CODEWITHSUNDEEP
phpmysql
Martin Beardmore
Martin Beardmore

Reputation: 75

PHP not displaying result from MYSQL query

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.

Upvotes: 2

Views: 131

Answers (1)

Alex
Alex

Reputation: 17289

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) );
}

Upvotes: 1

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