Warning: file_get_contents(/data/phpspider/zhask/data//catemap/3/sql-server-2005/2.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Arrays 如何使用嵌套数组创建JSON响应_Arrays_List_Nested_Salesforce_Apex - Fatal编程技术网

Arrays 如何使用嵌套数组创建JSON响应

Arrays 如何使用嵌套数组创建JSON响应,arrays,list,nested,salesforce,apex,Arrays,List,Nested,Salesforce,Apex,我有一个列表,如下所示: { "data1": "abcd", "data2": "efgh", } 我试图通过循环上面的数组并构建以下内容来创建这种格式的JSON响应: { "id": "123456", "nestedArray": [ { "data1": "abcd", "somedata": 1234 }, { "data2": "efgh", "somedata": 1234 } ]

我有一个列表,如下所示:

{
  "data1": "abcd",
  "data2": "efgh",
}
我试图通过循环上面的数组并构建以下内容来创建这种格式的JSON响应:

{
  "id": "123456",
  "nestedArray": [
    {
      "data1": "abcd",
      "somedata": 1234
    },
    {
      "data2": "efgh",
      "somedata": 1234
    }
  ]
}
我创建了一个模型来映射嵌套数组的结构:

public class nestedArray{
    public String data1;
    public Integer data2;

    public nestedArray(string data) {
        this.data1 = uri;
        this.data2 = 1234;
    }
}
但是我被困在如何构建最终响应上,非常感谢任何提示

试试这个:

JSON:

{
  "id": "123456",
  "nestedArray": [
    {
      "data1": "abcd",
      "somedata": 1234
    },
    {
      "data1": "efgh",
      "somedata": 1234
    }
  ]
}

Model:
public class Resp {

    public class NestedArray {
        public String data1;
        public Integer somedata;
    }

    public String id;
    public List<NestedArray> nestedArray;


    public static Resp parse(String json) {
        return (Resp) System.JSON.deserialize(json, Resp.class);
    }
}
JSON:
{
“id”:“123456”,
“Nestedaray”:[
{
“数据1”:“abcd”,
“somedata”:1234
},
{
“数据1”:“efgh”,
“somedata”:1234
}
]
}
型号:
公共类Resp{
公共类嵌套阵列{
公共字符串数据1;
公共数据;
}
公共字符串id;
公共列表嵌套阵列;
公共静态Resp解析(字符串json){
return(Resp)System.JSON.deserialize(JSON,Resp.class);
}
}