fastjson包將json字符串轉為map及轉為List 【我】

 

假設返回報文為：

 

 對應字符串：

{"customer": [{"controlLevel": "13013001","commonRegionName": "集團","createDate": "","custId": "4164275","partyNumber": "LN2018122110041351","extCustId": "","statusCd": "","controlType": "13013001","ChannelMemberId": "","commonRegionId": "100000","custNumber": "1000001994420000","custName": "華夏銀行股份有限公司鞍山分行","industryCd": "DD0302"}]}

 

下面代碼對其進行解析，其中包括轉為Map、轉為List<Map>：

import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.JSONObject;

public static void main(String[] args) {
    String s = "{\"customer\": [{\"controlLevel\": \"13013001\",\"commonRegionName\": \"集團\",\"createDate\": \"\",\"custId\": \"4164275\",\"partyNumber\": \"LN2018122110041351\",\"extCustId\": \"\",\"statusCd\": \"\",\"controlType\": \"13013001\",\"ChannelMemberId\": \"\",\"commonRegionId\": \"100000\",\"custNumber\": \"1000001994420000\",\"custName\": \"華夏銀行股份有限公司鞍山分行\",\"industryCd\": \"DD0302\"}]}";
    //將json字符串轉為map對象
    JSONObject jsonObject = JSONObject.parseObject(s);
    Map<String,Object> map = (Map<String,Object>)jsonObject;
    //將Map中符合數組結構的字符串轉為list對象，list對象中是Map
    String s2 = map.get("customer")+"";
    System.out.println(s2);
    List<Map> list = JSONObject.parseArray(s2, Map.class);//這里的第二個參數也可以用對應的自定義javaBean.class
    //遍歷其中的map，取出每個屬性值
    for (Map item : list) {
        System.out.println(item.get("custId"));
    }
}

控制台輸出：

[{"custId":"4164275","partyNumber":"LN2018122110041351","ChannelMemberId":"","controlType":"13013001","commonRegionName":"集團","industryCd":"DD0302","commonRegionId":"100000","extCustId":"","statusCd":"","custNumber":"1000001994420000","custName":"華夏銀行股份有限公司鞍山分行","createDate":"","controlLevel":"13013001"}]

4164275