Forum Moderators: coopster
I'm going to be using PHP/JavaScript/MySQL for it. The basic layout is this:
On the website, a user can click a link, JavaScript takes over and furnishes a pop-up window. In this window I would need information to be displayed according to what link they clicked.
So let's start like this...On "index.php" there are three links:
Link A
Link B
Link C
Link A - Contains information about cars, listed in MySQL database "info" as line 1.
Link B - Information on houses, MySQL database "info" as line 2.
Link C - Information on pets, MySQL database "info" as line 3.
I suppose it might help if I show you the coding I'm using..
<script>
var newwindow;
function box(url)
{
newwindow=window.open(url,'name','height=400,width=200');
if (window.focus) {newwindow.focus()}
}
</script>
The links are:
<a href="javascript:box('popup.php');">Link</a>
"popup.php" would contain the code for connecting to the MySQL database to retrieve the relevant information.
My only question is...How do I tell it what row in the table to display?
Let's say my table is setup like this:
Column 1 - ID
Column 2 - Name
Column 3 - Description
My only guess would be to use the ID as a reference, but I'm not sure how to do that..So..I think that's my question.
I know this post is probably really confusing, so..If anyone needs clarification, just ask.
Also, after I get some help on that, I was wondering if anyone could lend their knowledge on setting up a so-called "Admin Interface".
This would generally be a form to input data into the MySQL table for the links..I'm guessing it would be something like this:
<?php
$dbhost='localhost';
$dbusername='';
$dbuserpass='';
$dbname='';
mysql_connect ($dbhost, $dbusername, $dbuserpass);
if (!mysql_select_db($dbname)) die(mysql_error());
$id = "";
$name = "";
$desc = "";
$query1 = "INSERT INTO info (id,name,description) VALUES ('".$id."','".$name."','".$desc."')";
mysql_query($query1)or die(mysql_error());
?>
But I'm not sure how to make a web-interface to handle entering the variables. That script would just be run from a browser, but you have to fill in the variables by hand in the script.
Thanks in advance!
Regards,
Ken
It should just be a case of passing the id of each link to popup.php on the query string - just as form variables are submitted using a form with the "GET" method. Your links would look something like this:
<a href="javascript:box('popup.php?id=1');">Link A</a>
<a href="javascript:box('popup.php?id=2');">Link B</a>
<a href="javascript:box('popup.php?id=3');">Link C</a>
Then, in popup.php, you simply query the database using an SQL statement with a WHERE clause something like:
$sql = "SELECT * FROM info WHERE id=".mysql_escape_string($_GET["id"]);
Strictly speaking the mysql_escape_string function is not required but you should use it because the value of $id is untrusted.
P.S. - Thanks, I'll let you know how it goes.
Regards,
Ken
<?php
$dbhost='localhost';
$dbusername='#*$!';
$dbuserpass='#*$!';
$dbname='info';
mysql_connect ($dbhost, $dbusername, $dbuserpass);
if (!mysql_select_db($dbname)) die(mysql_error());
$sql = "SELECT * FROM info WHERE id=".mysql_escape_string($_GET["id"])";
$result = mysql_query($query) or die('Error, query failed');
while ($row = mysql_fetch_array($result)) {
echo "<table border='5'>
<tr>
<td>
'.$row['name'].'
</td>
</tr>
<tr>
<td>
'.$row['description'].'
</td>
</tr>
</table>";
}
?>
Just not seeing why it's not working...
Oh yeah..It's not giving errors, it's just...Nothing displays. :)
I guess it would help to let you know I'm running PHP 4.4.0, MySQL 4.1.14, Apache 2.0.54 On Win Server 2003
If so, view source on the resulting window and see if there's anything in there.
No source code.
<?php
$dbhost='localhost';
$dbusername='root';
$dbuserpass='password';
$dbname='database';mysql_connect ($dbhost, $dbusername, $dbuserpass);
if (!mysql_select_db($dbname)) die(mysql_error());
$query = "SELECT * FROM info WHERE id=".mysql_escape_string($_GET["id"]);
$result = mysql_query($query) or die('Error, query failed');
while ($row = mysql_fetch_assoc($result)) {
echo $row['id'];
echo $row['name'];
echo $row['description'];
}
?>
[edited by: coopster at 4:35 pm (utc) on May 8, 2006]
[edit reason] generalized information [/edit]
$dbusername='root';
Looks like a server login to me, be careful
