Question

在下面的代码我正在构建具有作者和标题的Book.groovy json。我的问题来自Book.groovy我如何删除storesfield ..我已经尝试了以下和渲染后我仍然看到我的商店json回应

 def book_custom_list(Integer max) {
    println "In book_custom_list"
    book_instance.authors.titles.remove('stores')
    book_instance.each { e ->
        println e.authors.titles

    }
    render .......
}

Domain Controllers

Book.groovy

class Book {
String name

static hasMany = [authors: Author]

static constraints = {
}

@Override
public String toString() {
    name
}
}

Author.groovy

class Author {
String name;

static hasMany = [titles : Title]
static constraints = {
    titles cascade: "all-delete-orphan"
}
}

Title.groovy

class Title {

Stores stores
Date dateCreated


 }

JSON：

{
"id": 1,
"name": "test",
"authors": [
  {
    "id": 1,
    "name": "sdsd",
    "titles": [
      {
        "id": 7,
        "stores": {
          "id": 8,
         }
          "dateCreated": "Sep 25, 2013 12:50:42 PM",
      }
      ]
   }]
   }

Answer 1

如果您只想一直包含特定属性，那么您真的想要使用ObjectMarshaller接口。见this article 了解更多详情。

Answer 2

有很多方法可以在2.3中自定义json渲染。请查看文档：{{3}}

从object / json中删除字段属性

2 个答案: