CODEWITHSUNDEEP
phpsqlmysqliprepared-statement
markasoftware
markasoftware

Reputation: 12662

prepared statement mysqli not returning rows

I have some php code with a prepared statement. It is called via Ajax in javascript. I am sure there is an account called mark (no caps) in the accounts table, but when I call this with $_POST['query'] set to m it finds $stmt->num_rows===0 to be true. No errors can be seen, it just doesn't work! As you can tell, it is echoing content into a table. I tried this query in phpMyAdmin: SELECT username FROM accounts WHERE username LIKE '%m%' and it worked fine. I created this code to search a mysql database for usernames. In case your wondering, $conn is a valid mysqli object defined in the include file.

<?php
require_once('./include.php');
$stmt=$conn->stmt_init();
$stmt->prepare('SELECT username FROM accounts WHERE username LIKE ?');
$compquery='%'.$_POST['query'].'%';
$stmt->bind_param('s',$compquery);
$stmt->execute();
echo '<tr><td>';
if($stmt->num_rows!==0){
    $stmt->bind_result($name);
    while($stmt->fetch()){
        echo "$name</td></tr><tr><td>";
    }
   echo '</td></tr>';
}
else
    echo 'No Results Found</td></tr>';

Upvotes: 4

Views: 1914

Answers (1)

Passerby
Passerby

Reputation: 10070

Extending from comment:

You have to use mysql_stmt::store_result() before you can use mysqli_stmt::num_rows:

$stmt=$conn->stmt_init();
$stmt->prepare('SELECT username FROM accounts WHERE username LIKE ?');
$compquery='%'.$_POST['query'].'%';
$stmt->bind_param('s',$compquery);
$stmt->execute();
$stmt->store_result();

Upvotes: 6

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