mysql - SELECT 查询检查数据库之外的内容

Question

我在下面有以下查询，它在大多数情况下都有效，直到我注意到它导致了一个小问题。基本上，现有查询的作用是从我的“home_comments”表中选择所有信息，然后匹配 profile_img 表中的最大 ID 并获取该图像。

我遇到的问题是，并非我的所有用户都在该数据库中拥有个人资料图片。对于默认图像，我使用静态图像。那么，无论如何我可以检查用户是否有 profile_img，如果没有则使用它？

$default_profile_img = '<img class="home-profile-pic" src="profile_images/default.jpg">';

查询:

$select_comments_sql = "
        SELECT c. *, p.user_id, p.img
        FROM home_comments AS c
        INNER JOIN (SELECT max(id) as id, user_id 
                    FROM profile_img 
                    GROUP BY user_id) PI
          on PI.user_id = c.user_id
        INNER JOIN profile_img p
          on PI.user_id = p.user_id
         and PI.id = p.id
        ORDER BY c.id DESC

编辑:更新的 SQL

$select_comments_sql = "
    //SELECT c. *, p.user_id, p.img
    SELECT c. *, PI.user_id, case when PI.img <> '' and PI.img is not null then PI.img else 'profile_images/default.jpg' end img
    FROM home_comments AS c
    LEFT JOIN (SELECT max(id) as id, user_id 
                FROM profile_img 
                GROUP BY user_id) PI
      on PI.user_id = c.user_id
    LEFT JOIN profile_img p
      on PI.user_id = p.user_id
     and PI.id = p.id
    ORDER BY c.id DESC
";

错误:

[27-Oct-2016 13:29:56 America/Chicago] PHP Fatal error:  Uncaught exception 'PDOException' with message 'SQLSTATE[42000]: Syntax error or access violation: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '//SELECT c. *, p.user_id, p.img
            SELECT c. *, PI.user_id, case when PI.img <> ' at line 1' in /home4//public_html/.com/account/ajax-php/comment-retrieve.php:36`

Answer 1

您可以使用 case 语句:

$select_comments_sql = "
    SELECT c. *, p.user_id, case when p.img <> '' and p.img is not null then p.img else 'my-default.png' end img
    FROM home_comments AS c
    INNER JOIN (SELECT max(id) as id, user_id 
                FROM profile_img 
                GROUP BY user_id) PI
      on PI.user_id = c.user_id
    INNER JOIN profile_img p
      on PI.user_id = p.user_id
     and PI.id = p.id
    ORDER BY c.id DESC