从一个表中查找另一个表中不存在的记录

时间:2023-04-02
本文介绍了从一个表中查找另一个表中不存在的记录的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

我有以下两个表(在 MySQL 中):

I've got the following two tables (in MySQL):

Phone_book
+----+------+--------------+
| id | name | phone_number |
+----+------+--------------+
| 1  | John | 111111111111 |
+----+------+--------------+
| 2  | Jane | 222222222222 |
+----+------+--------------+

Call
+----+------+--------------+
| id | date | phone_number |
+----+------+--------------+
| 1  | 0945 | 111111111111 |
+----+------+--------------+
| 2  | 0950 | 222222222222 |
+----+------+--------------+
| 3  | 1045 | 333333333333 |
+----+------+--------------+

如何查明哪些电话是由 phone_number 不在 Phone_book 中的人拨打的?所需的输出是:

How do I find out which calls were made by people whose phone_number is not in the Phone_book? The desired output would be:

Call
+----+------+--------------+
| id | date | phone_number |
+----+------+--------------+
| 3  | 1045 | 333333333333 |
+----+------+--------------+

推荐答案

有几种不同的方法可以做到这一点,效率各不相同,具体取决于您的查询优化器的性能以及两个表的相对大小:

There's several different ways of doing this, with varying efficiency, depending on how good your query optimiser is, and the relative size of your two tables:

这是最短的语句,如果您的电话簿很短,则可能是最快的:

This is the shortest statement, and may be quickest if your phone book is very short:

SELECT  *
FROM    Call
WHERE   phone_number NOT IN (SELECT phone_number FROM Phone_book)

或者(感谢 Alterlife)

SELECT *
FROM   Call
WHERE  NOT EXISTS
  (SELECT *
   FROM   Phone_book
   WHERE  Phone_book.phone_number = Call.phone_number)

或(感谢 WOPR)

SELECT * 
FROM   Call
LEFT OUTER JOIN Phone_Book
  ON (Call.phone_number = Phone_book.phone_number)
  WHERE Phone_book.phone_number IS NULL

(忽略这一点,正如其他人所说,通常最好只选择您想要的列,而不是*")

(ignoring that, as others have said, it's normally best to select just the columns you want, not '*')

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