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Home/ Questions/Q 6581849
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T16:12:22+00:00 2026-05-25T16:12:22+00:00

Using the following simple code: $aQuery = INSERT into myTable (a, b, c) VALUES

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Using the following simple code:

$aQuery = "INSERT into myTable (a, b, c) VALUES (1, 'Hello', 'Goodbye')";
$result = sqlsrv_query($myConn, $aQuery ); //where $myConn is valid and defined elsewhere
if (!$result) {
    echo 'FAIL';
}
else {
    echo 'It worked BABY!';
}

What is the value of $result if the query executes successfully, and how do I access the resource properly? echo $result; will print: Resource id #8.

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  1. Editorial Team
    Editorial Team
    2026-05-25T16:12:22+00:00Added an answer on May 25, 2026 at 4:12 pm

    When the query fails $result == false

    So in your code:

    if (!$result) { //if result is NOT true
        echo 'FAIL'; //it failed
    }
    else { //otherwise
        echo 'It worked BABY!'; //it did NOT fail
    }
    

    Any value that is not false or 0 is considered true


    Addendum based on comments below.

    Only SELECT returns content.

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