TalkPHP - time out issues with this script

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- - time out issues with this script (http://www.talkphp.com/advanced-php-programming/4824-time-out-issues-script.html)

time out issues with this script

Code:

$quary = @$_GET['q'];



$query = "SELECT COUNT(id) as num FROM ph WHERE sitename LIKE '%".$quary."%' OR keywords LIKE '%".$quary."%' OR title LIKE '%".$quary."%' OR body LIKE '%".$quary."%' OR writtenby LIKE '%".$quary."%'";



while($row = mysql_fetch_array($result))

                {

        

        $post = substr($row['body'],$position,1); // Find what is the last character displaying. We find it by getting only last one character from your display message.



        if($post !=" "){ // In this step, if last character is not " "(space) do this step .



        // Find until we found that last character is " "(space) 

        // by $position+1 (14+1=15, 15+1=16 until we found " "(space) that mean character 20) 

        while($post !=" "){

        $i=1;

        $position=$position+$i;



        $post = substr($row['body'],$position,1); 

        }



        }



        $post = substr($row['body'],0,$position); // Display your message

        

        print $row['sitename'];

        echo "<br/>";

        echo $post;

        echo "...";

        echo "<br/><br/>";

        }

        ?>



<?=$pagination?>

However for some reason when I type run the following "Development Blog" in q=Development Blog i get this error

Fatal error: Maximum execution time of 90 seconds exceeded in /home/example/public_html/search.php on line 202

In the ph table the column sitename has the word Development Blog, however I am getting that time out error, can any one help?

Maximum execution time is set in the php.ini file. But 90sec is quite a long time to display a blog list!

I have worked out that there is an error with this part of the script

Code:

$position=11; // Define how many characters you want to display.



$message="Jooria Refresh Your Website";

$post = substr($message,$position,1); // Find what is the last 

character displaying.

We find it by getting only last

 one character from your display message.



if($post !=" "){ // In this step,

 if last character is not " "(space) do this step .



// Find until we found that last character is " "(space)

// by $position+1 (14+1=15, 15+1=16 until

 we found " "(space) that mean character 20)

while($post !=" "){

$i=1;

$position=$position+$i;



$message="Jooria Refresh Your Website";

$post = substr($message,$position,1);

}



}



$post = substr($message,0,$position); // Display your message

echo $post;

echo "...";

?>

What it should do, is find if the last character is a space if not it moves one more till it finds that space.

Is there an error with this?
As it works, expect when searching for a sitename for some reason.

Fixed the error, however need some help

Code:

<?php



  

                while($row = mysql_fetch_array($result))

                {

                $sitename = $row['sitename'];

                $title = $row['title'];

                $body = $row['body'];

                }

                

                $position=300; // Define how many character you want to display.



        $post = substr($body,$position,1); // Find what is the last character displaying. We find it by getting only last one character from your display message.



        if($post !=" "){ // In this step, if last character is not " "(space) do this step .



        // Find until we found that last character is " "(space) 

        // by $position+1 (14+1=15, 15+1=16 until we found " "(space) that mean character 20) 

        while($post !=" "){

        $i=1;

        $position=$position+$i;



        $post = substr($body,$position,1); 

        }



        }



        $post = substr($body,0,$position); // Display your message

        

        print $sitename;

        echo "<br/>";

        echo $post;

        echo "...";

        echo "<br/><br/>";

        

        

        ?>

Now my issue is this part of the code needs to be duplicated

Code:

        print $sitename;

        echo "<br/>";

        echo $post;

        echo "...";

        echo "<br/><br/>";

Is there any way to do this?

Thanks