MySQL，PHP：从最小到最大的回显ID

Question

I have simple code in PHP and actual code throws all items from MySQL table with even ID and under it all items with odd ID.

And I need to all IDs with content will be listed from smallest to largest under themselves.

if( $db-> connect_error ) {
    die( "Not connected" . $db-> connect_error );
}

$sqlEven = "SELECT * FROM documents WHERE MOD (id,2)=0";
$resultEven = $db-> query($sqlEven);

if( $resultEven-> num_rows ) {
    while( $rowEven = $resultEven-> fetch_assoc() ) {
        echo $rowEven["id"] . "<li class=\"dark\"><a href=\"" . $rowEven["url"] . "\">" . $rowEven["name"] . "</a></li>";
    }
}

if( $db-> connect_error ) {
    die( "Not connected" . $db-> connect_error );
}

$sqlOdd = "SELECT * FROM documents WHERE MOD (id,2)<>0";
$resultOdd = $db-> query($sqlOdd);

if( $resultOdd-> num_rows ) {
    while( $rowOdd = $resultOdd-> fetch_assoc() ) {
        echo $rowOdd["id"] . "<li class=\"light\"><a href=\"" . $rowOdd["url"] . "\">" . $rowOdd["name"] . "</a></li>";
    }
}

Answer 1

You can reduce this to one query and check in PHP if the number is even or odd. Using a ternary operator, it can be done inline. Using ORDER BY MOD(id, 2), id (sort by even first, then ascending order), you get the data sorted by even and odd (see live fiddle: http://sqlfiddle.com/#!9/af25e5/2).

This way, you only need one query and one loop.

if ($db->connect_error) {
    die("Not connected: " . $db->connect_error);
}

$sql = "SELECT * FROM documents ORDER BY MOD(id, 2), id";
$result = $db->query($sql);

if ($result->num_rows) {
    while ($row = $result->fetch_assoc()) {
        echo $row["id"] . '<li class="'.($row['id'] % 2 == 0 ? 'dark' : 'light').'"><a href="'.$row["url"].'">'.$row["name"]."</a></li>";
    }
}