File Upload With <INPUT TYPE="FILE"

How do you handle a file being uploaded from the browser's machine with the html form element <INPUT TYPE
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Are you prepared to program Perl to do this?
icacheAuthor Commented:
Got the Llama book next to me, we're prepared....
First get a copy of the library from

The latest version of this provides for file upload. The above URL will give a description of the program.

Second. You need an input form. This is one I use.

<form method='POST' enctype='multipart/form-data' action=''>
<input type="file" name="upfile" size=25>
<INPUT TYPE="image" NAME="submit" value="submit" src="/img/upload.gif" border=0 alt="upload" hspace=2>

Third. You need to produce your own cgi-script to process the input. The file contents will be in one of the input variables. This is the segment of code I use to save an uploaded file.

require "";

if (open (FUPL, ">../files/uploadfilename.tmp")) {
  print FUPL $in{'upfile'};
  close (FUPL);
} else {
  print "<b>$in{filename}</b> upload failed<br>\n";

And that is all there is to it!

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I'm afraid that works on text files only, doesn't it?
You need some extra processing to handle encoded binary data.
No. It works on binary files as well. (At least I have had no problems so far!).
icacheAuthor Commented:
You've stated that the file's contents will be in the the variable of the element $in('file'}. What I'm getting as the content of that element is just the file's name, not the contents. I haven't had a chance to install and give your suggestion a try, but does findthe contents in that element while stdin alone cannot
Actually it should be in $in{'upfile'} since that is the name of the <input type="file" input field.

Are you sure you used the correct 'enctype'?

Interpretation of stdin without can be quite complex. I suggest you load a copy of

Another possible problem could be that ISTR that Microsoft IE does not support the input type="file". (I have not checked this recently).

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