有沒有更好的方法來編寫此mysql選擇查詢？

Question

SELECT jobs.*, user_table.* 
FROM jobs 
INNER JOIN user_table 
ON jobs.userid = user_table.userid WHERE 
(user_table.userid = ".$_SESSION['userid']." AND approved = 1) OR 
(user_table.userid = ".$_SESSION['userid']." AND approved = 2)

我以前擁有它的方式不起作用是：

SELECT jobs.*, user_table.* 
FROM jobs INNER JOIN user_table 
ON jobs.userid = user_table.userid WHERE 
user_table.userid = ".$_SESSION['userid']." AND approved = 1 OR approved = 2

第一個有效，但想知道是否有更短的方法。 第二個較短，但不起作用，因為它會拉取已批准等於2但用戶ID不等於會話userid的記錄。

Answer 1

您可以嘗試IN語法：

WHERE user_table.userid = ".$_SESSION['userid']." 
  AND approved IN (1,2)

如果您在usertable(userid, approved)上創建復合索引， usertable(userid, approved)該查詢將利用該索引（如果approved在user_table approved ，則無法從您的問題中得知）。

Answer 2

更改

user_table.userid = ".$_SESSION['userid']." AND approved = 1 OR approved = 2

至

user_table.userid = ".$_SESSION['userid']." AND (approved = 1 OR approved = 2)

在第二個查詢中

Answer 3

SELECT jobs.*, user_table.* 
FROM jobs JOIN user_table 
ON jobs.userid = user_table.userid WHERE 
user_table.userid = ".$_SESSION['userid']." AND (approved = 1 OR approved = 2)