Question

我有两个mysql表，其中一个内容称为请愿书

{"success":1,"petitions":[{"id":"6","name":"should he go","timestamp":"2013-10-26 03:02:44"},{"id":"3","name":"Olara Otunu should get married","timestamp":"2013-10-24 14:33:53"},{"id":"4","name":"Teachers deserve 30 not 20 salary rise","timestamp":"2013-10-24 14:33:53"},{"id":"5","name":"Prostitution should be banned","timestamp":"2013-10-24 14:33:53"},{"id":"1","name":"Has Jennifer Musisi done great work for Kampala","timestamp":"2013-10-24 14:32:58"},{"id":"2","name":"Do lecturers deserve 100% salary increase","timestamp":"2013-10-24 14:32:58"}]}

，另一个表名为petition_response

{"success":1,"petition_response":[{"id":"2","petitionID":"2","yes":"0","no":"1","memberID":"14","timestamp":"2013-11-02 08:36:20"},{"id":"1","petitionID":"1","yes":"1","no":"0","memberID":"14","timestamp":"2013-11-01 21:26:02"}]}

我需要选择在petition_response中没有回复的请愿书。但是如果他们在petition_response表中有任何响应，那么memberID不应该等于14

我在下面尝试过这段代码，但它无效

$result = mysql_query("select * from petition_response where memberID='14' order by timestamp DESC", $db->connect());
while($row = mysql_fetch_assoc($result)){
    $id = $row['id'];

    $result2 = mysql_query("select * from petitions where id != '$id' order by timestamp DESC", $db->connect());            
}
return $result2;

Answer 1

$query = "
SELECT * 
  FROM petition_response pr
  LEFT
  JOIN petitions p
    ON p.id = pr.id
 WHERE pr.memberID = 14 
   AND p.id IS NULL
 ORDER
    BY pr.timestamp DESC;
";

Answer 2

您希望实现反连接，在MySQL中有三种可能性：

使用NOT IN：

SELECT *
FROM   petitions
WHERE  id NOT IN (
         SELECT petitionID
         FROM   petition_response
         WHERE  memberID = 14
       )

使用NOT EXISTS：

SELECT *
FROM   petitions p
WHERE  NOT EXISTS (
         SELECT *
         FROM   petition_response r
         WHERE  r.petitionID = p.id
            AND r.memberID   = 14
         LIMIT  1
       )

使用OUTER JOIN：

SELECT p.*
FROM   petitions p
         LEFT OUTER JOIN petition_response r
                      ON r.petitionID = p.id AND r.memberID = 14
WHERE  r.petitionID IS NULL

在sqlfiddle上查看。

根据@Quassnoi's analysis：

摘要

MySQL可以优化所有三种方法来进行NESTED LOOPS ANTI JOIN。

它将从t_left中获取每个值，并在t_right.value的索引中查找。如果索引命中或索引未命中，相应的谓词将分别立即返回FALSE或TRUE，并且将立即从t_left返回行的决定而不检查t_right中的其他行。

然而，这三种方法生成三个不同的计划，由三个不同的代码执行。执行EXISTS谓词的代码 30％效率低于执行index_subquery和LEFT JOIN优化使用Not exists方法的代码。

这就是为什么在 MySQL 中搜索缺失值的最佳方法是使用LEFT JOIN / IS NULL或NOT IN而不是NOT EXISTS。

如何根据mysql中另一个表的结果从一个表中获取内容

2 个答案:

摘要