使用php和MySQL設置多個變量

Question

所以我有一個要使用的代碼

mysql_select_db("website", $con);
    $result=mysql_query("SELECT * FROM characters where online='1'");
    while ($row=
                mysql_fetch_array($result))
                {
                echo $row['name'];
                }
                if ($row['race'] == "1");
                {
                echo '<img src="img/8-0.gif" />';
                }
                if ($row['class'] == "3");
                {
                echo '<img src="img/3.gif" ?>';
                }
            mysql_close($con);
        ?>

現在，我只希望顯示在線域是否為1的兩個圖像，但是無論如何它們都顯示。 有誰知道我該如何解決？ 謝謝。

Answer 1

為了論證：

mysql_select_db("website", $con);
$result=mysql_query("SELECT * FROM characters where online='1'");
while ($row= mysql_fetch_array($result))
{
    echo $row['name'];
//} This bracket would immediately close your query processing and only display the last images. Or is that the desired behaviour?
    if($row['isOnline'] == '1') { //Makes sure, that 'isOnline' is set before displaying.
            if ($row['race'] == "1");
            {
            echo '<img src="img/8-0.gif" />';
            }
            if ($row['class'] == "3");
            {
            echo '<img src="img/3.gif" ?>';
            }
    }
} //This bracket closes the actual query result handling
mysql_close($con);
?>

Answer 2

            if ($row['race'] == "1" AND $row['online'] == 1);
            {
            echo '<img src="img/8-0.gif" />';
            }
            if ($row['class'] == "3" AND $row['online'] == 1);
            {
            echo '<img src="img/3.gif" ?>';
            }