Postgresql插入多个记录的表值

gj3fmq9x  于 2023-04-29  发布在  PostgreSQL
关注(0)|答案(2)|浏览(108)

我需要在一个表(groupL)中插入两行(custId,aName)。custId是常量,而aName是基于嵌套子查询的可变大小的多个记录。
该子查询是:

SELECT aName 
FROM artwork 
WHERE artwork.title IN (SELECT title 
                        FROM classify 
                        WHERE NEW.g_name = classify.g_name);

我试过很多SQL,但我真的不明白。我正在寻找这样的东西,如果我们在aName(name1,name2)中有两行,它会做类似的事情

INSERT INTO groupL(custId, aName) 
    SELECT custIdConstant, name1 
    WHERE NOT EXISTS (SELECT custIdConstant, aName 
                      FROM groupL 
                      WHERE custId = custIdConstant AND aName = name1);

INSERT INTO groupL(custId, aName) 
    SELECT custIdConstant, name2 
    WHERE NOT EXISTS (SELECT custIdConstant, aName 
                      FROM groupL 
                      WHERE custId = custIdConstant AND aName = name2);

我该怎么做?任何指导将不胜感激。

nfg76nw0

nfg76nw01#

你可以在一个Go中完成,但是如果你想比较的话,你需要先有id,当然还有FROM子句

CREATE tABLe groupL(custId int , aName TEXT)
CREATE TABLE
CREATE TABLe aName (name1 TEXT, name2 TEXT)
CREATE TABLE
INSERT INTO aName VALUEs ('A','B')
INSERT 0 1
DO $$
DECLARE  
   custIdConstant integer := 10;  
BEGIN  
INSERT INTO groupL(custId, aName) 
SELECT custIdConstant, name1 FROM aName  WHERE NOT EXISTS (SELECT 1 FROM groupL WHERE custId = custIdConstant AND aName = name1);

INSERT INTO groupL(custId, aName) 
SELECT custIdConstant, name2 FROM aName WHERE NOT EXISTS (SELECT 1 FROM groupL WHERE custId = custIdConstant AND aName = name2);

END $$;
DO
SELECT * FROM groupL
阿纳梅
10个一个
10个B
DO $$
DECLARE  
   custIdConstant integer := 10;  
BEGIN  
INSERT INTO groupL(custId, aName) 
SELECT custIdConstant, name1 FROM aName  WHERE NOT EXISTS (SELECT 1 FROM groupL WHERE custId = custIdConstant AND aName = name1);

INSERT INTO groupL(custId, aName) 
SELECT custIdConstant, name2 FROM aName WHERE NOT EXISTS (SELECT 1 FROM groupL WHERE custId = custIdConstant AND aName = name2);

END $$;
DO
SELECT * FROM groupL
阿纳梅
10个一个
10个B
SELECT 2

fiddle

2nbm6dog

2nbm6dog2#

我想你只是想通过第二个子查询来检查这些行的存在:

INSERT INTO groupL (custId, aName)
SELECT <custId>, a.aName 
FROM artwork as a
WHERE a.title IN (SELECT c.title 
                  FROM classify as c INNER JOIN NEW ON NEW.g_name = c.g_name)
  AND NOT EXISTS (SELECT 1
                  FROM groupL as g
                  WHERE g.custId = <custId> AND g.aName = a.aName);

您可能还想检查merge

相关问题