time out issues with this script

russellharrower · 08-07-2009, 10:04 AM

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?

JaoudeStudios · 08-07-2009, 10:37 AM

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

russellharrower · 08-07-2009, 01:56 PM

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.

russellharrower · 08-07-2009, 02:44 PM

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