<?php
echo date(“Y”); ?>

easy!

I always use this bit of code for my navigation links that are part of a php include file.

Whenever the recordset on the particular page (rs_page) is equal to the unique id from the row (in this case 1) it echoes “class=”selected”

<?php $pageon = $row_rs_page['page_id']; $pagenumber = “1″; if ($pageon == $pagenumber) {echo ” class=\”selected\”";} ?>

… or another way would be to declare the name of each page using this at the top of your HTML (example ‘servicespage’) :

<?php $page_name = ‘servicespage’; ?>

and use this in your menu file:

<?php if($page_name == ‘servicespage’) echo ‘class=”selected”‘; ?>

With a neat bit of php of course! Using the code below you can reveal a section of your page when your recordset’s (yourrecordset) table row (tablerow) is equal to a certain result. I used this to reveal a ‘New Product’ image over the top of a product if the content of the recordset result matched what was‘yourconditionhere’.

<?php if ($row_yourrecordset['tablerow']==”yourconditionhere“){ ?>
Hello World!
<?php } ?>

To display a date retrieved from a recordset in a new format (In this case from a recordset ‘rs_news’):

• <?php echo date(‘d M Y’, strtotime($row_rs_news['date'])); ?>

When inserting the date in your MySQL database from an update form it needs to be in the MySQL format yyyy-mm-dd. To update a record in this format use;

• GetSQLValueString(date(‘Y-m-d’, strtotime($_POST['date'])), “date”));

A simple bit of code like the one below is all you need. In this case the recordset name is ‘rs_page’ and the table column that is being tested is ‘image’. If there is nothing returned from the query then the area ‘HTML here’ is hidden.

<?php if (!empty($row_rs_page['image'])) {?>

HTML here

<?php }?>

The answer is a simple line of SQL. Simply run this SQL query on the offending table (in this case ‘warranty_customers’ is the table and the bad entries range from id 279 to 1400). Simples!

DELETE FROM  `warranty_customers` WHERE  `warranty_customers`.`id` >279 LIMIT 1400 ;

If you are using WordPress you can use the line in PhpMyAdmin:

DELETE FROM wp_comments WHERE comment_approved = 0

or if you are using Askimet (where the value of approved may be set to ‘spam’) use this line in phpMyAdmin:

DELETE FROM wp_comments WHERE comment_approved = ‘spam’