PostgreSQL 通过子查询返回多行的操作详解

作为一个表达式使用的子查询返回了多列:

在查询中,我们需要以第2条查询语句作为第一条查询语句的条件,但是第一条根据这个条件查询出来是多个数据,这时候又需要保留多个数据,运用子查询就会报错,

以下就为解决这种多对多关系查询,且没有关联关系的表的解决方案:

select c.rain_value,c.ad_nm from
(
select *, json::json->t2.lon_lat as rain_value from actual_time_model_json t1,
(SELECT DISTINCT lon || ‘_’ || lat as lon_lat,ad_nm from grid_all_points_null)t2 where section=’0′ and t1.filename = ‘Z_NWGD_C_BCCD_20180711022009_P_RFFC_SPCC-ER01_201807110800_02401.GRB2’
)c where c.rain_value is not null

补充:PostgreSQL 的子查询 相关的知识 ,exists,any,all

Subquery

SELECT
film_id,
title,
rental_rate
FROM
film
WHERE
rental_rate > (
SELECT
AVG (rental_rate)
FROM
film
);

有了子查询,在设定 需要查询表才能得到 查询条件时,就可以 直接 在一条语句中 写,不用分开多条写了,方便了许多。

子查询返回多条时,可以在 where 子句中 用 IN,来匹配查询条件。

SELECT
film_id,
title
FROM
film
WHERE
film_id IN (
SELECT
inventory.film_id
FROM
rental
INNER JOIN inventory ON inventory.inventory_id = rental.inventory_id
WHERE
return_date BETWEEN ‘2005-05-29’
AND ‘2005-05-30’
);

EXISTS 操作符

在 where 子句的 查询条件中,exists 操作符,会在子查询有返回行时,返回true;不论返回几行。

因此,子查询中的查询字段仅写1就好;标准的写法:EXISTS (SELECT 1 FROM tbl WHERE condition)

SELECT
first_name,
last_name
FROM
customer
WHERE
EXISTS (
SELECT
1
FROM
payment
WHERE
payment.customer_id = customer.customer_id
);

NO EXISTS ,与之相反,当子查询返回0行时,返回true

SELECT first_name,
last_name
FROM customer c
WHERE NOT EXISTS
(SELECT 1
FROM payment p
WHERE p.customer_id = c.customer_id
AND amount > 11 )
ORDER BY first_name,
last_name;

当子查询返回 NULL,会返回true, 也就是返回所有行。

SELECT
first_name,
last_name
FROM
customer
WHERE
EXISTS( SELECT NULL )
ORDER BY
first_name,
last_name;

ANY

与任何子查询返回的 值 匹配就 返回 true

expresion operator ANY(subquery)

表达式一般为 字段

操作符为 >,<,=,<>,>=,<=

ANY 可以与 SOME 替换

子查询 返回的 必须是 一列,

SELECT title
FROM film
WHERE length >= ANY(
SELECT MAX( length )
FROM film
INNER JOIN film_category USING(film_id)
GROUP BY category_id );

The = ANY is equivalent to IN operator.

Note that the <> ANY operator is different from NOT IN. The following expression:

x <> ANY (a,b,c)

is equivalent to

x <> a OR x <> b OR x <> c

ALL

所有子查询返回的值 匹配 则 返回 true

也就是 大于最大,小于最小

SELECT
film_id,
title,
length
FROM
film
WHERE
length > ALL (
SELECT
ROUND(AVG (length),2)
FROM
film
GROUP BY
rating
)
ORDER BY
length;

以上为个人经验,希望能给大家一个参考,也希望大家多多支持。如有错误或未考虑完全的地方,望不吝赐教。


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