Re: Variable post as array

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