Question

我有一个用户表：

user_id | firstname | last name | picUrl ...
    1        bob       smith       null  
    2        bobby     smith       null  
    3        rob       smithies    null

我还有一个朋友表：

friend1_id | friend2_id | status | date_established
      1          2        accepted   2018-02-06 10:23:01
      2          1        accepted   2018-02-06 10:23:01

我有一个简单的查询来根据条件（$ searchString）搜索用户，并根据＆＃34; best＆＃34;进行排序。匹配：

$query = "SELECT user_id, firstname, lastname, picUrl, LOWER(CONCAT(firstname, ' ', lastname)) as fullname  
          FROM user
          WHERE LOWER(CONCAT(firstname, ' ', lastname)) LIKE LOWER('%" . $searchString . "%') AND user_id != '{$userId}'
          ORDER BY
             CASE WHEN fullname = LOWER('" . $searchString . "') THEN 0
                  WHEN fullname = LOWER('" . $searchString . "%') THEN 1
                  WHEN fullname = LOWER('%" . $searchString . "%') THEN 2
                  WHEN fullname = LOWER('%" . $searchString . "') THEN 3
                  ELSE 4
                END, fullname DESC
          LIMIT 5";

查询将忽略当前用户：

user_id != '{$userId}'

因为你无法将自己添加为朋友。

如何忽略/不显示已经是朋友的用户？

示例

我是用户1。

我搜索＆＃34;史密斯＆＃34;。

=＆GT;我是用户1，所以我不是显示。

=＆GT;用户2已经是我的朋友，所以他们没有显示。

=＆GT;只显示 Rob Smithies 。

Answer 1

只需添加一个与friends'ids匹配的条件。更改您的查询如下

$query = "SELECT user_id, firstname, lastname, picUrl, LOWER(CONCAT(firstname, ' ', lastname)) as fullname  
          FROM user
          WHERE LOWER(CONCAT(firstname, ' ', lastname)) LIKE LOWER('%" . $searchString . "%') AND (user_id != '{$userId}' AND user_id not in (SELECT friend1_id from friends where friend2_id = '{$userId}') AND user_id not in (SELECT friend2_id from friends where friend1_id = '{$userId}'))
          ORDER BY
             CASE WHEN fullname = LOWER('" . $searchString . "') THEN 0
                  WHEN fullname = LOWER('" . $searchString . "%') THEN 1
                  WHEN fullname = LOWER('%" . $searchString . "%') THEN 2
                  WHEN fullname = LOWER('%" . $searchString . "') THEN 3
                  ELSE 4
                END, fullname DESC
          LIMIT 5";

MySQL搜索朋友查询

1 个答案: