 'skip empty line
                              
                              oUploadFile.Start = nCurPos-1
                              oUploadFile.Length = FindToken(vDataSep, nCurPos) - 2 - nCurPos
                              
                              If oUploadFile.Length > 0 Then UploadedFiles.Add LCase(sFieldName), oUploadFile
                        End If
                  Else
                        Dim nEndOfData
                        nCurPos = FindToken(tNewLine, nCurPos) + 4 'skip empty line
                        nEndOfData = FindToken(vDataSep, nCurPos) - 2
                        If Not FormElements.Exists(LCase(sFieldName)) Then
                              FormElements.Add LCase(sFieldName), String2Byte(MidB(VarArrayBinRequest, nCurPos, nEndOfData-nCurPos))
                        else
                    FormElements.Item(LCase(sFieldName))= FormElements.Item(LCase(sFieldName)) & ", " & String2Byte(MidB(VarArrayBinRequest, nCurPos, nEndOfData-nCurPos))
                end if

                  End If

                  'Advance to next separator
                  nDataBoundPos = FindToken(vDataSep, nCurPos)
            Loop
            StreamRequest.Write(VarArrayBinRequest)
      End Sub

      Private Function SkipToken(sToken, nStart)
            SkipToken = InstrB(nStart, VarArrayBinRequest, sToken)
            If SkipToken = 0 then
                  Response.write "Error in parsing uploaded binary request."
                  Response.End
            end if
            SkipToken = SkipToken + LenB(sToken)
      End Function

      Private Function FindToken(sToken, nStart)
            FindToken = InstrB(nStart, VarArrayBinRequest, sToken)
      End Function

      Private Function ExtractField(sToken, nStart)
            Dim nEnd
            nEnd = InstrB(nStart, VarArrayBinRequest, sToken)
            If nEnd = 0 then
                  Response.write "Error in parsing uploaded binary request."
                  Response.End
            end if
            ExtractField = String2Byte(MidB(VarArrayBinRequest, nStart, nEnd-nStart))
      End Function

      'String to byte string conversion
      Private Function Byte2String(sString)
            Dim i
            For i = 1 to Len(sString)
               Byte2String = Byte2String & ChrB(AscB(Mid(sString,i,1)))
            Next
      End Function

      'Byte string to string conversion
      Private Function String2Byte(bsString)
            Dim i
            String2Byte =""
            For i = 1 to LenB(bsString)
               String2Byte = String2Byte & Chr(AscB(MidB(bsString,i,1)))
            Next
      End Function
End Class

Class UploadedFile
      Public ContentType
      Public Start
      Public Length
      Public Path
      Private nameOfFile

    ' Need to remove characters that are valid in UNIX, but not in Windows
    Public Property Let FileName(fN)
        nameOfFile = fN
        nameOfFile = SubstNoReg(nameOfFile, "\", "_")
        nameOfFile = SubstNoReg(nameOfFile, "/", "_")
        nameOfFile = SubstNoReg(nameOfFile, ":", "_")
        nameOfFile = SubstNoReg(nameOfFile, "*", "_")
        nameOfFile = SubstNoReg(nameOfFile, "?", "_")
        nameOfFile = SubstNoReg(nameOfFile, """", "_")
        nameOfFile = SubstNoReg(nameOfFile, "<", "_")
        nameOfFile = SubstNoReg(nameOfFile, ">", "_")
        nameOfFile = SubstNoReg(nameOfFile, "|", "_")
    End Property

    Public Property Get FileName()
        FileName = nameOfFile
    End Property

    'Public Property Get FileN()ame
End Class


' Does not depend on RegEx, which is not available on older VBScript
' Is not recursive, which means it will not run out of stack space
Function SubstNoReg(initialStr, oldStr, newStr)
    Dim currentPos, oldStrPos, skip
    If IsNull(initialStr) Or Len(initialStr) = 0 Then
        SubstNoReg = ""
    ElseIf IsNull(oldStr) Or Len(oldStr) = 0 Then
        SubstNoReg = initialStr
    Else
        If IsNull(newStr) Then newStr = ""
        currentPos = 1
        oldStrPos = 0
        SubstNoReg = ""
        skip = Len(oldStr)
        Do While currentPos <= Len(initialStr)
            oldStrPos = InStr(currentPos, initialStr, oldStr)
            If oldStrPos = 0 Then
                SubstNoReg = SubstNoReg & Mid(initialStr, currentPos, Len(initialStr) - currentPos + 1)
                currentPos = Len(initialStr) + 1
            Else
                SubstNoReg = SubstNoReg & Mid(initialStr, currentPos, oldStrPos - currentPos) & newStr
                currentPos = oldStrPos + skip
            End If
        Loop
    End If
End Function
%>


What I need help with is when the files upload I need them named something other than the names selected.  The naming should be a client#_head.ext and client#_body.ext.

I only want them to input jpg or gif files.
0
 
hc0904pcdCommented:
for checking file extensions, you can try this snippet,

Ext = Upload("File1").FileExt
Select Case Ext
Case "GIF", "JPG"
    FileName = Upload("File1").FileName
    Folder = Server.MapPath("Uploads") & "\"
    Upload("File1").SaveAs Folder & FileName
Case Else
    Response.Write "File type not supported."
End Select
0
 
hc0904pcdCommented:
sorry, ignore that, you're using a different upload script now so that won't work.
0
 
pinkstonmAuthor Commented:
will you be able to help with this script?

The other one continually got permission errors and this one works
0
 
hc0904pcdCommented:
According to the documentation for this script, the sample upload page uploadTester.asp has a form validation JavaScript function ("onSubmitForm") where you can test for filetypes.

Could you replace the script in your html with this, and see if it works?

<script>
/* original onSubmitForm function
function onSubmitForm() {
    var formDOMObj = document.frmSend;
    if (formDOMObj.attach1.value == "" && formDOMObj.attach2.value == "" && formDOMObj.attach3.value == "" && formDOMObj.attach4.value == "" )
        alert("Please press the browse button and pick a file.")
    else
        return true;
    return false;
}
*/
function onSubmitForm() {
      var formDOMObj = document.frmSend;

      //set 2 flags, 1 for whether attachments are empty, 1 for whether they are correct format
      var checkFilesAreAttached = 0;
      var checkFileFormats = 0;

      //check that two files have been selected
      if ( formDOMObj.attach1.value == "" || formDOMObj.attach2.value == "" ){
            checkFilesAreAttached = 0;
      } else {            
            checkFilesAreAttached = 1;
      }

      //check that file 1 is in the correct format, either jpg or gif      
      if ( formDOMObj.attach1.value.indexOf("jpg") == -1 && formDOMObj.attach1.value.indexOf("gif") == -1 ) {      
            checkFileFormats = 0;
      } else {            
            //check that file 2 is in the correct format, either jpg or gif      
            if ( formDOMObj.attach2.value.indexOf("jpg") == -1 && formDOMObj.attach2.value.indexOf("gif") == -1 ) {      
                  checkFileFormats = 0;      
            } else {
                  checkFileFormats = 1;
            }
      }
            
      if(checkFilesAreAttached == 0) {
            //first check has failed, which means they did not select two files to be uploaded
            alert("Please press the browse button and pick a file.");
            return false;
      } else {
            if(checkFileFormats == 0) {      
            //second check has failed, which means that oine of the files is neither jpg nor gif
                  alert("Please ensure you select only JPG or GIF images to upload");
            return false;
                  } else {
                  return true;
                  }
            }
      }
}
</script>
0
 
hc0904pcdCommented:
To rename the files, you will have to do this after uploading.
The documentation makes it clear that this script does not have a saveas function, or a rename function.

Since you've got this script working I think it's worth sticking with it, so lets look at renaming the files using fso.

At the moment of uploading, do you know the client id?
0
 
hc0904pcdCommented:
here's the fuction for renaming the file

'renames a file using fso
Function fsoMoveFile(folderPath, fileName, newFileName)
      Set fso = CreateObject("Scripting.FileSystemObject")
      Set object = fso.GetFile(folderPath & "\" & fileName)
      object.Move(folderPath & "\" & newFileName)
      Set object = Nothing
      Set fso = Nothing
      fsoMoveFile = True
End Function
0
 
hc0904pcdCommented:
When you call that function, which I think will be after the upload process has finished? , you need to pass 3 parameters - the folder path, the uploaded filemame, and your desired new filename.  

0
 
hc0904pcdCommented:

to get your new filenames, first set out what it is currently, as uploaded.

uploadedFileName1 =
uploadedFileName2 =

'you could also add an additional check here, that the file suffix is either jpg or gif, if necessary
'if ( trim(lcase(right(uploadedFileName1,3))) <> "jpg" and trim(lcase(right(uploadedFileName1,3))) <> "gif" ) then ...
'if ( trim(lcase(right(uploadedFileName2,3))) <> "jpg" and trim(lcase(right(uploadedFileName2,3))) <> "gif" ) then ...

newFileName1 = client_id & "_head." & right(uploadedFileName1,3)
newFileName2 = client_id & "_body." & right(uploadedFileName2,3)

fsoMoveFile(folderPath, fileName1, newFileName1)
fsoMoveFile(folderPath, fileName2, newFileName2)

does that make sense?


once you get it working, you could also think about add some checks.
this code does not check that the file exists before attempting to rename it. in theory you wouldn't need to, you've just uploaded it so you know it's there, but still ...
this code also does not check that the new filename does not already exist, which could possibly occur if you load an image for the same client twice. if that is likely to ever happen, let me know and I'll help you add 2 extra steps using fso - fileExists and deleteFile - which effectively turn the rename function into an overwrite.





0
 
hc0904pcdCommented:
how did you go with this?
0
 
hc0904pcdCommented:
Hi,
From looking at Q_23169140.htm it looks like you've taken a new approach to uploading images?
If that's the case, could you close or delete this Q, and I'll stop monitoring this question.
(and if that's not the case, let me know and I'll stay on the case)

thanks.
0
 
pinkstonmAuthor Commented:
Definitley took a new approach but you assistance was great!
0
 
hc0904pcdCommented:
okay, good luck with the website
0

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