(函数.MySQL -result]:无法跳转到MySQL结果索引3的第0行

Question

我的时事通讯出错了，我不知道哪里出错了。请帮帮我。现在我在做通讯，我第一次在我的代码中使用MySQL。

错误如下:

警告:mysql_result() [function。

MySQL .php的第54行

下面是我的代码:

    if($mail == NULL){
    }
    else{
        $token = sha1(time());
        $result1 = @mysql_query("INSERT INTO newsletter (Address,Token) VALUES ("".$mail."", "".$token."") ");
        if ($result1) {
            sendmail($mail);
        }
        else{ /*This else */
            $result2 = mysql_query("SELECT Confirmed FROM newsletter where Address = "".$mail."" ");
            $confirm = mysql_result($result2,0);
            if($confirm == "y"){
            }
            else if($confirm == "n"){
            }
        }
    }
    ?>

我做错了什么?

Answer 1

你可以试试:

(...)
else{ /*This else */
    $result2 = mysql_query("SELECT Confirmed FROM newsletter where Address = "".$mail."" ");
    if($row = mysql_fetch_array($result2)) {
        $confirm = $row["Confirmed"];
    }
    else {
        $confirm = "n"; // Returned 0 rows
    }
    $confirm = mysql_result($result2,0);
    if($confirm == "y"){
}
else if($confirm == "n"){
(...)