Had a simple form script that suddenly stopped working
It was made about a year ago and had been working fine. Last time it was known to work for sure was
Need help PLEASE
ok i have this warning showing up
Warning: in_array() [function.in-array]: Wrong datatype for
retrieving images from mysql database using php
So I've been trying to figure out how to store images in a mysql database, and as far as i can tell
Problem assigning value to variable in "IF" function
Does this script makes sense? I am trying to take the value that is set to "authenticat" a
Just a white page
Okay so, my website, when I click SignUp on it it takes me to /join.php but its a complete white pag
drop-down with sub-category appear
Hello,
i know how to build a simple dro-down list, im looking for a code when im gonna choose
Why is this query failing?
Why is this not working?
$query = "SELECT * FROM `users` WHERE `userid` = " . $USERID
PHP Error
On my .php page I have a drop down box that has several names in it. When a user clicks the name &am
Hotlinking Picasa as the image folder of a website
Hi there PHP freaks, I would like to create a private album in Picasa to use it as the image folder
Select Rows as Columns..
is there a way to select COLUMN_NAME from user_tab_columns where table_name='TABLENAME';
and ha
