我的代码有什么问题？？？ 更新查询不起作用

Question

I'm posting this question after debugging it several times. Please have a look.

<?php
if(isset($_POST["g"]) && isset($_POST["c"]))
{
    $g = preg_replace('#[^a-z]#i', '', $_POST['g']);
    $c = preg_replace('#[^a-z ]#i', '', $_POST['c']);
    //echo $g.$c.$log_username;
    if($g == "" || $c == "")
    {
        echo "The form submission is missing values.";
        exit();  
    }
    else
    {
        $sql = "UPDATE users SET gender='$g',country='$c' WHERE username='$log_username'";
        $query = mysqli_query($db_conx,$sql);
        //echo $query;
        $numrows = mysqli_num_rows($query);
        //echo $numrows;
        if($numrows > 0)
        {
            echo "success";
            exit();
        }
        else
        {
            echo "failed";
            exit();
        }
        exit();
    }
}
?>

The error i'm getting is Warning: mysqli_num_rows() expects parameter 1 to be mysqli_result, boolean given in C:\\xampp\\htdocs\\sns2\\root\\account_config.php on line 36

I'm perfectly getting the values of $g,$c,$log_username (tested it using echo). I'm being connected to the database. I don't think there is something wrong with the query syntax because I have manually executed the same query using the values of $g,$c,$log_username . The query was executed. So could be wrong in this?? Please help!!

Answer 1

Citing the php documentation for mysqli_query :

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

So you're probably getting a TRUE return value and should use http://de3.php.net/manual/en/mysqli.affected-rows.php instead

Answer 2

在 UPDATE 查询的情况下使用 mysqli_affected_rows。