PHP not displaying result from MYSQL query - php

Question

I have the following code
$sql = "SET #uid := (SELECT ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);";
$sql = "UPDATE channels SET Used = 1 WHERE ID = #uid;";
$sql = "SELECT * FROM channels WHERE ID IN = #uid;";
$result = mysqli_multi_query($conn, $sql)
or die( mysqli_error($sql) );
if (mysqli_num_rows($result) > 0) {
$text = '';
while($row = mysqli_fetch_assoc($result)) {
$Channel_Location = $row['Channel_Location'];
$text = $text . $Channel_Location;
}
}
Now the issue i'm having is the php isnt displaying the result returned by the MYSQL query which is stored in a session later on in the code to be displayed on a dummy page it comes up with the following error
Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result
The my SQL query does exactly what I need it to I just need to, so I don't really want to change it. I just need some advice on how i'd get the PHP to echo the #uid
is there anyone willing to help me solve the issue? if so thankyou.

Answer 1

You have 3 queries in your $sql so you should use multi_query function
http://php.net/manual/en/mysqli.multi-query.php
And you can change your first query to:
SET #uid = 0;
SELECT #uid := ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);
Update You can try this fragment of your code modified with all commented improvements.
$sql = 'SET #uid = 0;';
$sql .= 'SELECT #uid:= ID FROM channels WHERE Used = 0 ORDER BY RAND() LIMIT 1);';
$sql .= 'UPDATE channels SET Used = 1 WHERE ID = #uid;';
$sql .= 'SELECT * FROM channels WHERE ID IN = #uid;';
if (mysqli_multi_query($conn, $sql)) {
do {
$result = mysqli_store_result($conn);
} while(mysqli_next_result($conn));
if (mysqli_num_rows($result) > 0) {
$text = '';
while($row = mysqli_fetch_assoc($result)) {
$Channel_Location = $row['Channel_Location'];
$text = $text . $Channel_Location;
}
}
} else {
die( mysqli_error($conn) );
}