$_FILES does not accept value

Asked May 13 '15 at 06:25

Active May 13 '15 at 06:45

Viewed 53 times

I have an input that keeps photos on itself . And I'm using it for a website panel . But when it saves the photo , in editing , I should put a value for it to read the last picture . but it doesn't work well. $_files is empty again ! What can I do for that ?

thanks

<input<?php if(!empty($target_file)) echo ' value="'.$target_file.'"';?> type="file" id="photo" name="photo" class="photo" />

<?php
    $photo= $_FILES['photo'];
    $target_dir = "uploads/";
    srand(time());
    $random_num = (rand()%99999);
    $target_file = $target_dir . $random_num . basename($_FILES["a_photo"]["name"]);
    $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
    move_uploaded_file($_FILES["photo"]["tmp_name"], $target_file);
?>

I hace checked stackoverflow for my answer but didn't find anything .

edited May 13 '15 at 06:45

asked May 13 '15 at 06:25

Majid Sadr

try to check that your form have enctype = 'multipart/form-data' ornot.if not then please put it and check – Alive to die - Anant May 13 '15 at 06:27
1

did you write ` enctype="multipart/form-data"` in your form tag – Bushra Shahid May 13 '15 at 06:28
shouldn't `$_FILES["photo"]` be `$_FILES['a_photo']`? – David R. May 13 '15 at 06:28
@anantkumarsingh the form has this attribute. – Majid Sadr May 13 '15 at 06:30
@BushraShahid the form has this attribute. – Majid Sadr May 13 '15 at 06:30
then do this $_FILES["photo"] be $_FILES['a_photo'] – Alive to die - Anant May 13 '15 at 06:30
@DavidR. I edited it for here . In my code that is p_photo and without any problem – Majid Sadr May 13 '15 at 06:31
See waht the name you give to the field in form that only name you need to put in $_FILES. if it is p_photo then it should be $_FILES['p_photo'] – Alive to die - Anant May 13 '15 at 06:32
See what's stored in with `var_dump($_FILES)` – pbaldauf May 13 '15 at 06:33
@anantkumarsingh this is correct in my php page . I edited it for here . I think the problem is for the `value=""` . because the value of the photo variable is an array and I don't know What to do for that. – Majid Sadr May 13 '15 at 06:34
@pbaldauf I got an echo from `var_dump($_FILES)` .result was: ` array(0) { } ` – Majid Sadr May 13 '15 at 06:39
you cannot send an array with $_FILSE, because when you put a file-name it itself create an array of$_FILES with name, size, tmp_name etc – Alive to die - Anant May 13 '15 at 06:42
@anantkumarsingh so what should be the value of the `$_FILES` input ? – Majid Sadr May 13 '15 at 06:43
it will be file name only.means just a file selected by this filed.As normaly you are doing in fileupload – Alive to die - Anant May 13 '15 at 06:45
1

1) `$_FILES` will indeed be empty, it only temporarily holds files uploaded *in the current request*. 2) You cannot pre-fill file input fields. – See the duplicate for how to handle this. – deceze May 13 '15 at 06:47

$_FILES does not accept value

0 Answers0