empty function in php not working right

Question

Hello for some reason I can't get my program to see if my file is empty or not if something is selected it should not go through but it is. Here is the code

if (empty($_FILES['file'] ) )
    {
        echo "testing";
        $seterror =1;
        returnBack();

    }

what exactly isn't working? try using "var_dump($_FILES);" & "echo $_FILES['file'];" to make sure that it's what you expect. — MatthewSot
– MatthewSot, Commented Sep 4, 2012 at 23:54
yeah, but what's not working? what are you having a problem with? is it echoing "testing" or not with that $_FILE array? — MatthewSot
– MatthewSot, Commented Sep 5, 2012 at 0:06

deanWombourne · Accepted Answer · 2012-09-04 23:55:49Z

2

What do you get if you output the files array? (i.e. print_r($_FILES);)

Your code won't tell you if the contents of the file called file is empty, it will tell you if the value of $_FILES['file'] is empty, which might not be what you want. You also might not want the logic of empty - have you tried this to see if there is a difference?

if (array_key_exists('file', $_FILES)) {
    ...
}

answered Sep 4, 2012 at 23:55

deanWombourne

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1 Comment

David Biga Over a year ago

if works but even if I don't enter a file it still goes through.

MatthewSot · Accepted Answer · 2012-09-05 00:36:57Z

2

From the var_dump($_FILES) you provided in the comments, it looks like it isn't echoing "testing" (i.e. it's not working) because $_FILE['file'] isn't empty.

from what you provided me:

$_FILES['file'] = array(5) { ["name"]=> string(0) "" ["type"]=> string(0) "" ["tmp_name"]=> string(0) "" ["error"]=> int(4) ["size"]=> int(0) }

I'm not completely sure how empty() handles arrays (the documentation just says it will return true for an empty array), but at the very least you'd need to either get rid of $_FILES['file']['error'] (which isn't empty since it's an int(4)), or use empty() to check something like $_FILES['file']['name']

so, using my suggestion, here's how the code would look:

if (empty($_FILES['file']['name'])) {
    echo "testing";
    $seterror = 1;
    returnBack();    
}

or (since there might be something wrong with my understanding of PHP arrays), try this:

if(empty($_FILES['file']['name'])) {
    echo "testing";
    $seterror = 1;
    returnBack();
}

edited Sep 5, 2012 at 0:36

answered Sep 5, 2012 at 0:15

MatthewSot

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4 Comments

David Biga Over a year ago

The problem is with using this It always see that the file is not empty and i have so clue why

MatthewSot Over a year ago

@DavidBiga - sorry, I made an error in my code (which I just fixed), try with it now :)

MatthewSot Over a year ago

o.o that should work...could you do var_dump($_FILES['file']['name']); ?

David Biga Over a year ago

just fixed it :) thanks for the help but I needed to use error instead of name

staticsan · Accepted Answer · 2012-09-05 00:36:17Z

1

The language construct empty() and arrays in PHP do not mix well. In addition, if you have a <input type='file' name='file' /> in your form, you will get a $_FILES[] array. You should be doing if( !array_key_exists('file', $_FILES) ) { // no file uploaded } to start with.

In fact, you should be doing several checks:

If the file element is populated in $_FILES
If there is no error (that is, $_FILES['file']['error'] == 0)
If there is a filename for the uploaded file's temporary location
If PHP believes it is an uploaded file and/or it can move it

Only of all of those are true will you have a file to process. Otherwise you don't.

I've found following these steps religiously on a file upload handler to be thoroughly reliable.

edited Sep 5, 2012 at 0:36

answered Sep 5, 2012 at 0:23

staticsan

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2 Comments

David Biga Over a year ago

Only when I do not select a file.

staticsan Over a year ago

$_FILES['file']['error'] will be 0 if there is a file uploaded. If it is not 0 at that step (or it doesn't exist in the array), you do not have a file.

Zoltan Toth · Accepted Answer · 2012-09-05 01:06:27Z

1

If you are testing for empty files here is the code

$file = 'test.txt';

if(filesize($file) == NULL) {
    echo "empty"; 
}

edited Sep 5, 2012 at 1:06

Zoltan Toth

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answered Sep 5, 2012 at 0:01

Sudeep

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1 Comment

David Biga Over a year ago

worked except even if it does have a file it still goes through

The Alpha · Accepted Answer · 2012-09-05 02:03:21Z

1

I think you can try this

if(empty($_FILES['file']['name']))
{
    // not uploaded
}
else
{
    // uploded
}

edited Sep 5, 2012 at 2:03

answered Sep 5, 2012 at 0:15

The Alpha

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Comments

David Biga · Accepted Answer · 2012-09-05 00:35:08Z

0

Okay so what I had to do was check to see if it error - so if it did it would not go through because it would mean its empty

code:

if (!empty($_FILES['file']['error'] ) )
    {

        echo "file";
        $seterror =1;
        returnBack();

    }

Thank you so much for the help guys!

answered Sep 5, 2012 at 0:35

David Biga

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1 Comment

staticsan Over a year ago

You should be checking that $_FILES['file']['error'] is or is not 0. The construct empty() just happens to work because a value of 0 is interpreted as "empty" by PHP.

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