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Editorial Team
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Asked: 2026-05-29T12:11:03+00:00

We have several pages generated using PHP on our website with the following titles

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We have several pages generated using PHP on our website with the following titles (for example):

http://www.mysite.com/project/category/1
http://www.mysite.com/project/category/2
http://www.mysite.com/project/category/3

Each one is created dynamically with the same page layout with each showing a different database result depending on the predefined conditions.

I would like an image to be displayed at the top of the page for just one of the results, let’s say for http://www.mysite.com/project/category/2 – how can I go about this?

The relevant code on our page is this:

$category=mysql_fetch_array(mysql_query("select * from project_category where project_category_id='".$project_category_id."'"));?>

If we go down the if statement route can you show an example of how to display an example image by modifying the above code to get me started?

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1 Answer

  1. Editorial Team

    In the end I opted for an if statement (as found here http://www.tizag.com/phpT/if.php).

    The original code above was modified in the following way:

    $category=mysql_fetch_array(mysql_query("select * from project_category where project_category_id='".$project_category_id."'"));

 if ( $project_category_id == "2" ) {
    echo '<a href="http://www.mywebsite.com/imagelink" target="_self"><img src="http://www.mywebsite.com/image.jpg" width="675" height="75" border="0" /></a>';
}?>
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