Thanks Colin this will help me, like i said im just trying to help its not my code and i dont work with photos like this, but i Do appreciate your patience with me. "Colin Guthrie" <gmane@xxxxxxxxxxxxxx> wrote in message news:fqm88a$q4n$1@xxxxxxxxxxxxxxxx > Pieter du Toit wrote: >> That is the problem, i can see the name of the file with >> $txtPhoto['name'] >> but when i use the same with temp_name, i get nothing, one weird thing is >> this [error] => 0 [size] => 17008 , what is that? >> >> this is the code that is giving the problem >> $txtPhotoData = addslashes(fread(fopen($txtPhoto, "r"), >> filesize($txtPhoto))); >> >> And this is the error >> >> Warning: fopen() expects parameter 1 to be string, array given in >>> /usr/www/users/zululr/marketplace/myzululand/specials_proc.php on line >>> 49 >> >> I did not do the code im just trying to help out, if never use this kind >> of >> code to work with photos, i normally use move_uploaded_file and put the >> path >> in the databse, but it look like this code puts the photo in the database >> as >> binary or something. > > OK, I'll try and spell it out seeing as the link to the PHP Manual > didn't take ;) > > The variable $txtPhoto exists because you have register global option set. > > It will be the same value as $_FILES['txtPhoto']. > > This variable *is an array*. It is not meant to contain just a filename. > > Your fopen() statement assumes that this variable *is a string* and this > is fundamentally wrong. > > The PHP Manula page cannot be more clear about what the various elements > in this array mean: > > http://uk.php.net/manual/en/features.file-upload.php > (here their element is called 'userfile', yours is called 'txtPhoto'. > > > $_FILES['userfile']['name'] > The original name of the file on the client machine. > > $_FILES['userfile']['type'] > The mime type of the file, if the browser provided this information. > An example would be "image/gif". This mime type is however not checked > on the PHP side and therefore don't take its value for granted. > > $_FILES['userfile']['size'] > The size, in bytes, of the uploaded file. > > $_FILES['userfile']['tmp_name'] > The temporary filename of the file in which the uploaded file was > stored on the server. > > $_FILES['userfile']['error'] > The error code associated with this file upload. This element was > added in PHP 4.2.0 > > > > The ['name'] is the original name of the file the person uploaded. Your > code tries to load the contents of the file into a variable. To do this, > this simplest way is to do: > > $photoData = file_get_contents($_FILES['txtPhoto']['tmp_name']); > > Your code calls addslashes() on it but that's pretty braindead as it's > binary data. If it's going into a database you'd be better formatting it > using a database formatting function (e.g. mysql_real_escape_string) and > not just relying on an ad-hoc method. > > HTHs > > Col > > > > > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
