php get_result() returns an empty result set

Question

So, I have a prepared statement that I have successfully prepared, bound, and executed:

SELECT * FROM users WHERE email = ? AND pass = SHA1(?)

It looks like one row is returned, as expected. However, why is the $result variable empty when I call @$stmt->get_result()? Thanks in advance.

$num_rows = mysqli_num_rows($stmt->get_result());

if ($num_rows == 1) {

    // Fetch the result set
    $result = $stmt->get_result();

    //if result empty echo false
    if(empty($result)) {
        echo "result is empty";
    }   
}

Silly question, but you call get_result() at the start of your code, and then call it again inside the if statement - are you sure that it'll return what you're expecting the second time it's called? — andrewsi
– andrewsi, Commented Mar 5, 2016 at 22:49

VolkerK · Accepted Answer · 2016-03-05 23:33:16Z

3

Just to put the two comments together and elaborate a litte....

<?php
$result = $stmt->get_result();
$num_rows = mysqli_num_rows($result);

if ($num_rows == 1) {
    // already fetched the mysqli_result
    // now lets fetch the one record from that result set
    $row = $result->fetch_assoc();
    // ....do something with row
}
else {
    // either 0   ...or more than 1 row
    foo();
}

But you can even get rid of the call to mysqli_num_rows() (so it also works in case of unbuffered queries)

$result = $stmt->get_result();
$row = $result->fetch_assoc();
if ( !$row ) {
    // no such record
    foo();
}
else {
    // ....do something with row

    // might want to check whether there are more matching records
    // given the context there shouldn't, but ...
}

answered Mar 5, 2016 at 23:33

VolkerK

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