I am a learner in PHP - I have been using Larry Ullman's Peachpit beginners
book which I have found useful to take me through the basics.
But the examples were written before the general applicaton of the new
$_POST, $_GET and similar arrays. It has been valuable learning to change
the code for that (I use version 4.2. (and a bit) and Larry Ullman is very
helpful on his own website forum with readers' queries on the book examples.
But I am having real trouble making a file upload programme work. It needs
substitution of $FileName given in the example with $_FILES array values
which I have done and use of isset instead of a raw if(). But I still
cannot get it to work
May I put the code here and ask if anyone can tell me the problme.
Incidentally I use Windows ME with Apache 2.0.36 and PHP 4.2.. The files
upload is on with a max size of 2M and no specific directory given for a
temp store (the INI says it will use the default directory.)
My test out script is as follows::-
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Uploader</title>
</head>
<body>
<?php
if(isset($_FILES['File'])) {
$File=$_FILES['File'];
$File_name=$_FILES['File']['name'];
$File_size=$_FILES['File']['size'];
echo ("<p>$File_name</p><br />\n");
echo ($_FILES['File']['name']);
print("File name: $File_name<p />\n");
print("File size: $File_size<p />\n");
if (copy ($_FILES["File"]["tmp_name"],"users/$File_name")) {
print("The file was sucessfully uploaded.<p />\n");
}else{
print("Your file could not be copied unfortunately.<p />\n");
}
unlink($File);
}
echo ("Upload a file to the server.");
echo ("<Form action=\"FileUpload.php\" method=POST
enctype=\multipart/form-data\>\n");
echo ("File: <input type=file name=\"File\"><br />\n");
echo ("<input type=submit name=\"submit\" value=\"Press here to
upload.\"><br /></ form>\n");
//test whether anything prints for the FILES array
echo ("<p><em><font color=\"green\">Can I get it to print the variable
outside the conditional? i.e. this one ----
{$_FILES['File']['name']}.</font></em></p>\n");
//set any old variable and print it to be sure basics work
$TestPrint="<p><big><font color=\"purple\">Print this out
then.</font></big></p>.\n";
echo($TestPrint);
?>
</body>
</html>
Is there something I've missed in the script. Or perhaps Apache 2 is the
problem?
Thanks
Regards
Adrian Greeman
Telephone +44 20 8672 9661
Mobile +44 780 329 7447
E-mail adresses
[EMAIL PROTECTED]
or
[EMAIL PROTECTED]
**Please note - there is a 400kB filter on my e-mail so if you send a large
item (eg a JPEG pic) please also send a separate smaller message to tell me
the other one is coming.
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