I am trying to use a page to view data via a link from another page.

I am trying to do it like this:

<a href="http://www.somewebsite.com/view.php?ID=$ID">CLICK HERE</a>

and depending on the table id you go to my view page ( example URL )

http://www.somewebsite.com/view.php?ID=2

and i am trying to pull the data fot that table id...can anyone see anything wrong with this. I know all my DB settings are good.

<?php
include ("include/header.inc.php");

include ("include/dbconnect.php");

if ($ID) {


$result = mysql_query("SELECT * FROM $table WHERE ID=$ID",$db);

   



   $links = mysql_fetch_array($result);
   
        $ID = $links["ID"];
	$paypal = $links["paypal"];
	$contact = $links["contact"];
        $mob_id = $links["mob_id"];
	$mob_name = $links["mob_name"];
   
	echo "<table width=100% border=0><tr><td valign=top width=50%>";
	

echo "<h3>MOBSTER Name: </b>$mob_name</h3><hr>
<b>$mob_name PayPal E-mail:</b> $paypal<br>
<b>$mob_name contact e-mail:</b>$contact<br>
<b>$mob_name MySpace MOBSTER ID: $mob_id<br>
<br><br><a href='edit.php?ID=$ID'>Edit $mob_name Information</a><br>";
	
	echo "</td><td valign=top width=50%></td></tr></table>";
	
	
} else {

	echo "You need to select customer";

	}

include ("include/footer.inc.php");

?>

If you don't have register_globals enabled (recommended) then the $ID variable won't be available. Try using $_GET['ID'] instead. Otherwise looks fine

Originally Posted by jason_alan

If you don't have register_globals enabled (recommended) then the $ID variable won't be available. Try using $_GET['ID'] instead. Otherwise looks fine

Yep, you have to declare that "$ID" gets its value from the URL.

So change this view.php?ID=$ID

to this view.php?ID=$_GET['ID']


or this if ($ID)

to if ($_GET['ID'])


I am still learning

Originally Posted by pappasaa

So change this view.php?ID=$ID

to this view.php?ID=$_GET['ID']


or this if ($ID)

to if ($_GET['ID'])


I am still learning

Second choice, keep the url to view.php?ID=12345

Remember you should access all url parameters with the $_GET['var'] syntax, and post variables with $_POST['var'] and so on...

Use print_r to debug variables for values when you're stuck, e.g. print_r($_REQUEST)

Originally Posted by pappasaa

So change this view.php?ID=$ID
to this view.php?ID=$_GET['ID']
or this if ($ID)
to if ($_GET['ID'])
I am still learning

example.com/?id=$_GET['ID'] - Will Not be a valid URL
example.com/?id=12 - Is Valid
* Using the URL From Above *

PHP Code:

$id = $_GET['id'];
print $id; // Prints 12 


Does it make scene now?

i got this and now it loads the page...i.e. all my little titles but its not pulling the data and giving me an error

here is what i got

<?php
include ("include/header.inc.php");

include ("include/dbconnect.php");


if ($_GET['ID'])  

{

$result = mysql_query("SELECT * FROM $table WHERE ID=$ID",$db);

   



   $links = mysql_fetch_array($result);
   
        $ID = $links["ID"];
	$paypal = $links["paypal"];
	$contact = $links["contact"];
        $mob_id = $links["mob_id"];
	$mob_name = $links["mob_name"];

   
	echo "<table width=100% border=0><tr><td valign=top width=50%>";
	

echo "<h3>MOBSTER Name: </b>$mob_name</h3><hr>
<b>$mob_name PayPal E-mail:</b> $paypal<br>
<b>$mob_name contact e-mail:</b>$contact<br>
<b>$mob_name MySpace MOBSTER ID: $mob_id<br>
<br><br><a href='edit.php?ID=$ID'>Edit $mob_name Information</a><br>";
	
	echo "</td><td valign=top width=50%></td></tr></table>";
	
	
} else {

	echo "You need to select customer";

	}

include ("include/footer.inc.php");

?>

here is the error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/biginfoo/public_html/gethitlisted/admin/view.php on line 17

i got it

I had to change this

$result = mysql_query("SELECT * FROM $table WHERE ID=$ID",$db);

to this

$result = mysql_query("SELECT * FROM $table WHERE ID",$db);

thanks for the help

I can't see why this works, because you now don't specify what the ID field should equal to.

To prevent errors you have to access arrays in double-quoted strings by using curly brackets:

PHP Code:

$result = mysql_query("SELECT * FROM $table WHERE ID={$_GET['ID']}",$db); 


Yeah I have to agree, this won't work right..

$result = mysql_query("SELECT * FROM $table WHERE ID",$db);

I think "this works" is because it is still a valid mysql query.

You need to understand some things about first) sql, then second) php, as the process should go like this:

1) Process posted/url variables with a security check. As in you're expecting a record id which should be a number. So if $_GET['id'] is valid then do a mysql query to get that row.

2) If query returned a row, then display results, otherwise display some kind of error like "No results found". This can be done with:

if (!$result) {
    print('No Results Found.');
}

3) Also, there are libraries that help with this sort of thing... as in I can construct a sql query and send variables to it without worrying about sql injection, etc... I use pear db. A sample of this is...

// From url: http://mysite.com/user/?userid=1234
$sql = "SELECT id FROM users WHERE username = ?";
$res = $db->getRow($sql, array($_GET['userid']), DB_FETCHMODE_ASSOC);
// Now $res['username'] = 'username'

4) There are other kinds of php mysql database abstractions you can use as well.., but this one I find easy to use once I have learned it.

Hope that helps!

I did update my code but forgot to post the code here. I am reading the php and sql sites trying to learn the hows and whys so I can do this without help but i keep getting stuck and need to "phone a friend" on some things. I am sorry for not posting an update. I have learned alot from the two threads I have posted so far. I am going to mark this one a solved.

Thank you for getting back to me... your the tips are very useful.