Reputation: 25
When running a while script on my local xampp it just returns the same result over and over again, here's the PHP I'm using:
<?php
$id = $_GET['id'];
$getmembers = mysql_query("SELECT * FROM group_memberships WHERE groupid = $id LIMIT 10");
$getids = mysql_fetch_assoc($getmembers);
while($id = $getids['userid']){
$getusers = mysql_fetch_assoc(mysql_query("SELECT look, username FROM users WHERE id = '$id'"));
echo ($getusers['username']);
echo('<br />');
}
?>
All that I get when I execute that is: http://prntscr.com/12nh6g My database contains 4 different users registered to that group. I'm not sure about joining tables in MySQL as I have no idea how to do it.
If I add a break; then it only returns one result, instead of timing out.
Any help is much appreciated!
Also I'm sorry about how messy the PHP is, I am new to this.
Upvotes: 1
Views: 471
Reputation: 26421
while($id = $getids['userid']){ #This condition will always be true.
Use your mysql_fetch_assoc
statement in a loop like below.
while ($getids = mysql_fetch_assoc($getmembers)) {
$id = $getids['userid']
$getusers = mysql_fetch_assoc(mysql_query("SELECT look, username FROM users WHERE id = '$id'"));
echo $getusers['username'];
echo '<br />';
}
For sub query you can try,
SELECT look, username FROM users
WHERE id IN
(SELECT userId FROM group_memberships
WHERE groupid = $id
LIMIT 10)
Using JOIN,
SELECT u.look,u.username
FROM users as u
LEFT JOIN group_memberships as gm
on gm.userId = u.id
WHERE gm.groupid = $id
LIMIT 10
Note: Please, don't use mysql_*
functions in new code. They are no longer maintained and are officially deprecated. See the red box? Learn about prepared statements instead, and use PDO, or MySQLi - this article will help you decide which. If you choose PDO, here is a good tutorial.
Upvotes: 1
Reputation: 364
<?php
$id = $_GET['id'];
$getmembers = mysql_query("SELECT * FROM group_memberships WHERE groupid = $id LIMIT 10");
while($getids = mysql_fetch_assoc($getmembers)){
$getusers = mysql_fetch_assoc(mysql_query("SELECT look, username FROM users WHERE id = '$id'"));
echo ($getusers['username']);
echo('<br />');
}
?>
you had the while loop wrong , you need to do the feching inside while loop , so it goes till the fetching stop.
Upvotes: 1
Reputation: 5520
In your code:
<?php
$id = $_GET['id'];
$getmembers = mysql_query("SELECT * FROM group_memberships WHERE groupid = $id LIMIT 10");
$getids = mysql_fetch_assoc($getmembers);
while($id = $getids['userid']){
$getusers = mysql_fetch_assoc(mysql_query("SELECT look, username FROM users WHERE id = '$id'"));
echo ($getusers['username']);
echo('<br />');
}
?>
tells PHP that "Do the loop" while $id is ASSIGNED to the value of $getids['userid']
Therefore you you will have a infinite loop because $id is assigned all the time (and always with the same value) $getids are never changed inside of the while-loop.
But even if $getids was about to change it would still be in an infinite loop because the loop isn't checking if $id IS value of $getids['userid'], it's just assigning the value.
Upvotes: 1
Reputation: 2600
<?php
$id = (int) $_GET['id']; // very important part here, I cast the id to int, else you have a SQL injection leak
$getmembers = mysql_query("SELECT * FROM group_memberships WHERE groupid = $id LIMIT 10");
while ($membership = mysql_fetch_assoc($getmembers)) // on each pass we take a new result off first query
{
$user = mysql_fetch_assoc(mysql_query("SELECT look, username FROM users WHERE id = '".$membership['user_id']."'"));
echo ($user['username']);
echo('<br />');
}
?>
Upvotes: 1