dsz1966 2013-04-04 22:01
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已采纳

MySql / PHP查询返回空

Here is my code:

$result = mysqli_query($dbconnection, Data::followUser($user_id, $followUser_id));

$result returns empty here.

followUser method in class Data

public static function followUser($user_id, $followUser_id) {
    global $database;

    $query = "
        SELECT * 
        FROM profile_follow
        WHERE user_id = '{$user_id}' 
            AND follow_id = '{$followUser_id}';";

    $result = $database -> query($query);
    $num = mysqli_num_rows($result);

    if ($num  < 1) {
        $toast = "Follow";

        $query = "
        INSERT INTO profile_follow (user_id, follow_id)
            VALUES ('{$user_id}', '{$followUser_id}');";

        $result = $database -> query($query);


    } elseif ($num > 0) {
        $toast = "Unfollow";

        $query = "
        DELETE FROM profile_follow
        WHERE user_id = '{$user_id}' 
            AND follow_id = '{$followUser_id}';";


        $result = $database -> query($query);

    }

    return $toast;
}

I have verified the function works correctly in echoing out $toast. It is either Follow or Unfollow based on condition. I don't think I am handling it right when it comes out?

Supplemental:

Here is what I am doing with $result:

if ($result == "Follow") {
            $output["result"] = "Follow";
            echo json_encode($output);
    } elseif ($result == "Unfollow") {
            $output["result"] = "Unfollow";
            echo json_encode($output);
    }
  • 写回答

1条回答 默认 最新

  • duanmo5724 2013-04-04 22:05
    关注

    What does this all accomplish? You've basically got:

    mysqli_query($dbconnection, 'Unfollow');
    

    which is NOT a valid query in any way. $result is NOT empty. It's a boolean false, indicating a failed query...

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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