File upload PHP problem

Question

Got this working

$i = 0;     
    foreach ($_FILES["image"]["error"] as $key => $error) {
        if ($error == UPLOAD_ERR_OK) {
            $tmp_name = $_FILES["image"]["tmp_name"][$key];
            $image_name = $_FILES["image"]["name"][$key];
            $image.$i = move_uploaded_file($tmp_name, "uploads/$image_name");
            $i ++;
        }
    }

I cant seem to get the directory storing into the variables $image# any ideas?

johannes · Accepted Answer · 2009-12-15 11:34:39Z

2

Why are oyu mixing usage of $_FILES and $HTTP_POST_FILES? Usage of the later suggests that you're using an old and outdated tutorial.

You'RE also not checking whether multiple files have been successfully transefered and using copy() for this purpose is not encouraged.

See move_uploaded_files() which has an example about handling multiple uploads.

answered Dec 15, 2009 at 11:34

johannes

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1 Comment

user195257 Over a year ago

Thanks, i have this code now, uploading works, but having trouble storing the destination in some variables, any ideas? $i = 0; foreach ($_FILES["image"]["error"] as $key => $error) { if ($error == UPLOAD_ERR_OK) { $tmp_name = $_FILES["image"]["tmp_name"][$key]; $image_name = $_FILES["image"]["name"][$key]; $image.$i = move_uploaded_file($tmp_name, "uploads/$image_name"); $i ++; } }

user195257 · Accepted Answer · 2009-12-17 12:32:31Z

1

I got this working in the end by creating an array and storing the values in this array

answered Dec 17, 2009 at 12:32

user195257

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Comments

Mathew · Accepted Answer · 2009-12-15 11:35:30Z

0

Shouldn't it be

$_FILES['image'][$i]['name']

Rather than

$_FILES['image']['name'][$i]

answered Dec 15, 2009 at 11:35

Mathew

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2 Comments

hanse Over a year ago

Nope, the files array has a somewhat weird behaviour, so it is correct.

Mathew Over a year ago

Doh, and I thought that was easy points! ;) Gotta love PHP's internal consistency...

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