MySQL从table01中选择table02上不存在的行[重复]

由一场梦而已发布于 2020-04-09 06:05:48

我有两个表，table03有10行，table01有21行，现在我想从table03中获取行，而table01中不存在这些行，到目前为止，我编写了此查询，但它显示了table03的所有行，甚至有些行也没有。 t存在于table01上。

SELECT T3.`DateAdded`, T3.`Query_name`, T3.`Fund`, T3.`Ticker` 
FROM `table_name03` T3 LEFT JOIN `table_name01` T1 
ON T3.`DateAdded` = T1.`DateAdded` 
AND `T3`.`Query_name` = `T1`.`Query_name` 
AND `T3`.`Fund` = `T1`.`Fund`
AND `T3`.`Ticker` = `T1`.`Ticker`

Table03:

Table01:

菊Oo

2020-04-09 06:05:49

You are on the right track. You can just add a where condition to filter on unmatched rows:

SELECT T3.DateAdded, T3.Query_name, T3.Fund, T3.Ticker 
FROM table_name03 T3 
LEFT JOIN table_name01 T1 
    ON  T3.DateAdded = T1.DateAdded 
    AND T3.Query_name = T1.Query_name 
    AND T3.Fund = T1.Fund
    AND T3.Ticker = T1.Ticker
WHERE T1.DateAdded IS NULL

You can also use not exists:

SELECT T3.DateAdded, T3.Query_name, T3.Fund, T3.Ticker 
FROM table_name03 T3 
WHERE NOT EXISTS (
    SELECT 1
    FROM table_name01 T1 
    WHERE
        T3.DateAdded = T1.DateAdded 
        AND T3.Query_name = T1.Query_name 
        AND T3.Fund = T1.Fund
        AND T3.Ticker = T1.Ticker
)

一场梦而已

这家伙很懒，什么都没留下

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