当我的查询返回1 [关闭]时，我的查询返回0结果

Question

The whole query skips my first if statement to else - "NE VRNE NIČ"... Here's my LINK TO DATABASE STRUCTURE: http://ibin.co/1j1bFYAswzXI

PS: $slug is the same as $naslov, but it get's modified throught the function.

    $id = "SELECT id FROM novica
           where slug = '$slug'";

    $result = $conn->query($id);

    if ($result->num_rows > 0) 
    {
        while($row = $result->fetch_assoc()) 
        {
        $idat = $row["id"];
        }
    }
    else //IT SKIPS DIRECTLY TO THE ELSE - SO IT HAS TO BE WITH QUERY $id
    {
        echo "NE VRNE NIČ!";
    }




    $sql = "UPDATE novica 
            SET naslov = '$naslov', podnaslov = '$podnaslov', vsebina = '$vsebina', slug = '$slug', posodobljeno = '$posodobljeno'
            where id = '$idat'";

    if($conn->query($sql) === true)
    {
        echo "POSODOBO SI";
    }
    else
    {
        $conn->error;
    }

    $conn->close();

Thank you in advance kind programmers.

LINK TO DATABASE STRUCTURE: http://ibin.co/1j1bFYAswzXI

Answer 1

You are setting $idat inside an if scope, to fix it, declare the variable outside the if statement, and try again.

$id = "SELECT id FROM novica
       where slug = '$slug'";

$result = $conn->query($id);

if ($result->num_rows > 0) 
{
    while($row = $result->fetch_assoc()) 
    {
    $idat = $row["id"];
    }
}

Would be... You have wrap a variable in a query with curly brackets. Please see the results now.

$id = "SELECT id FROM novica
       where slug = '{$slug}'";

$result = $conn->query($id);
$idat = 0;

if ($result->num_rows > 0) 
{
    while($row = $result->fetch_assoc()) 
    {
    $idat = $row["id"];
    }
}