how to check condition php file input image array value is empty or not empty

Question

Array name value is empty but print else condition. how to solve this.

this is my html file input code for image.

<input id="editimagefile" name="editimagefile[]" type="file" multiple=true>

this is my php value assigned code

$imgsfiles = $_FILES['editimagefile']['name'];

if(empty($imgsfiles))
{
    echo "empty";
}
else
{
    echo "value assigned";
}

i checked alredy print_r($_FILES) but else part are executed.... — Aravinth E
– Aravinth E, Commented Nov 9, 2016 at 8:52
than use trim maybe some extra space here if(empty(trim($imgsfiles))) — devpro
– devpro, Commented Nov 9, 2016 at 8:53

Muhammed Imran Hussain · Accepted Answer · 2016-11-09 09:10:03Z

3

Here name="editimagefile[]"is array field thus $_FILES['editimagefile']['name'] is an array too. Count function will return greater then 0 and it is not empty array but contains empty element.

So if you are sure you have more then one element you can use following:

if(in_array("", $_FILES['editimagefile']['name'])){
      echo 'contains empty elements';
}else{
      echo 'correct';
}

if only one element then you can use following:

if (empty($_FILES['editimagefile']['name'][0])) { 
    echo 'contains empty elements';
}else{
    echo 'correct';
}

answered Nov 9, 2016 at 9:10

Muhammed Imran Hussain

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1 Comment

Muhammed Imran Hussain Over a year ago

@jeroen if there are X file fields then count($_FILES['editimagefile']['name']) will return X no matter file selected or not

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