我的json字符串看起来像这样,
{
"userList" : {
"user" : [{
"Id" : "33",
"userId" : "raji2",
"customerId" : "35",
"username" : "raji2",
"status" : "1",
"locked" : "0",
"customAttributes" : "{\"uiPageId\":\"\",\"uiPageName\":\"\",\"uiPageViewMode\":\"Normal\"}",
"addressList" : {
"address" : ""
},
"contactList" : {
"contact" : {
"contactno" : "+919526346097",
"email" : "raji@gmail.com"
}
},
"roleList" : {
"roleId" : "7"
},
"privilegeList" : {
"privilegeId" : ["1", "10", "11", "12", "13", "14", "15", "16", "17", "23", "24", "25", "26", "27"]
},
"entityList" : {
"entityId" : ""
}
}, {
"Id" : "34",
"userId" : "raji3",
"customerId" : "35",
"username" : "raji3",
"status" : "1",
"locked" : "0",
"customAttributes" : "{\"uiPageId\":\"\",\"uiPageName\":\"\",\"uiPageViewMode\":\"Normal\"}",
"addressList" : {
"address" : ""
},
"contactList" : {
"contact" : {
"contactno" : "+919526346097",
"email" : "raji@gmail.com"
}
},
"roleList" : {
"roleId" : "7"
},
"privilegeList" : {
"privilegeId" : ["1", "10", "11", "12", "13", "14", "15", "16", "17", "23", "24", "25", "26", "27"]
},
"entityList" : {
"entityId" : ""
}
}
]
}
}当我尝试使用下面给出的示例解析它,并尝试接收电子邮件或customerId时,数据是空的,我尝试的代码片段是:
JSONObject obj = (JSONObject) new JSONParser().parse(data);
JSONObject jsonObject = (JSONObject) obj;
System.out.println(jsonObject);
String x=(String) jsonObject.get("customerId");
System.out.println("Check Data"+x);我使用了简单的json库。我收到的响应为空。
发布于 2014-04-26 19:05:06
正如其他评论者所说,您不能直接从根元素访问"customerId“。你得走了
root -> userList -> user -> user[0] -> customerId下面的代码实现了这一点
JSONObject obj = (JSONObject) new JSONParser().parse(data);
System.out.println(obj);
JSONObject userList = (JSONObject) obj.get("userList");
JSONArray user = (JSONArray) userList.get("user");
JSONObject userObj = (JSONObject) user.get(0);
String customerId = (String) userObj.get("customerId");
System.out.println("Check Data " + customerId);因为每个用户对象都有一个customerId属性,所以必须选择要使用的用户对象。如果希望使用第二个user对象,可以将user.get(1);放在第五行。
编辑要获得每个用户的customerId,请将最后三行替换为循环:
JSONObject obj = (JSONObject) new JSONParser().parse(data);
System.out.println(obj);
JSONObject userList = (JSONObject) obj.get("userList");
JSONArray user = (JSONArray) userList.get("user");
for (Object userObj : user) {
JSONObject userJSONObject = (JSONObject) userObj;
String customerId = (String) userJSONObject.get("customerId");
System.out.println("Check Data " + customerId);
}发布于 2014-04-26 18:45:57
我没有使用您的库,但我猜您需要首先获取"userList“字段,然后获取"user”数组并迭代数组中的JSONObjects,以获得它们各自的"customerId“。
发布于 2014-04-26 18:56:51
您将需要进行2次解析。下面的代码将不起作用。
jsonObject.get("userList").get("user");
尝试一个接一个地使用by访问最内部的元素。
https://stackoverflow.com/questions/23309710
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