no dice

RenardMF — Fri, 18 Mar 2005 20:11:12 +0000

Hmm it still makes the code work but doesnt catch the error if and id is manually entered.

Warning: mysql_result(): Unable to jump to row 3 on MySQL result index 4

I guess its not too big of a deal since within the popup people don't have access to the address bar. But on that note what do you put in the onclick= part of the hyperlink to make it so the location isn't displayed in the title bar?

dk01 — Fri, 18 Mar 2005 06:41:35 +0000

<?php
include("./config.inc.php");
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT *
    FROM films WHERE id='".$_GET['f']."' ORDER BY id DESC LIMIT 3";
$result=mysql_query($query);
if( !$result )
{
    die('SQL Statement Error: '. mysql_error());
    exit();
}

$line = array();
while( $row = @mysql_fetch_array($result) )
{
    $line[] = $row;
}

if(@mysql_num_rows($result) > 0){
    $film="$_GET[f]";
    $name=mysql_result($result,$film,"name");
    $synopsis=mysql_result($result,$film,"synopsis");
    echo"<strong>";
    echo"$name";
    echo"</strong>";
    echo"<br>";
    echo"$synopsis";
}else{

    echo "A film was not selected.";

}

mysql_close();
?>

Try that out. It's not tested but should work.

-dk

RenardMF — Fri, 18 Mar 2005 02:40:49 +0000

dk01 wrote: The easiest thing to do is to query the database with the id and if it returns less than 1 result have it throw an error. You will need to put an @ in front of mysql_query() to supress the system error messages.

-dk

Ok do you mean like move the rest of the code up and have the query come once the id is recieved? What would be the code for that?

dk01 — Fri, 18 Mar 2005 00:16:46 +0000

The easiest thing to do is to query the database with the id and if it returns less than 1 result have it throw an error. You will need to put an @ in front of mysql_query() to supress the system error messages.

-dk

I've Got it!

RenardMF — Thu, 17 Mar 2005 22:59:35 +0000

Well i got the code to work:

<?php
include("./config.inc.php");
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT *
    FROM films ORDER BY id DESC LIMIT 3";
$result=mysql_query($query);
if( !$result )
{
    die('SQL Statement Error: '. mysql_error());
    exit();
}

$line = array();
while( $row = mysql_fetch_array($result) )
{
    $line[] = $row;
}

if(isset($_GET['f'])){
    $film="$_GET[f]";
    $name=mysql_result($result,$film,"name");
    $synopsis=mysql_result($result,$film,"synopsis");
    echo"<strong>";
    echo"$name";
    echo"</strong>";
    echo"<br>";
    echo"$synopsis";
}else{

    echo "A film was not selected.";

}

mysql_close();
?>

But I'm not sure how to catch the error when an id over the amount that actually exists is given which gives "Mysql ...cant jump to table..etc"

Any ideas?

RenardMF — Thu, 17 Mar 2005 17:36:52 +0000

ok sounds good for not found thing. But what i what i was trying to do was basically have a page that recieved this "id=x" numbers and obtained the correspoding information from the database entry with the same id number. I understand how to connect to the database and get the table information I'm after and Id assume the best way to do what i want is create and array and have it echo the correct information but hats what im not sure how to do.

ex. site.com/page.php?id=1 would grab the information in the array marked with 1 as the id and display a particular field within the table.

Thanks for your help.

Renegade — Thu, 17 Mar 2005 10:32:07 +0000

I don't get what you mean but... you can try this:

if(strlen($_GET['id']) == "0"){
echo "Not found";
}