HTML adding an if statement to <option><select>

Question

I have an image uploader and I'm trying to make the site display what type of option has been selected when the file is uploaded. I will later add this functionality to connect to different tables in a database, but for now I just want it printing out what option has been selected.

Here is my upload. html:

<body>
<form enctype="multipart/form-data" action="uploader.php" method="POST">
<input type="hidden" name="MAX_FILE_SIZE" value="400000" />
Choose a file to upload:
<br />
<br />
<input name="uploadedfile" type="file" />
<br />
<select name = "uploadType">
<option value="Picture">Picture</option>
<option value="Video">Video</option>
<option value="Fact">Fact</option>
</select>
<br />
<br />
<input type="submit" value="Upload File" />
</form>
</body>

And here is my uploader. php:

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ".  basename( $_FILES['uploadedfile']['name']). 
" has been uploaded";

if (uploadType == "Picture"){
echo "\nYou have uploaded a picture.";

} if (uploadType == "Video"){
echo "\nYou have uploaded a video.";
} if (uploadType == "Fact"){
echo "\nYou have uploaded a fact.";
} else{
echo "\nuploadType fail!";
}

} else{
    echo "There was an error uploading the file, please try again!";
}

Every time I try to upload it just comes up with the "uploadType fail!" message. What am I doing wrong?

Any help would be greatly appreciated!

Thank you so much!

And that's why you turn on error reporting during development, because that would complain about an undefined constant "uploadType"... — deceze
– deceze ♦, Commented May 8, 2012 at 23:59
Just a heads up 400000 is 400kb or just under 1/2 a megabyte, dont confuse it for 4mb or 40mb. MAX_FILE_SIZE is in bytes — Lawrence Cherone
– Lawrence Cherone, Commented May 9, 2012 at 0:08

Evan Mulawski · Accepted Answer · 2012-05-09 00:59:21Z

2

$uploadType = $_POST["uploadType"];

Then use $uploadType in your if statements. Variables in PHP always begin with a dollar sign.

In addition, use else if for the other conditions.

edited May 9, 2012 at 0:59

answered May 8, 2012 at 23:58

Evan Mulawski

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1 Comment

DommyCastles Over a year ago

Works great, I should have known, I just wasn't sure on the exact syntax. Thank you so much!

marapet · Accepted Answer · 2012-05-09 00:09:04Z

1

You are missing two 'else' statements - before the video and before the fact 'if' statement.

answered May 9, 2012 at 0:09

marapet

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1 Comment

marapet Over a year ago

Unless you choose "Fact", the fail message will always appear.

Dan · Accepted Answer · 2012-05-09 00:00:20Z

0

uploadType is being used as a constant here. Try replacing uploadType with $_POST['uploadType'].

answered May 9, 2012 at 0:00

Dan

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