I'm trying to make a interactive page and made a function to query a string from my database.

code:

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title_query($id)

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code:

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Database   ----->    title_query($id)
          <-----
          ------>  Output on the website

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hhmm i dont really foolow, what has this got to do with str_replace?

AH, right, well... forgot to say, I was trying to work with str_replace to let it work, but I thought maybe someone here knows another way.

from the look of that little picture you will need 3 quierys, are you using MYSQL?

i dont get why you need to insert it and get it back out again, if youre inserting it you obviously have it in a variable so why not just output it normally?

That's not really what I mean, and yes I use MySql.

I'll try to explain better

In my database I type title_query(3), in my code I query the info from the database to the page, but before it ouputs the data to the website I want title_query($id) to be executed.

EDIT:

quote:

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Originally posted by Lee Jeffery
i dont get why you need to insert it and get it back out again, if youre inserting it you obviously have it in a variable so why not just output it normally?

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quote:

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Originally posted by Ezra
That's not really what I mean, and yes I use MySql.

I'll try to explain better

In my database I type title_query(3), in my code I query the info from the database to the page, but before it ouputs the data to the website I want title_query($id) to be executed.

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code:

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$link = mysql_connect('mysql_host', 'mysql_user', 'mysql_password')
   or die('Could not connect: ' . mysql_error());
echo 'Connected successfully';
mysql_select_db('my_database') or die('Could not select database');
$query = 'SELECT * FROM my_table';
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
mysql_close($link);

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Well... I have a basic lay-out and all the data is in the database. And if you change the id number the data changes but not the lay-out.

That's why I need to add the query to the database.

http://imdb.tsdme.nl/profile.php?id=1
http://imdb.tsdme.nl/profile.php?id=2


That's my function

code:

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function title_query($id){
global $title;
$title = mysql_fetch_array(mysql_query("SELECT `title` FROM `imdb_data_title` WHERE id=".$id));
echo $title['0'];
}

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I tried and tried and tried, and couldn't get it working...

So I thought about another thing a function that looks for special tags and changes them to data from another table.

Example:

Ezra made a great movie called [[t13]] which is made in the year [[y13]] and is the next big hit next to [[t11]] made in [[y13]].

This means that the T stands for title of row 13, and Y for the year from row 13

Does anyone think he knows how to make this or help me to make this

Any help is appreciated

code:

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<?php
$id = $HTTP_GET_VARS['id'];
function query_db(){
global $id;
$query_q = mysql_query("SELECT * FROM imdb_data_title WHERE id = '$id'");
while($query_row = mysql_fetch_array($query_q)){
$output = ('<b>Title/b> <i>' . $query_row['title'] . '</i><b> Yearb> <i>' . $query_row['year'] . '</i>');
}
return $output;
}

echo('' . query_db() . '');

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I had that already working , but my problem was that I had to store a part of the php code in the database and it wouldn't execute that part.

My version:

code:

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function title_query($id){
global $title;
$title = mysql_fetch_array(mysql_query("SELECT * FROM `imdb_data_title` WHERE id=".$id));
echo "<b>Title</b>: <i>".$title['title']."</i> <b>Year</b>: <i>".$title['year']."</i><br>";
}

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eval()