Mysql - 使用一个查询从多个表中删除

Question

我有4个表,用于存储每个用户的不同信息.每个表都有一个带有user_id的字段,用于标识哪一行属于哪个用户.如果我想删除用户,这是从多个表中删除该用户信息的最佳方法吗？我的目标是在一个查询中完成.

查询:

"DELETE FROM table1 WHERE user_id='$user_id';
DELETE FROM table2 WHERE user_id='$user_id';
DELETE FROM table3 WHERE user_id='$user_id';
DELETE FROM table4 WHERE user_id='$user_id';";

Answer 1

显然,这是可能的.从手册:

您可以在DELETE语句中指定多个表,以根据WHERE子句中的特定条件从一个或多个表中删除行.但是,您不能在多表DELETE中使用ORDER BY或LIMIT.table_references子句列出了连接中涉及的表.其语法在第12.2.8.1节"JOIN语法"中描述.

手册中的示例是:

DELETE t1, t2 FROM t1 INNER JOIN t2 INNER JOIN t3
WHERE t1.id=t2.id AND t2.id=t3.id;

应适用1:1.

Answer 2

您可以使用ON DELETE CASCADE选项在表上定义外键约束.

然后从父表中删除记录将从子表中删除记录.

请访问此链接:http://dev.mysql.com/doc/refman/5.5/en/innodb-foreign-key-constraints.html

Answer 3

您还可以使用以下查询:

DELETE FROM Student, Enrollment USING Student INNER JOIN Enrollment ON Student.studentId = Enrollment.studentId WHERE Student.studentId= 51;

Answer 4

在这种情况下,连接语句不必要地复杂化.原始问题仅涉及同时从多个表中删除给定用户的记录.直觉上,你可能会期望这样的东西起作用:

DELETE FROM table1,table2,table3,table4 WHERE user_id='$user_id'

当然,它没有.但是,不是编写多个语句(冗余和低效),使用连接(对于初学者来说很难)或外键(对于新手来说更难以及在所有引擎或现有数据集中都不可用),您可以使用LOOP简化代码!

作为使用PHP的基本示例(其中$ db是您的连接句柄):

$tables = array("table1","table2","table3","table4");
foreach($tables as $table) {
  $query = "DELETE FROM $table WHERE user_id='$user_id'";
  mysqli_query($db,$query);
}

希望这有助于某人!

Answer 5

您可以使用以下查询从多个表中删除行,

DELETE table1, table2, table3 FROM table1 INNER JOIN table2 INNER JOIN table3 WHERE table1.userid = table2.userid AND table2.userid = table3.userid AND table1.userid=3

Answer 6

从两个带外键的表中你可以试试这个查询:

DELETE T1, T2
FROM T1
INNER JOIN T2 ON T1.key = T2.key
WHERE condition

Answer 7

DELETE 的文档告诉您多表语法。

DELETE [LOW_PRIORITY] [QUICK] [IGNORE]
    tbl_name[.*] [, tbl_name[.*]] ...
    FROM table_references
    [WHERE where_condition]

或者：

DELETE [LOW_PRIORITY] [QUICK] [IGNORE]
    FROM tbl_name[.*] [, tbl_name[.*]] ...
    USING table_references
    [WHERE where_condition]