用java中的键对Json响应进行分组 - android studio
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【中文标题】用java中的键对Json响应进行分组 - android studio【英文标题】：Grouping Json response with keys in java - android studio 【发布时间】：2017-01-05 09:04:02 【问题描述】：我是 java 和 android 开发的新手。我正在学习 json 解析。我得到了类似这样的 json 响应。
"code":"0","data":
[ "chrDesigName":"Developer","chrempName":"Test Employee1",
"chrDesigName":"Developer","chrempName":"Test Employee2",
"chrDesigName":"Tester","chrempName":"Test Employee3",
"chrDesigName":"Analyst","chrempName":"Test Employee4",
"chrDesigName":"Developer","chrempName":"Test Employee5",
"chrDesigName":"Tester","chrempName":"Test Employee6"]
我想把这个 Json 解析成这样的东西。
"Developer" : ["chrempName" : "Test Employee1",
"chrempName" : "Test Employee2",
"chrempName" : "Test Employee5"],
"Tester" : ["chrempName" : "Test Employee3",
"chrempName" : "Test Employee6"],
"Analyst" : ["chrempName" : "Test Employee4"]
我怎样才能做到这一点？我的动机是根据他们的指定对所有员工进行分组。
【问题讨论】：
【参考方案1】：作为一个快速破解，我使用了 2 个不同的库来完成它。您也可以使用 Jackson 来“统一”。
import java.io.StringWriter;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import org.codehaus.jackson.map.ObjectMapper;
import com.google.gson.JsonArray;
import com.google.gson.JsonElement;
import com.google.gson.JsonParser;
public class JsonReformatter
public static void main(String[] args) throws Exception
String jsonString = "\"code\": \"0\",\"data\": [\"chrDesigName\": \"Developer\",\"chrempName\": \"Test Employee1\",\"chrDesigName\": \"Developer\",\"chrempName\": \"Test Employee2\",\"chrDesigName\": \"Tester\",\"chrempName\": \"Test Employee3\",\"chrDesigName\": \"Analyst\",\"chrempName\": \"Test Employee4\",\"chrDesigName\": \"Developer\",\"chrempName\": \"Test Employee5\",\"chrDesigName\": \"Tester\",\"chrempName\": \"Test Employee6\"]";
ObjectMapper mapper = new ObjectMapper();
// Map<String, Object> dataMap = mapper.readValue(jsonString,
// new TypeReference<Map<String, Object>>()
// );
//
// System.out.println(dataMap.get("data") +
// ((ArrayList)dataMap.get("data")).get(0).getClass().getName()) ;
JsonArray dataArray = new JsonParser().parse(jsonString).getAsJsonObject().getAsJsonArray("data");
HashMap<String, List<String>> designationsMap = new HashMap<String, List<String>>();
for (JsonElement element : dataArray)
String designation = element.getAsJsonObject().get("chrDesigName").getAsString();
String empName = element.getAsJsonObject().get("chrempName").getAsString();
if (designationsMap.containsKey(designation))
designationsMap.get(designation).add(empName);
else
ArrayList<String> emptyList = new ArrayList<String>();
emptyList.add(empName);
designationsMap.put(designation, emptyList);
StringWriter result = new StringWriter();
mapper.writeValue(result, designationsMap);
System.out.println(result);
打印出来
"Analyst":["Test Employee4"],"Tester":["Test Employee3","TestEmployee6"],"Developer":["Test Employee1","Test Employee2","Test Employee5"]
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
<version>2.8.1</version>
</dependency>
<dependency>
<groupId>com.google.code.gson</groupId>
<artifactId>gson</artifactId>
<version>2.7</version>
</dependency>
【讨论】：
你是如何导入这两个库的？我试过“编译'com.fasterxml.jackson.core:jackson-core:2.7.3'编译'com.fasterxml.jackson.core:jackson-annotations:2.7.3'编译'com.fasterxml.jackson.core:jackson -databind:2.7.3' 在 gradle 中编译 'com.google.code.gson:gson:2.2.4'" 并在库中添加了 "jackson-mapper-lgpl-1.2.1.jar"。出现“错误：无法访问 org.codehaus.jackson.ObjectCodec 的 ObjectCodec 类文件未找到”，无法解析符号 objectmapper 并且无法解析方法 writevalue 添加了我在 Maven 中使用的依赖项。如果您手动添加 jar，可能会很困难，因为在运行时您可能会遇到 ClassNotFound 异常 很好。我现在只得到 mapper.writefile 的未处理异常 你能说出完整的堆栈跟踪和你得到的错误吗？另外，如果答案有效，请务必接受答案：) 错误:(16, 32) 错误: 包 org.codehaus.jackson.map 不存在, 错误:(28, 32) 错误: 包 org.codehaus.jackson.map 不存在,错误:(52, 9) 错误: 找不到符号类 ObjectMapper, 错误:(52, 35) 错误: 找不到符号类 ObjectMapper, 错误:(85, 16) 错误: 找不到符号类 JsonMappingException【参考方案2】：代码如下：
JsonParser parser = new JsonParser();
JsonElement jsonElement = parser.parse(json);
JsonObject jsonObj = jsonElement.getAsJsonObject();
JsonArray data = new JsonArray();
data = jsonObj.getAsJsonArray("data");
JsonArray developer = new JsonArray();
JsonArray tester = new JsonArray();
JsonArray analyst = new JsonArray();
for (int i=0; i < data.size; i++)
JsonObject itemArr = (JsonObject)data.get(i);
JsonObject temp = new JsonObject();
if(itemArr.get("chrDesigName").getAsString().equals("Developer"))
temp.addProperty("chrempName",itemArr.get("chrempName").getAsString());
developer.add(temp);
else if(itemArr.get("chrDesigName").getAsString().equals("Tester"))
temp.addProperty("chrempName",itemArr.get("chrempName").getAsString());
tester.add(temp);
else if(itemArr.get("chrDesigName").getAsString().equals("Analyst"))
temp.addProperty("chrempName",itemArr.get("chrempName").getAsString());
analyst.add(temp);
JsonObject resultObject = new JsonObject();
resultObject.add("Developer",developer);
resultObject.add("Tester",tester);
resultObject.add("Analyst",analyst);
这样您就可以根据需要创建一个新的resultObject 对象。这是一种静态方式。您可以将其进一步采用动态方式。您需要添加com.google.gson 库来编译此代码。
【讨论】：
我不知道会有多少名称或所有名称。所以我必须以动态的方式做到这一点。你能帮忙吗？【参考方案3】：您可以通过以下方式简单地实现此目的：首先将所有 'chrDesigName' 值存储在一个数组中，然后遍历该数组的每个元素以创建所需的 JSON 对象。
e.g. //using JSON.simple module for decode/encode into JSON object.
JSONObject obj1 = "code":"0","data":
[ "chrDesigName":"Developer","chrempName":"Test Employee1",
"chrDesigName":"Developer","chrempName":"Test Employee2",
"chrDesigName":"Tester","chrempName":"Test Employee3",
"chrDesigName":"Analyst","chrempName":"Test Employee4",
"chrDesigName":"Developer","chrempName":"Test Employee5",
"chrDesigName":"Tester","chrempName":"Test Employee6"];
//Now, store all 'chrDesigName' in a local arrray.
List<String> arr = new ArrayList<String>();
for(JSONObject json : obj1.get("data"))
//avoid to add duplicate entry. e.g. there are multiple Developer so avoid adding more than 1 times.
boolean isDuplicate = false;
for(int i=0; i<arr.size();i++)
if(arr[i]==json.get("chrDesigName"))
isDuplicate = true;
if(isDuplicate==false)
arr.add(json.get("chrDesigName"));
//Now finally create your desired parsed JSON object.
JSONObject result= new JSONObject(); //final Result JSON object.
for(String chrDesigName : arr)
//first adding empty list.
result.put(chrDesigName, new ArrayList<JSONObject>());
//now, checking on each data obj1 if it has chrDesigName and if yes then getting its chrEmpName and adding in list.
for(JSONObject json : obj1.get("data"))
if(chrDesigName==json.get("chrDesigName")
JSONObject sub_json = new JSONObject();
sub_json.put("chrempName",json.get("chrempName");
result.get(chrDesigName).add(sub_json);
【讨论】：
在“result.get(chrDesigName).add(sub_json);”上获取“无法解析方法 add(org.json.JSONObject)”【参考方案4】：代码如下：
public class TestJSONGroup
public static void main(String[] args) throws ParseException
JSONParser parser = new JSONParser();
String jsonString = "\"code\":\"0\",\"data\":"
+ "[ \"chrDesigName\":\"Developer\",\"chrempName\":\"Test Employee1\","
+ " \"chrDesigName\":\"Developer\",\"chrempName\":\"Test Employee2\","
+ " \"chrDesigName\":\"Tester\",\"chrempName\":\"Test Employee3\","
+ " \"chrDesigName\":\"Analyst\",\"chrempName\":\"Test Employee4\","
+ " \"chrDesigName\":\"Developer\",\"chrempName\":\"Test Employee5\","
+ " \"chrDesigName\":\"Tester\",\"chrempName\":\"Test Employee6\"]";
//JSONObject obj1 = (JSONObject) parser.parse( jsonString );
JSONObject obj1 = new JSONObject(jsonString);
//Now, store all 'chrDesigName' in a local arrray.
List<String> arr = new ArrayList<String>();
JSONArray json = obj1.getJSONArray("data");
for (int j = 0; j < json.length(); j++)
//avoid to add duplicate entry. e.g. there are multiple Developer so avoid adding more than 1 times.
boolean isDuplicate = false;
String chrDesigName = new String();
for(int i=0; i<arr.size();i++)
JSONObject jsonObject1 = json.getJSONObject(j);
chrDesigName = jsonObject1.optString("chrDesigName");
if(arr.get(i).equalsIgnoreCase(chrDesigName.trim()))
isDuplicate = true;
if(isDuplicate==false)
arr.add(chrDesigName);
arr.remove(0);
//Now finally create your desired parsed JSON object.
JSONObject result= new JSONObject(); //final Result JSON object.
for(String chrDesigName : arr)
//first adding empty list.
result.put(chrDesigName, new ArrayList<JSONObject>());
//now, checking on each data obj1 if it has chrDesigName and if yes then getting its chrEmpName and adding in list.
String chrDesigName2 = new String();
JSONArray array = new JSONArray();
for (int j = 0; j < json.length(); j++)
JSONObject jsonObject1 = json.getJSONObject(j);
chrDesigName2 = jsonObject1.optString("chrDesigName");
if(chrDesigName.equalsIgnoreCase(chrDesigName2.trim()))
JSONObject sub_json = new JSONObject();
sub_json.put("chrempName",json.get(j));
array.put(json.get(j));
result.put(chrDesigName, array);
System.out.println("result: "+ result.toString());
结果会是，只是在其他一些地方修改
"Analyst": [
"chrDesigName": "Analyst",
"chrempName": "Test Employee4"
],
"Tester": [
"chrDesigName": "Tester",
"chrempName": "Test Employee3"
,
"chrDesigName": "Tester",
"chrempName": "Test Employee6"
],
"Developer": [
"chrDesigName": "Developer",
"chrempName": "Test Employee1"
,
"chrDesigName": "Developer",
"chrempName": "Test Employee2"
,
"chrDesigName": "Developer",
"chrempName": "Test Employee5"
]
【讨论】：
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