你可以做这样的事情(这没有任何错误处理,只是返回null,我不希望被调用的方法):
// First define our class to hold our created 'Emp' objects
@groovy.transform.Canonical
class Emp {
String id
String value
List<Emp> children = []
}
class EmpBuilder extends BuilderSupport{
def children = []
protected void setParent(Object parent, Object child){
parent.children << child
}
protected Object createNode(Object name){
if(name == 'root') {
this
}
else {
null
}
}
protected Object createNode(Object name, Object value){
null
}
protected Object createNode(Object name, Map attributes){
if(name == 'emp') {
new Emp(attributes)
}
else {
null
}
}
protected Object createNode(Object name, Map attributes, Object value){
null
}
protected void nodeCompleted(Object parent, Object node) {
}
Iterator iterator() { children.iterator() }
}
那么,如果我们把这个与你的要求建设者代码如下所示:
b = new EmpBuilder().root() {
emp(id: '3', value: '1')
emp(id:'24') {
emp(id: '1', value: '2')
emp(id: '6', value: '7')
emp(id: '7', value: '1')
}
emp(id: '25') {
emp(id: '1', value: '1')
emp(id: '6', value: '7')
}
}
我们可以打印出 '树',像这样
b.each { println it }
和s ee值,我们得到我们要求的结构:
Emp(3, 1, [])
Emp(24, null, [Emp(1, 2, []), Emp(6, 7, []), Emp(7, 1, [])])
Emp(25, null, [Emp(1, 1, []), Emp(6, 7, [])])
打印树我一定要发出db.each {调用println它} – anish
烨...我忘了粘贴在该位到/ –
当应该代码会来其他如果(名称=='根'){ 这 } – anish