gettign variable from multipart form

[treds1]

how can i get the location of the file i wish to add to db ive seen examples using just e.g $bin_data(being the name of the file on form) but doesnt seem to work for me. i can get the name of the file using

$fileinput = $_FILES['bin_data']['name'];

but not its location

$image = addslashes(fread(fopen($fileinput,"r"), 1000000));

hope thats enough description

[treds1]

dont worry ive solved the problem

by using

$fileinput = $_FILES['bin_data']['tmp_name'];
as tmp_name holds the full location of the file not just the name.

[cgraz]

please post the solution and mark the thread resolved (see my sig). This will help others with the same problem who run across your post.

Cgraz