PHP: move_upload_file() is not working

Question

I am trying to insert an image in one folder, but I guess i am not doing it correctly, can someone correct my code?

//Place the image in the folder  
$newname = "$menu_name-01.jpg";    
move_uploaded_file($_FILES['fileField']['tmp_name'],"/product_images/$brand/$newname");

Where,

$menu_name = trim(mysql_prep($_POST['menu_name']));  
$brand = trim(mysql_prep($_POST['brand']));

$menu_name-01.jpg is not a valid variable name. They cannot have "." in it and numerics. So "$nename" is probably NULL or FALSE. var_dump that so see whats going on. — paskl
– paskl, Commented Aug 9, 2014 at 17:15
Also I'd suggest you to enable error mode, so you would've seen this during development. — paskl
– paskl, Commented Aug 9, 2014 at 17:16
are you sure your file is uploading correctly? if you are sure the file is uploading, then check for destination directory permissions (if you are using unix OS) — Ehsan Khodarahmi
– Ehsan Khodarahmi, Commented Aug 9, 2014 at 17:19
No, I am using Windows. I think the fault is in the destination path only. I will check it. — user3832642
– user3832642, Commented Aug 9, 2014 at 17:22

Daniel McAssey · Accepted Answer · 2014-08-09 17:15:07Z

1

You should change:

$newname = "$menu_name-01.jpg";

to

$newname = $menu_name.'-01.jpg';

Because its looking for the variable as a whole ($menu_name-01.jpg) instead of concatenating $menu_name variable with -01.jpg.

answered Aug 9, 2014 at 17:15

Daniel McAssey

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3 Comments

user3832642 Over a year ago

Its still not working, but your answer is very much beneficial. This was a mistake, but it is still not inserting the image in the specified folder. What is your suggestion?

Charlotte Dunois Over a year ago

@MrTree You're wrong. He used double quotes so PHP replaced $menu_name with the assigned value.

Daniel McAssey Over a year ago

@CharlotteDunois No, it would take $menu_name-01 as a variable, but there is no $menu_name-01 variable, only $menu_name, the intention was to concatenate $menu_name and -01.jpg, you would either have to have a space in-between $menu_name and -01.jpg or join the string to the variable.

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