Code: $filepaths[] = $_FILES['new_image'];
foreach ($filepaths as $filepath)
{
$imagename = strtolower($filepath['name']);
$charref = substr($imagename, 0, strrpos($imagename, '.'));
if(ctype_alnum($charref))
{this is for a form with multiple image uploads that need to be resized. im trying to loop them thrue the resize script but it over writes as it goes untill the only img when its done is the final image field.
question about stripslashes and real_escape_string
im cleaning up an old app that I wrote fixing some of the vulernabilities from attacks.I have roughly 30 files. I want to be able to edit every $_POST and $_GETCode:
A WBS element with the name XXX already exists in version FREI_VERSION
Hi.
Print out contents of to Excel
I have got this script that gathers all the data that I need but I need it to send it to excel instead back to the screen in the web browser. How can I do this? Should I use something
Adding to the next element in a multidimensional array
Hi, I'm trying to add a value to $node->field_spaces['nid'] where x is the next available spot in the field_spaces array. Always adding to the next spot. I'll be calling this in a for loop to
Javascript or not?
How many people prefer javascript/ajax sites? How many prefer the good old fashion straight php sites?
Object Interfaces
EDIT: Never mind, I just updated to php 5.Hey all,I'm currently experimenting with php object interfaces. However, whenever I try to implement one, I get a php error.interface iTemplate{ public
Error in Configure System Landscape Directory phase
Hi
Login Issue's
Code: <?php $file = fopen('user.txt', 'r');/* Set login to false initially */$login = false/* Break out of loop if login becomes true OR EOF is encountered */while(($login == false)
Using insert variable
need a way to inert variable data to mysql database $acc = "212121212";$nok = "Nokia1100";$db_link = mysql_connect("localhost","root",
Error in SQL Syntax HELP!!!
I have this page:Code: <?phpsession_start();//connect to server and select database$conn = mysql_connect("localhost", "root", "") or die(mysql_error());$db =