Question

在我的项目中，我试图将所选元素作为JSON从请求的参数返回。

域类：

class Component{
    String name
    String level
    .
    .
    .
}

我有像

这样的http请求

http://localhost:8080/myapp/component/showJson?name=name

所以我应该只返回

{
   name:xyz
}

如果我的要求是

http://localhost:8080/myapp/component/showJson?name=name&level=level

然后我应该回来

{
     name:xyz
     level:1
}

任何建议都表示赞赏。

更新了JSON（多级）

[
   {"name":"one","level":0," 
         componentTypes":[
                {"name":"one one","level":1,
                    "componentTypes":[
                      {"name":"one one one","level":2,"componentTypes":[]}
                    ]
                 },
                 {"name":"one two","level":1,"componentTypes":[]}
         ]
   }, 
   {"name":"two","level":0,"componentTypes"[]},
   {"name":"three","level":0,"componentTypes":[]}
]

class ComponentType {
    String name
    Integer level
    static hasMany = [componentTypes:ComponentType]
    ComponentType parent
    static constraints = {
        parent nullable:true
    }
    static mapWith = "mongo"
}

控制器操作

componentTypeList = ComponentType.createCriteria().list(){
            eq("level", 0)
        }

Answer 1

您可以将params贴图与对象属性贴图相交并返回结果。我没有尝试过这个，但我无法想到它无法运作的原因。

def properties = component.properties;
def result = properties.subMap(params.keySet())
render result as JSON

<强>更新

class ComponentType {

    .
    .
    .



    def toJSON(def params) {
        def properties = this.properties
        def result = properties.subMap(params.keySet())
        if(this.componentTypes) {
            result.componentTypes = componentTypes*.toJSON(params)
        }
        result
    }
}


def componentTypeList = ComponentType.createCriteria().list(){ eq("level", 0) } 
render componentTypeList*.toJSON(params) as JSON

Grails - 从请求参数返回选定的json元素

1 个答案: