假设我有两个简单的域类:
class A {
String name
static hasMany = [bs: B]
}
class B {
String title
}
现在,我想生成一个像这样结构化的项目列表:
// id of A instance, name of A instance, comma separated list of Bs titles associated to A istance
1, "A1", "B1, B2"
2, "A2", "B2, B5"
...
这是我的标准:
def list = A.withCriteria {
createAlias 'bs', 'bs', CriteriaSpecification.LEFT_JOIN
projections {
property 'id'
property 'name'
property 'bs.title' // this is the interesting line
}
}
这显然只检索与我的A实例相关的B元素的第一个标题。像这样:
1, "A1", "B1"
2, "A2", "B2"
...
现在,现实生活中的场景有点复杂,我已经简化到了这一点,即:我怎样才能获得与bs游戏相同的mysql group_concat效果?
我试图在一个标准中做到这一点,但如果不可能,我很乐意讨论不同的解决方案。
答案 0 :(得分:0)
与您的订购相同的实施方式。
def list = A.withCriteria {
createAlias 'bs', 'bs', CriteriaSpecification.LEFT_JOIN
projections {
property 'id'
property 'name'
property 'bs.title' // this is the interesting line
}
order "id", "asc"
order "bs.title", "asc"
}
//Bootstrap
def a = new A(name: "TestA1").save()
def a1 = new A(name: "TestA2").save()
def b1 = new B(title: "TitleB1")
def b2 = new B(title: "TitleB2")
def b3 = new B(title: "TitleB3")
def b4 = new B(title: "TitleB4")
def b5 = new B(title: "TitleB5")
[b1, b2, b3].each{a.addToBs(it)}
[b2, b4, b5].each{a1.addToBs(it)}
[a, a1]*.save(flush: true, failOnError: true)
您可以groupBy为每个组合获取一个键值对。
//This can be optimized
list.groupBy({[it[0], it[1]]})
//Would give
[[1, TestA1]:[[1, TestA1, TitleB1], [1, TestA1, TitleB2], [1, TestA1, TitleB3]],
[2, TestA2]:[[2, TestA2, TitleB2], [2, TestA2, TitleB4], [2, TestA2, TitleB5]]
]
答案 1 :(得分:0)
这是一种替代方式
class A {
String name
static hasMany = [bs: B]
def childrenString() {
B.findAllByParent(this).collect{ it.title }.join(',')
}
}
class B {
static belongsTo = A
A parent
String title
static constraints = {
}
}
A.list().each { it ->
println "${it.name}, ${it.childrenString()}"
}