allydm

Hi there. I have a select query pulling information from my table correctly on most pages apart from a few, and my problem has really confused me. Basically, the pages contain a sports league tables and the info is pulled out on every page load. To best explain the problem, i've taken a screen shot of it... [img]http://www.allydm.co.uk/images/php-problem.gif[/img] Now, the problem is that it is falsly executing, as i specify that the home_team_score should be greater than the away team score...HOWEVER as you can see in the result that it gets, the home_team_score is 3 whereas the away_team_score is infact 12. Any ideas why this is happening?

Yes, the first bit would go on the login page....in the part of the code where you telling the php what to do if the user logs in correctly.

Have you set the variable $Courriel, because i don't see it anywhere in the code above.

Well as you havn't mentioned if you have built a login form or not, i'm going to leave that stage out for the moment. Basically, you need a user to enter there information, check it and then if its true set session variables. So, presuming that you know how to check that a users password matches the one they've entered, the next stage would be [code]<?php session_start(); if ($submitted_password = $_POST['password']) { $_SESSION['user_id'] = $login_info['user_id']; } ?> [/code] That would then set a session variable. Remember: you should [b]never[/b] store a users password in a session variable. Then, at the start of each page you need to check if $_SESSION['user_id'] is set or not, and if its not then tell them to login. Such as this: [code] <?php session_start(); if (empty($_SESSION['user_id'])) { header("Location: login.php"); } ?> [/code] That would send a user to login.php if they havn't logged in. Its not 100% secure, but its a good start for you.

You could use a combination of [b]$_SERVER['PHP_SELF'][/b] and [b]preg_replace()[/b]. The $_SERVER['PHP_SELF'] would tell you the url, excluding the www.phpfreaks.com (for example), and then you could just tell the preg_replace() function, to look for /somefolder/thispage.php in the $_SERVER['PHP_SELF'] result, and then replace if with nothing. The result would be that you have the anotherpage.php?stuff here Just incase i wasn't too clear (i got lost with myself above  :)) It would be something like. [code]$full_file_name = $_SERVER['PHP_SELF']; $file_name = preg_replace('/somefile/, '', $full_file_name);[/code] That would remove any proceeding folder name (which i have called somefile) and leave you with the file name. That won't work, but it hopefully shows you a process that you could go through.

Hi people. Basically, i'm building a gallery which pulls 9 images per page. I've coded the pages so that 9 results are displayed per page and thats all working marvelous. However at the moment they're all one after the other going vertically down the page via a loop. I'd like to make it look nicer, by creating a 3 x 3 table and having the results display in the box. Heres an image of what i'd like to achieve. [img]http://www.allydm.co.uk/images/demo1.jpg[/img] I'm not sure how to go about doing this, have any of you got any ideas? I started off with two loops, one going horizontally while the other goes down vertically however that got messy and didn't work. Thanks for any suggestions. Alex

Hi. Unfortunatly it is still not showing even when the 'Other' option is selected. Any ideas? Thanks.

Hi there. I have a form with a drop down selection box, and if someone selects the option in it called 'Other' then i would like a text entry box to appear. So far i have... [code]<select name="subject" id="subject">     <option value="Purchasing Artwork">Purchasing Artwork</option>     <option value="Viewing Artwork">Viewing Artwork</option>     <option value="Commissions">Commissions</option>     <option value="Other" onSelect="document.getElementById('iAmHidden').style='visibility:visible;';" />Other</option> </select> <div style="visibility:hidden; display:none;" id="iAmHidden"> <input name="reason" type="text" id="reason" value="Please Specify"> </div>[/code] However this isn't achieving the effect i'm after (the box won't display!). Can anyone see why? Or suggest a better code. Thanks for any help guys/gals. Alex.

You could use Javascript to do it instantly on the page as it's entered. Depending on how you 'define' the $_POST variables on the form you could have it do the calculations on loading the page (which i presume is a shopping basket or something). If you could use the code for the page that posts the data we may be able to give you some code.

Hi. I have a table, which contains 'setting' information for my website. The table has the following columns: setting_id, setting_name and setting_value. Now there are currently 30 rows stored in this table, however i am having trouble pulling rows out on a 'as needed' basis. At the moment i have this select query:     $settings_query = "SELECT * FROM `robin_settings`";     $settings_result = mysql_query($settings_query); And from that, i would like to pull individual results, for example... I have a setting which has a 'setting_id' of "4", a 'setting_name' called "banner_on" and a 'setting_value' of  "1". How am i able to pull that information from the results i get? I know i could pull it by putting a WHERE clause on the SQL query, however that would mean having 30 select queries per page and i don't really fancy that! Thanks for any help! (i'm new at this :) ) Alex.

Here is the section of code: [code] $i = 1; $product_query1 = "SELECT * FROM `robin_basket` WHERE session_id = '".session_id()."'"; $product_result1 = mysql_query($product_query1) or die (mysql_error() ); $product_info1 = mysql_fetch_assoc($product_result1); while($product_info1 = mysql_fetch_assoc($product_result1)) {   echo "<input type='hidden' name='item_name_$i' value='{$product_info1['product_name']}' />         <input type='hidden' name='amount_$i' value='{$product_info1['product_price']}' />";   $i++;   } [/code] I have changed the name of the variables from $product_info to $product_info1,  just so that i could differenciate between them. I really do appreciate your help with this

We have progressed! Okay, now i'm getting two of the three items information sent through, however one of the items isn't. Items: item_name_1 item_name_2 i am getting, but i'm not getting: item_name_3 Now i've tried all sorts of combinations (been up to item_name_11), tried item_name, item_name_, item_name_0 however the third item just dosn't seem to appear. Any ideas why? There are 3 rows fetched by the query, and 3 items are shown by the same query earlier in the page, but not here.

[quote author=Barand link=topic=106879.msg428089#msg428089 date=1157410728] Are you looping through the query results in the code before getting to this bit? [/quote] I am, as it displays the current cart contents in a table further up the page...is this a problem?

That outputs... "There are 3items"

Hi. $product_query is being executed correctly, and the query is returning results. Is there any more information i should provide to help with this?