SQL PostgreSQL:在插入中进行子选择

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时间:2020-09-01 11:44:40  来源:igfitidea点击:

PostgreSQL: Sub-select inside insert

sqlpostgresqlinsertsubquery

提问by Bantha Fodder

I have a table called map_tags:

我有一张桌子叫map_tags

map_id | map_license | map_desc

And another table (widgets) whose records contains a foreign key reference (1 to 1) to a map_tagsrecord:

另一个表 ( widgets) 的记录包含对记录的外键引用(1 到 1)map_tags

widget_id | map_id | widget_name

Given the constraint that all map_licenses are unique (however are not set up as keys on map_tags), then if I have a map_licenseand a widget_name, I'd like to perform an insert on widgetsall inside of the same SQL statement:

考虑到所有map_licenses 都是唯一的约束(但是没有设置为 上的键map_tags),那么如果我有 amap_license和 a widget_name,我想widgets在同一个 SQL 语句的所有内部执行插入:

INSERT INTO
    widgets w
(
    map_id,
    widget_name
)
VALUES (
    (
        SELECT
            mt.map_id
        FROM
            map_tags mt
        WHERE
            // This should work and return a single record because map_license is unique
            mt.map_license = '12345'
    ),
    'Bupo'
)

I believeI'm on the right track but know right off the bat that this is incorrect SQL for Postgres. Does anybody know the proper way to achieve such a single query?

相信我在正确的轨道上,但马上就知道这对于 Postgres 来说是不正确的 SQL。有人知道实现这种单一查询的正确方法吗?

回答by gahooa

Use the INSERT INTO SELECTvariant, including whatever constants right into the SELECTstatement.

使用INSERT INTO SELECT变体,包括SELECT语句中的任何常量。

The PostgreSQL INSERTsyntax is:

PostgreSQLINSERT语法是:

INSERT INTO table [ ( column [, ...] ) ]
 { DEFAULT VALUES | VALUES ( { expression | DEFAULT } [, ...] ) [, ...] | query }
 [ RETURNING * | output_expression [ [ AS ] output_name ] [, ...] ]

Take note of the queryoption at the end of the second line above.

请注意上面第二行末尾的查询选项。

Here is an example for you.

这是一个例子。

INSERT INTO 
    widgets
    (
        map_id,
        widget_name
    )
SELECT 
   mt.map_id,
   'Bupo'
FROM
    map_tags mt
WHERE
    mt.map_license = '12345'

回答by Clodoaldo Neto

INSERT INTO widgets
(
    map_id,
    widget_name
)
SELECT
    mt.map_id, 'Bupo'
FROM
    map_tags mt
WHERE
    mt.map_license = '12345'

回答by J.M.I. MADISON

Quick Answer:You don't have "a single record" you have a "set with 1 record" If this were javascript: You have an "array with 1 value" not "1 value".

快速回答:您没有“单个记录”,您有一个“设置 1 个记录”如果这是 javascript:您有一个“具有 1 个值的数组”而不是“1 个值”。

In your example, one record may be returned in the sub-query, but you are still trying to unpack an "array" of records into separate actual parameters into a place that takes only 1 parameter.

在您的示例中,子查询中可能会返回一条记录,但您仍在尝试将记录的“数组”解包为单独的实际参数,并将其放入仅包含 1 个参数的位置。

It took me a few hours to wrap my head around the "why not". As I was trying to do something very similiar:

我花了几个小时才把头放在“为什么不”上。当我试图做一些非常相似的事情时:

Here are my notes:

以下是我的笔记:

tb_table01: (no records)
+---+---+---+
| a | b | c | << column names
+---+---+---+

tb_table02:
+---+---+---+
| a | b | c | << column names
+---+---+---+
|'d'|'d'|'d'| << record #1
+---+---+---+
|'e'|'e'|'e'| << record #2
+---+---+---+
|'f'|'f'|'f'| << record #3
+---+---+---+

--This statement will fail:
INSERT into tb_table01
    ( a, b, c )
VALUES
    (  'record_1.a', 'record_1.b', 'record_1.c' ),
    (  'record_2.a', 'record_2.b', 'record_2.c' ),

    -- This sub query has multiple
    -- rows returned. And they are NOT
    -- automatically unpacked like in 
    -- javascript were you can send an
    -- array to a variadic function.
    (
        SELECT a,b,c from tb_table02
    ) 
    ;

Basically, don'tthink of "VALUES" as a variadicfunction that can unpack an array of records. There is no argument unpacking here like you would have in a javascript function. Such as:

基本上,不要将“ VALUES”视为 可以解压缩记录数组的可变参数函数。这里没有像在 javascript 函数中那样解包的参数。如:

function takeValues( ...values ){ 
    values.forEach((v)=>{ console.log( v ) });
};

var records = [ [1,2,3],[4,5,6],[7,8,9] ];
takeValues( records );

//:RESULT:
//: console.log #1 : [1,2,3]
//: console.log #2 : [4,5,7]
//: console.log #3 : [7,8,9]

Back to your SQL question:

回到你的 SQL 问题:

The reality of this functionality not existing does not change just because your sub-selection contains only one result. It is a "set with one record" not "a single record".

不存在此功能的现实不会因为您的子选择只包含一个结果而改变。它是一个“单条记录”,而不是“单条记录”。