BetterPHP

No, it doesn't. Everything looks like it supposed to be.

Here's the page code. <?php include('../example/other.php'); $users_info = fetch_users_info($_SESSION['username']); if(!isset($_SESSION['username'])){ header('Location: http://example.com'); die(); } $errors = array(); if(isset($_REQUEST['mode']) && $_REQUEST['mode'] == "send"){ if...

Hi there, I'm populating a combobox with values stored in a database. When I try to submit the form again the combobox gets no value at all. <?php $errors = array(); if(isset($_REQUEST['mode']) && $_REQUEST['mode'] == "send"){ if (empty($_POST['touser'])) $errors[] = "Seleccio...

So, I did what you said and I still get no error. But I noticed that when I write something on the fields called 'of' (Second Query) it does work and create the row on the database. If not, it won't create it. This creates a doubt on me. Isn't PHP supposed to create the row with the values that are ...

Ok. Here is the complete code. Hope you can help me. :) <?php include('init.inc.php'); $b_name = mysql_real_escape_string(htmlentities($_POST['b_name'])); $b_phone_number = mysql_real_escape_string(htmlentities($_POST['b_phone_number'])); $b_address = mysql_real_escape_string(htmlentities($_POST['b_...

I cover every column name with ` but it doesn't work. And by using

echo mysql_error();

I get no error

I'm using this code in the same page. The last query works. Does anyone sees something wrong in the code? The table name and the fields are ok. $b_name = mysql_real_escape_string(htmlentities($_POST['b_name'])); // Updates Offers $of1 = mysql_real_escape_string(htmlentities($_POST['of1'])); $of2 = m...

Sorry! It was a Typo.

It's working.

Here I'm again. I'm using this code (From line 142 - 196): function fetch_business_info($name){ $sql = "SELECT `b_name`, `b_address`, `b_city`, `b_country`, `b_category`, `b_email`, `b_website` FROM `business_information` WHERE `b_name` = '{$name}'"; $result = mysql_query($sql); return mys...

Thanks for all your replies.

@EcazS '{$name}' solve it
Thank You!

I tried using the following code: <?php //fetches profile information for the given user function fetch_user_info($cname){ $sql = "SELECT `cname`, `address`, `tel`, `email`, `wsite` FROM `cInfo` WHERE `cname` = {$cname}"; $result = mysql_query($sql); return mysql_fetch_assoc($result); } ?>...

Hi there, Im using this code: <?php function fetch_user_info($name){ $sql = "SELECT 'bname' AS 'name', 'email' AS 'email', 'website' AS 'website', FROM 'cInfo' WHERE 'name' = {$name}"; $result = mysql_query($sql); return mysql_fetch_assoc($result); } ?> but I'm getting "Warning: mysql...