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9 Posted Topics
 | I have literally been banging my head on my keyboard out of frustration. I'm trying to display PHP sessions as values in my php form, but it's not going so well. here's what I have: [CODE]<?php session_start();?><html> <head> <title>Create An Event</title> <style> label {float: left; width: 100px;} input[type=text],textarea {width: 210px;} … |
| | 1.Store the absolute src of the image in your database.(eg: [url]http://www.randomsite.com/randomimage.jpg[/url]) 2.Fetch the url using mysql and bind results to $url. 3 Create a img src in your html using a php varible as the src. (EG:<img src='<?php echo"$url";?>' /> 4. Make sure you sanitize the values/or use prepared statements.  |
| | it just combines things together [CODE]eg: $b .= "There!"; // sets $b to "Hello There!", just like $b = $b . "There!";[/CODE] you can add strings using the "." as well eg:[CODE] $home="USA"; echo "Theres"."No"."place"."like"."$home"."!";[/CODE] would output, "Theres No place like USA!" [url]http://www.php.net/manual/en/language.operators.assignment.php[/url] |
| | I'm very stressed out right now and cant seem to fix the problem with this php/mysqli script. I will pay $10 through paypal to anyone who duplicates my code, (and its fixed) as a reply to this thread. My deadline is very close, and I need it within a day. … |
| | right now my php script is vulnerable to anyone putting in a random member_id into the url and having it excute sucessfully how can I encrypt the id="id#" in the url, so a guest is unable to type in there own id in the posted id retrieved through the url? … |
| | i've successfully developed a page where i can echo a `$_SESSION['SESS_MEMBER_ID']` and display the member id of a person who is logged in. however i need to display all the data of two mysql databases that are relational. $sql = "SELECT * FROM `members`.`players` WHERE `member_id` = 3 LIMIT 0, … |
| | Re: JS to PHP javascript is not a serverside its a client side. it cannot query a database. where is il_pageparameters saved at? if its using mysql(i) like you've defined its not going to work with js variables, just put the field of your MySQL table your calling for. if your selecting it with … |
| | I've been trying to get this code to work for hours can anyone spot the mistake i made? its a code to edit the contents of a database field calling it by its primary key. my error code is Parse error: syntax error, unexpected T_VARIABLE in E:\xampp\htdocs\php\databaseview\records.php on line 103 …  |
| | mysql_query("UPDATE tuck SET id=id--1 WHERE id >= '$id'"); you only had one "-" if you put two it automatically decreases by one each time regardless of the present state. |
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