how to print a selected photo name from input type

Question

Can anyone tell me, how I can print a photo name? The photo will be selected from input type=file.

Here is what I have tried so far:

$photo = $_FILES['photo']['tmp_name'];
//$photo_name = basename($photo);
echo $photo;

This code doesn't show actual file name. It print as like this (php3515.tmp);

`echo $_FILES['photo']['name']`. RTM: http://php.net/manual/en/features.file-upload.post-method.php — M. Eriksson, Jul 01 '17 at 09:06
Possible duplicate of [Isn't Using the basename function with $\_FILES\['userFile'\]\['name'\] Redundant?](https://stackoverflow.com/questions/2347056/isnt-using-the-basename-function-with-filesuserfilename-redundant) The question itself answers this very basic question. — mickmackusa, Jul 01 '17 at 11:46

score 4 · Accepted Answer · answered Jul 01 '17 at 09:07

4

Print the file name instead of tmpname

Try this:

echo $_FILES['photo']['name'];

answered Jul 01 '17 at 09:07

Ankit Singh

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score 2 · Answer 2 · answered Jul 01 '17 at 09:10

php3515.tmp in your case is a temporary file generated by PHP. You have to move this file to your desired location with move_uploaded_file() function, which takes your $_FILES['photo']['tmp_name'] as first argument, and path and new filename as second argument.

Here is an example taken from PHP manual:

$uploaddir = '/var/www/uploads/';
$uploadfile = $uploaddir . basename($_FILES['userfile']['name']);

echo '<pre>';
if (move_uploaded_file($_FILES['userfile']['tmp_name'], $uploadfile)) {
    echo "File is valid, and was successfully uploaded.\n";
} else {
    echo "Possible file upload attack!\n";
}

echo 'Here is some more debugging info:';
print_r($_FILES);

print "</pre>";

Note $_FILES['userfile']['name'] in this example, which contains the original file name.

This doesn't even answer the OP's question, which simply was _"how can i print a photo name"_. — M. Eriksson, Jul 01 '17 at 09:12
@Magnus Yes, my mistake. I already edited the answer to highlight the `name` index in `$_FILES` array. I leave the answer here as it could provide more useful information about file upload in PHP. Especially the `print_r($_FILES);` line would be priceless for education reasons. — Lukas, Jul 01 '17 at 09:17

timiTao · Answer 3 · 2018-05-14T05:31:11.423

You need to know, that you have 5 parameters in $_FILES['userfile']:

name - original name of uploaded file
tmp_name - system name that is physically writen on disk by system
size - size of file
error - code of error - 0 mean no problem
type - type of saved file

If you want you original name, to display, use simply:

$_FILES['userfile']['name']

You also have function move_uploaded_file(), that allows you to safely move the file from a temporary place to your own. Another way, after the end of PHP process, you uploaded file will be deleted.

$path = __DIR__  . '/upload/' . basename($_FILES['userfile']['name']);
$result = move_uploaded_file($_FILES['userfile']['tmp_name'], $path);

$result show you, is system correctly save a file in given place.

how to print a selected photo name from input type

3 Answers3