Help with argument's on a anchor

kuyote

I am need of some help. Here is what I am trying to do:

I have a dynamic table generated with information for a mySQL query.

This isn't the entire file, only snippet's that I think are relevant.

//issues.php
echo "<tr><td>$myrow[21]</td><td>$myrow[0]</td><td>$myrow[22]</td><td>$myrow[3]</td><td>$myrow[4]</td><td>
<a href=\"ticket.php?name=$myrow[21]\">edit</a></td></tr>";

When the edit button is selected, this open a file named ticket.php, the URL is ticket.php?name=02-0822-5 or whatever the ticket ID is that I select.

Here is a snippet from ticket.php
echo "select * from tickets where ID = $name";

This is just a test to see if the argument is even passed. I am planning on putting a SQL query once I can see it is working.

This is the query I tried but it doesn't work.

$result = mysql_query("select * from tickets where ID = $_POST['name'];", $db);

For some reason I can't get the $name from the URL, can someone help me out with this?

Thanks,

Josh

djroki

If it's along the URL, it's using GET variables

Even then, you shouldn't have any need for using $POST or $GET, can you not simply use:

$result = mysql_query("select * from tickets where ID = '$name'", $db);

Good luck, roki

Edited 'cause I spotted an extra ";" in your query

the_Igel

May kuyote write simply $name or may not depends on system settings.

And never forget to put $name in quotes, or MySQL would evaluate it:
02-0822-5 = 2 - 822 - 5 = -825.

kuyote

I have the server setup so you must pass $POST's or $GET's.

I have done some test's on this,

$result = mysql_query("select * from tickets where ID = '$name'", $db);

will not work,

however I can substitute the value that is supposed to be passed to get it to work

$result = mysql_query("select * from tickets where ID = '02-0822-2'", $db);

This will work great. It looks like I am missing something with the passed "name" variable.

Thanks again for your help

kuyote

OK I figured it out....

here is how I did it. I added this before the query

$name = $_GET['name'];

and here is the query

$query = "select * from tickets where ID = '$name'";
$result = mysql_query($query, $db);

Thanks for your help I appreciate it.