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File Upload With <INPUT TYPE="FILE"

Posted on 1997-12-16
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171 Views
Last Modified: 2013-12-25
How do you handle a file being uploaded from the browser's machine with the html form element <INPUT TYPE
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Question by:icache
7 Comments
 
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Expert Comment

by:haystor
ID: 1831666
Are you prepared to program Perl to do this?
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Author Comment

by:icache
ID: 1831667
Got the Llama book next to me, we're prepared....
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Accepted Solution

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icd earned 200 total points
ID: 1831668
First get a copy of the cgi-bin.pl library from

http://cgi-lib.stanford.edu/cgi-lib/

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='upload.pl'>
<input type="file" name="upfile" size=25>
<INPUT TYPE="image" NAME="submit" value="submit" src="/img/upload.gif" border=0 alt="upload" hspace=2>
</form>

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 "cgi-bin.pl";
&ReadParse;

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

And that is all there is to it!
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LVL 5

Expert Comment

by:julio011597
ID: 1831669
I'm afraid that works on text files only, doesn't it?
You need some extra processing to handle encoded binary data.
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Expert Comment

by:icd
ID: 1831670
No. It works on binary files as well. (At least I have had no problems so far!).
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Author Comment

by:icache
ID: 1831671
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 cgi-bin.pl and give your suggestion a try, but does cgi-bin.pl findthe contents in that element while stdin alone cannot
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LVL 5

Expert Comment

by:icd
ID: 1831672
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 cgi-bin.pl can be quite complex. I suggest you load a copy of cgi-bin.pl.

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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