Custom Parshing help

Rayzor

I'm looking to parse out data that is stored in a database such as

"This is some data [limg=51] and I want to be able to do something cool [limg=250] and parse things out"

Now for each of the limg= I want to be able to pull data from the database based on what follows the = sign, and then replace the [limg=51] with something like

I understand all of the pulling of information from the database, my problem is how do I extract the information from the text i.e. the how do I find each and from those replace that code with the appropriate html tag while at the same time extracting the #54 so I can query the datasource and do the fetch_array.

I've gotten to this point:
Ok so now I have this:

function getimg($id)
{
$sql = "SELECT a.image_id, a.id, a.name, a.image_active, a.thumb_file, a.comments
FROM images a, categories b
WHERE a.image_active=1
AND a.id = b.id
AND a.image_id = ".$id;
echo $sql;
return $sql;
}
$test ="[limg=59] test of something [limg=60]";
$output_string = preg_replace( '/[limg=([0-9]+)]/i', "<a href='gallery.php?img=\1 '><img src='http://www.mythicarealms.com/gallery/".getimg('$1')."'></a>", $test );
echo $output_string;
?>

The "echo $sql"; in the function prints out

"And a.image_id = $1"

and when it returns the $sql in the getimg('$1') call it prints
"And a.image_id = 60"

So the value of $1 isn't interpreted as it passes out to the function, is there someway to force the function to evaluate the $1 before it calls the function.

Any help would be great.

Daniel

leatherback

You end your querystring too early:;

$sql = "SELECT a.image_id, a.id, a.name, a.image_active, a.thumb_file, a.comments
FROM images a, categories b
WHERE a.image_active=1
AND a.id = b.id
AND a.image_id = $id";
echo $sql;

Anyway.. Hope you realize that this is not giving you result, but that you have to use a statement like mysql_fetch_array to get the actual results?

Rayzor

Thanks for the response, but the sql query ends up the same way in the function it places: id = $1 (not the value parsed out). I was printing it so I could see why I wasn't getting any results and passing it back in so I could see some of the mechanics. After I run it this is the page that is generated. You'll see the SQL query is echoed twice. Once in the function and once when $sql is returned.

In the function you'll see "id=$1" as the actual output and in the regular expression it appears "id=60" which means the $1 is only being parse IN the expression and is not being passed out into the function.

I also just updated the function to put a tag around the query it outputs, and what is concerning is it only outputs the data once not twice New Code:

function getimg($id)
{
$sql = "SELECT a.image_id, a.id, a.name, a.image_active, a.thumb_file, a.comments
FROM images a, categories b
WHERE a.image_active=1
AND a.id = b.id
AND a.image_id = ".$id;
echo "INFUNCTION<b>$sql</b>";
return $sql;
}
$test ="[limg=59] test of something [limg=60]";
$output_string = preg_replace( '/[limg=([0-9]+)]/i', "<a href='gallery.php?img=\1 '><img src='http://www.mythicarealms.com/gallery/".getimg('$1')."'></a>", $test );
echo $output_string;

HTML GENERATED:

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html>
<head>
	<title>Untitled</title>
</head>

<body>
INFUNCTION<b>SELECT a.image_id, a.id, a.name, a.image_active, a.thumb_file, a.comments 
FROM images a, categories b 
WHERE a.image_active=1 
AND a.id = b.id 
AND a.image_id = $1</b><a href='gallery.php?img=59 '><img src='http://www.mythicarealms.com/gallery/SELECT a.image_id, a.id, a.name, a.image_active, a.thumb_file, a.comments 
FROM images a, categories b 
WHERE a.image_active=1 
AND a.id = b.id 
AND a.image_id = 59'></a> test of something <a href='gallery.php?img=60 '><img src='http://www.mythicarealms.com/gallery/SELECT a.image_id, a.id, a.name, a.image_active, a.thumb_file, a.comments 
FROM images a, categories b 
WHERE a.image_active=1 
AND a.id = b.id 
AND a.image_id = 60'></a> 

</body>
</html>

Thanks for your reply 🙂

leatherback

Oooh.. Shoiop!

I must be sleeping.

But..

When you call the function, you call it with quotes around the variable. So it takes the variablename as a string -> searches for the var name.
Remove the quotes in the function call, and you should be fine (?)

Rayzor

sadly I tried that one alredy too:

Parse error: parse error, unexpected T_LNUMBER, expecting T_VARIABLE or '$' in parsetest.php on line 27

leatherback

How about changing the variable name:

Variable names follow the same rules as other labels in PHP. A valid variable name starts with a letter or underscore, followed by any number of letters, numbers, or underscores.

Rayzor

didn't think I could change it from 1 to something else since its in the preg_replace statement.