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#1 |
a.k.a. Sparky
Join Date: Sep 2004
Location: West Palm Beach, FL, USA
Posts: 2,396
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HTML:
Code:
<form method="post" enctype="multipart/form-data"> <input name="uploadedfile" type="file"> <input type="submit" name="submit" value="Upload"> </form> Code:
if (!(file_exists($_FILES['uploadedfile']['tmp_name']))) { $error .= "Please try again, I didn't get a file. "; } else { // do something with the file }
__________________
SnapReplay.com a different way to share photos - iPhone & Android |
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#2 |
They have the Internet on computers, now?
Join Date: Apr 2005
Posts: 148
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Howling Wulf,
I have tried your input field and submitted a form but the resulting output gives me a local file that I can't open. ie: E:\\My Pictures\\2006-03-23, Rumbelinas II\\Rumbelinas II 001.jpg since this file is on my computer only how could I open an image if somebody submitted one to me and it was located on their local harddrive, not a server? cd34, I do not understeand where I would put your code(s) in my form |
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#3 |
Me fail English? That's unpossible!
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The input field is put between the <form> tags, and you'll need code to handle the uploaded file. cd34's post has full details you need.
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