Check if specific input file is empty

Question

In my form I have 3 input fields for file upload:

<input type=file name="cover_image">
<input type=file name="image1">
<input type=file name="image2">

How can I check if cover_image is empty - no file is put for upload?

Rudi Visser · Accepted Answer · 2023-02-03 11:33:02Z

94

You can check by using the size field on the $_FILES array like so:

if ($_FILES['cover_image']['error'] == 4 || ($_FILES['cover_image']['size'] == 0 && $_FILES['cover_image']['error'] == 0))
{
    // cover_image is empty (and not an error), or no file was uploaded
}

(I also check error here because it may be 0 if something went wrong (ie. a file was selected, but there's no data received). I wouldn't use name for this check since that can be overridden). error with a value of 4 is UPLOAD_ERR_NO_FILE, so we can check for that too.

edited Feb 3, 2023 at 11:33

answered Jan 22, 2013 at 12:29

Rudi Visser

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3 Comments

Bentleyo Over a year ago

An error code of "4" can be returned if no file is uploaded. Here is the PHP Reference

Sandhu Over a year ago

I use it like... if ($_FILES['cover_image']['name'] == true){ ... }

Rudi Visser Over a year ago

@Sandhu that's not even remotely valid, and is not what I suggested.

Techie · Accepted Answer · 2013-01-22 12:29:10Z

32

Method 1

if($_FILES['cover_image']['name'] == "") {
// No file was selected for upload, your (re)action goes here
}

Method 2

if($_FILES['cover_image']['size'] == 0) {
// No file was selected for upload, your (re)action goes here
}

answered Jan 22, 2013 at 12:29

Techie

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2 Comments

Touhid Over a year ago

Appropriate method to me. :)

dnns Over a year ago

Instead of if($_FILES['cover_image']['name'] == "") { you can use if(empty($_FILES['cover_image']['name'])) {.

BenM · Accepted Answer · 2013-01-22 12:30:54Z

13

You can check if there is a value, and if the image is valid by doing the following:

if(empty($_FILES['cover_image']['tmp_name']) || !is_uploaded_file($_FILES['cover_image']['tmp_name']))
{
   // Handle no image here...
}

answered Jan 22, 2013 at 12:30

BenM

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Comments

Rakesh Sharma · Accepted Answer · 2013-01-22 12:29:49Z

7

if (empty($_FILES['cover_image']['name']))

answered Jan 22, 2013 at 12:29

Rakesh Sharma

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1 Comment

BenM Over a year ago

That doesn't check the validaty of the file. Better to use is_uploaded_file in combination.

Bruno Ribeiro · Accepted Answer · 2017-12-13 19:44:20Z

5

simple :

if($_FILES['cover_image']['error'] > 0)
    // cover_image is empty

edited Dec 13, 2017 at 19:44

answered Jun 26, 2015 at 0:10

Bruno Ribeiro

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2 Comments

WhiteHorse Over a year ago

Empty File means Error 4. Above fails.

Bruno Ribeiro Over a year ago

@WhiteHorse that's right, if the error code is more than 0, the upload not worked

user7282 · Accepted Answer · 2013-01-22 12:29:14Z

3

check after the form is posted the following

$_FILES["cover_image"]["size"]==0

answered Jan 22, 2013 at 12:29

user7282

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Comments

Johny · Accepted Answer · 2013-01-22 12:32:53Z

3

 if( ($_POST) && (!empty($_POST['cover_image'])) )    //verifies  if post exists and cover_image is not empty
    {
    //execute whatever code you want
    }

answered Jan 22, 2013 at 12:32

Johny

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2 Comments

Wayfarer Over a year ago

I did this one first and then check for empty on $_FILES, otherwise the non existent $_FILES threw an error.

user31782 Over a year ago

This does not work with <input type=file>. Use $_FILES instead and moreover !empty($_FILES["cover_image"]) is not viable.

Arpico ICT · Accepted Answer · 2020-08-28 08:48:29Z

3

    if (!$_FILES['image']['size'][0] == 0){ //}

answered Aug 28, 2020 at 8:48

Arpico ICT

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Comments

Prabhagaran · Accepted Answer · 2015-08-19 06:49:18Z

2

if($_FILES['img_name']['name']!=""){
   echo "File Present";
}else{
  echo "Empty file";
}

answered Aug 19, 2015 at 6:49

Prabhagaran

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Comments

Parth Dhorda · Accepted Answer · 2016-03-24 11:18:15Z

2

if ($_FILES['cover_image']['size'] == 0 && $_FILES['cover_image']['error'] == 0)
{ 
      // Code comes here
}

This thing works for me........

answered Mar 24, 2016 at 11:18

Parth Dhorda

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Comments

kmoser · Accepted Answer · 2020-08-11 05:58:36Z

2

if(!empty($_FILES)) { // code if not uploaded } else { // code if uploaded }

edited Aug 11, 2020 at 5:58

kmoser

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answered Aug 10, 2020 at 21:53

my_name_is_njuno

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Comments

Pavan Poojary · Accepted Answer · 2020-12-25 09:40:47Z

<input type="file" class="custom-file-input" id="imagefile" name="imagefile[]"  multiple lang="en">
<input type="hidden" name="hidden_imagefile[]" value="<?=$row[2]; ?>" class="form-control border-input" >

    if($_FILES['imagefile']['name'] == '')
        {
          $img = $_POST['hidden_imagefile'];
        }
        else{
          $img = '';
          $uploadFolder = 'uploads/gallery/';
          foreach ($_FILES['imagefile']['tmp_name'] as $key => $image) {
            $imageTmpName = time() .$_FILES['imagefile']['tmp_name'][$key];
            $imageName = time() .$_FILES['imagefile']['name'][$key];
            $img .= $imageName.',';
            $result = move_uploaded_file($imageTmpName, $uploadFolder.$img);
          }
          
        }

Samuel Ramzan · Accepted Answer · 2021-06-17 02:50:50Z

2

$_FILES is an associative POST method array, if You want to check anything about $_FILES You must take into account the index... I tried a lot of suggested options, and the only method that worked for me, was when I included an index in my verification method.

$_FILES['Your_File']['name'][0];

So bye doing this:

 if(empty($_FILES['Your_File']['name'][0])){
    print('this thing is empty');
 }else{
    print('Something, something, something');
 }

There's nothing like good old experimentation and lots of reading.

answered Jun 17, 2021 at 2:50

Samuel Ramzan

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Comments

user17837751 · Accepted Answer · 2022-01-05 05:54:02Z

2

if ($_FILES['cover_image']['error'] == 4){
    // the user did not choose any file
}else{
   // the user chose a file to be uploaded
}

answered Jan 5, 2022 at 5:54

user17837751

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Comments

hsnfirdaus · Accepted Answer · 2022-05-04 16:29:40Z

2

This will work

if ($_FILES['cover_image']['size'] == 0 && $_FILES['cover_image']['error'] == 0)

// checking if file is selected and not an error

{ 
      
// file is not selected and it is not an error

}

edited May 4, 2022 at 16:29

hsnfirdaus

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answered Apr 24, 2021 at 4:19

Raj

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Comments

Bahman World · Accepted Answer · 2022-01-12 16:53:24Z

1

UPDATED: Use this method:

First check if 'cover_image' key exists in $_FILES then check other file errors

if (in_array('cover_image', array_keys($_FILES) && $_FILES['cover_image']['error'] == 0) {
  // TODO: write your code
} else {
  // return error
}

edited Jan 12, 2022 at 16:53

answered Jan 12, 2022 at 16:42

Bahman World

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