Question

如何离开外连接2表？

class Person {

    String firstName
    String lastName
    String gender

    //static hasMany = [votes: Vote]

    static mapping = {
        version false
    }

    static constrains = {
    }

}
class Vote {

    Person voter;
    Person subject;

    static mapping = {
        version false
    }

    static constraints = {
        voter nullable: false
        subject nullable: false
    }


}

我需要让每个未经投票的人为特定的人。让我说1人投票给5人中的3人，我需要另外2人他没有投票为他出现。查询应该如何？

修改

def personInstance1 = new Person(firstName: "Antonio", lastName: "Vivaldi", gender: "m")
def personInstance2 = new Person(firstName: "Dennis", lastName: "Rodman", gender: "m")
def personInstance3 = new Person(firstName: "Marc", lastName: "Oh", gender: "m")
def personInstance4 = new Person(firstName: "Gudrun", lastName: "Graublume", gender: "w")
def personInstance5 = new Person(firstName: "Hilde", lastName: "Feuerhorn", gender: "w")
def personInstance6 = new Person(firstName: "Mandy", lastName: "Muller", gender: "w")

personInstance1.save()
personInstance2.save()
personInstance3.save()
personInstance4.save()
personInstance5.save()
personInstance6.save()

def voteInstance1 = new Vote(voter: personInstance1, subject: personInstance2)
def voteInstance2 = new Vote(voter: personInstance1, subject: personInstance3)
def voteInstance3 = new Vote(voter: personInstance1, subject: personInstance4)
def voteInstance4 = new Vote(voter: personInstance1, subject: personInstance5)
def voteInstance5 = new Vote(voter: personInstance2, subject: personInstance1)

voteInstance1.save()
voteInstance2.save()
voteInstance3.save()
voteInstance4.save()
voteInstance5.save()

这是我的grails bootstrap文件，Antonio和Dennis已投票，每个人都需要出示他们没有投票的人员名单。

修改这样我似乎得到丹尼斯的结果，因为他只投了一次，但如果我把v.voter_id = 1，为了得到安东尼奥的结果，结果是根据他做了多少票而加倍。

 SELECT first_name FROM vote as v 
 LEFT OUTER JOIN person as p 
 ON v.subject_id != p.id AND v.voter_id = 2 
 WHERE p.id IS NOT NULL

Answer 1

试试这个：

SELECT * FROM Person P
WHERE NOT EXISTS(
    SELECT 'Vote' FROM Vote V
    WHERE V.subject = P
)

通过这种方式，您可以提取所有没有投票的人

修改

在SQL中，您可以通过以下方式检索矩阵：

CREATE TABLE #person (nome varchar(30)) CREATE TABLE #vote (votante varchar(30), candidato varchar(30)) INSERT INTO #person values ('Antonio Vivaldi'), ('Dennis Rodman'), ('Marc Oh'), ('Gudrun Graublume'), ('Hilde Feuerhorn'), ('Mandy Muller') INSERT INTO #vote values ('Antonio Vivaldi', 'Dennis Rodman'), ('Antonio Vivaldi', 'Marc Oh'), ('Antonio Vivaldi', 'Gudrun Graublume'), ('Antonio Vivaldi', 'Hilde Feuerhorn'), ('Dennis Rodman', 'Antonio Vivaldi') SELECT * FROM #person p CROSS JOIN #person c WHERE NOT EXISTS( SELECT 'X' FROM #vote v WHERE v.votante = p.nome AND v.candidato = c.nome ) AND p.nome <> c.nome ORDER BY p.nome

grails hql左外连接

1 个答案: