Converting Array in PHP ($_FILES)

Question

I would like to handle the multiple file uploads $_FILES in an array like this

foreach ($_FILES as $file) {
 // do stuff...
 // $file['name'] and as such
}

However the array seems something like this

Array ( 
    [name] => Array ( 
        [0] => 2010-10-04_205047.jpg 
        [1] => 
        [2] => 
    ) 

    [type] => Array (
        [0] => image/jpeg 
        [1] => 
        [2] => 
    ) 

    [tmp_name] => Array ( 
        [0] => E:\localhost\tmp\php118.tmp 
        [1] => 
        [2] => 
    ) 

    [error] => Array ( 
        [0] => 0 
        [1] => 4 
        [2] => 4 
    )   

    [size] => Array ( 
        [0] => 92127 
        [1] => 0 
        [2] => 0 
    ) 
)

How should I make it into the array of the format that I want ?

Thanks

Here is an example of how to do it that I wrote to similar question. — Danijel
– Danijel, Commented May 8, 2013 at 6:31

Marc B · Accepted Answer · 2010-11-17 10:30:42Z

7

This is kludgey, but

$_MYFILES = array();
foreach(array_keys($_FILES['name']) as $i) { // loop over 0,1,2,3 etc...
   foreach(array_keys($_FILES) as $j) { // loop over 'name', 'size', 'error', etc...
      $_MYFILES[$i][$j] = $_FILES[$j][$i]; // "swap" keys and copy over original array values
   }
}

answered Nov 17, 2010 at 10:30

Marc B

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3 Comments

Atif Over a year ago

I think creating keys manually will take less resources compared to using array_keys since its always the same ?

Marc B Over a year ago

True, but this is PHP... no guarantees that a key won't be renamed, or new ones added, etc... later on. Not likely, but not impossible. The inner array_keys call could be done once and cached in a var before the first foreach loop, though.

Atif Over a year ago

I completely agree to it .. Thanks Marc

heximal · Accepted Answer · 2010-11-17 10:24:37Z

0

why don't you want to declare new array var and to fill it in desirable format? eg

$myarr = array();
foreach ($_FILES as $file) {
  $myarr[] = array($file['name'][0], $file['type'][0], $file['tmp_name'][0], $file['size'][0]);
}

answered Nov 17, 2010 at 10:24

heximal

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2 Comments

Marc B Over a year ago

should be 'tmp_name', not 'temp_name' (isn't PHP's consistent naming convention a wonderful thing? :p)

Grimbot Over a year ago

@heximal In all seriousness I think you've misunderstood the problem.

Varshaan · Accepted Answer · 2017-04-14 13:10:55Z

0

this worked for me

$files = array();
for($i =0;$_FILES['name'][$i] != NULL;$i++){
      $files[] = array($_FILES['name'][$i], $_FILES['type'][$i], $_FILES['tmp_name'][$i], $_FILES['size'][$i]);
 }

answered Apr 14, 2017 at 13:10

Varshaan

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Comments

Emre Gemici · Accepted Answer · 2020-05-25 19:16:54Z

0

foreach ($_FILES['photo']['name'] as $key => $value){
    $photo[$key]['name'] = $value;
}

foreach ($_FILES['photo']['type'] as $key => $value) {
    $photo[$key]['type'] = $value;
}

foreach ($_FILES['photo']['tmp_name'] as $key => $value) {
    $photo[$key]['tmp_name'] = $value;
}

foreach ($_FILES['photo']['error'] as $key => $value) {
    $photo[$key]['error'] = $value;
}

foreach ($_FILES['photo']['size'] as $key => $value) {
    $photo[$key]['size'] = $value;
}

if ($photo[0]['error'] == 4)
    $photo = [];



echo '<pre>';
print_r($photo);
echo '</pre>';

answered May 25, 2020 at 19:16

Emre Gemici

1

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