PHP $stmt->num_rows doesnt work by prepared statements

Question

I want to check if ($numRows >= 1) then it should return something.

When I use $con->mysql("{QUERY}"), it works.

But when I use $stmt = $con->prepare("{QUERY}"), it doesn't work.

Does anyone have a clue?

Working method

<?php
if ($result = $con->query("SELECT username FROM users WHERE username = 'test'")) {
    $numRows = $result->num_rows;

    echo $numRows;
}
?>

Result: 1

Not working method

<?php
$name = 'test';

$stmt = $con->prepare("SELECT username FROM users WHERE username = ?");
$stmt->bind_param('s', $name);

$name = 'test';

$stmt->execute();

$numRows = $stmt->num_rows;

echo $numRows;
?>

Result: 0

`WHERE username = 'tst'` and you're assigning `$name = 'test';` or is that just a typo? — Funk Forty Niner, Oct 31 '16 at 12:37
@Fred-ii- had to edit somethings, did edited. can you check it now? — , Oct 31 '16 at 12:38
You need to store and bind the results. See the answers in the duplicate question it was closed with (it contains a prepared statement method) and Rajdeep's answer. — Funk Forty Niner, Oct 31 '16 at 12:41

score 5 · Accepted Answer · answered Oct 31 '16 at 12:40

You need to transfer the result set from the SELECT query before calling the ->num_rows() method.

// your code    

$stmt->execute();
$stmt->store_result();  
$numRows = $stmt->num_rows;

echo $numRows;

Here's the reference:

http://php.net/manual/en/mysqli.store-result.php

PHP $stmt->num_rows doesnt work by prepared statements

Working method

Not working method

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