Question

我有这个SQL查询：

SELECT users.*, users_oauth.* FROM users LEFT JOIN users_oauth ON users.user_id = users_oauth.oauth_user_id WHERE (
(MATCH (user_email, user_firstname, user_lastname) AGAINST ('"+smith "+john"' IN BOOLEAN MODE)) ) 
ORDER BY user_date_accountcreated DESC LIMIT 0,50

有没有办法让此查询的COUNT(*)忽略同一查询中的LIMIT 0,50？或者我必须做2个查询，一个用于结果，一个用于COUNT（*）？

感谢。

Answer 1

如果您需要在每行上重复计数（*），您可以使用相关查询添加列

，则限制不是问题

SELECT users.*, users_oauth.* , (select count(*) FROM users 
        LEFT JOIN users_oauth ON users.user_id = users_oauth.oauth_user_id 
        WHERE (
          (MATCH (user_email, user_firstname, user_lastname) AGAINST ('"+smith "+john"' IN BOOLEAN MODE)) ) )
FROM users 
LEFT JOIN users_oauth ON users.user_id = users_oauth.oauth_user_id 
WHERE (
  (MATCH (user_email, user_firstname, user_lastname) AGAINST ('"+smith "+john"' IN BOOLEAN MODE)) ) 
ORDER BY user_date_accountcreated DESC LIMIT 0,50

但是如果你只想在某些mysql版本中选择的列中添加count（*），你只能得到一行而在其他情况下，你有错误，因为mixin聚合函数和单列没有声明group by子句

Answer 2

如果您的MariaDB版本是10.2.0+，则支持窗口功能，您可以使用COUNT(*) OVER ()：

SELECT 
  users.*, 
  users_oauth.*,
  COUNT(*) OVER () AS countall
FROM users 
LEFT JOIN users_oauth ON users.user_id = users_oauth.oauth_user_id 
WHERE 
  MATCH (user_email, user_firstname, user_lastname) AGAINST ('"+smith "+john"' IN BOOLEAN MODE)
ORDER BY user_date_accountcreated DESC 
LIMIT 0,50

如果您的MariaDB版本是10.2.0-（在此之前），您可以在列中使用嵌套选择：

SELECT 
  users.*, 
  users_oauth.*,
  (SELECT count(*)
   FROM users 
   LEFT JOIN users_oauth ON users.user_id = users_oauth.oauth_user_id 
   WHERE MATCH (user_email, user_firstname, user_lastname) AGAINST ('"+smith "+john"' IN BOOLEAN MODE)
  ) AS countall
FROM users 
LEFT JOIN users_oauth ON users.user_id = users_oauth.oauth_user_id 
WHERE 
  MATCH (user_email, user_firstname, user_lastname) AGAINST ('"+smith "+john"' IN BOOLEAN MODE)
ORDER BY user_date_accountcreated DESC 
LIMIT 0,50

具有LIMIT的查询的计数（*）

2 个答案: