java 在java中解析嵌套的json数组
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Parsing nested json array in java
提问by vick_4444
I have json file in below format.
我有以下格式的json文件。
{
"data":[
{
"prjId": 1,
"name" : "Forj1",
"issue": [
{
"id": 00001,
"status" : "Closed"
},
{
"id": 00002,
"status" : "Open"
}
]
},
{
"prjId": 2,
"name" : "Forj2",
"issue": [
{
"id": 00003,
"status" : "Closed"
},
{
"id": 00004,
"status" : "Open"
}
]
}],
"issueCounter": 7,
"success": true
}
Here "data" is array of projects, and within project attribute there is array of "issue".
这里的“数据”是项目数组，在项目属性中有“问题”数组。
So far if I remove "issue" array, I am able to traverse the json to one level down in "data" attribute, If this json has "issue" array I get an error saying missing comma.
到目前为止，如果我删除“问题”数组，我可以将 json 遍历到“数据”属性中的下一级，如果此 json 有“问题”数组，我会收到一个错误，提示缺少逗号。
javax.json.stream.JsonParsingException: Invalid token=NUMBER at (line no=15, column no=14, offset=242) Expected tokens are: [COMMA]
Below is the code that I have right now. I have two problems with this, one is the error while reading if I place the "issue" attribute, and secondly a way to read the "issue" array and traverse all attributes within.
下面是我现在拥有的代码。我对此有两个问题，一个是如果我放置“问题”属性时读取的错误，其次是读取“问题”数组并遍历其中的所有属性的方法。
InputStream fis = new FileInputStream(pathToFile+"data3.json");
JsonReader jsonReader = Json.createReader(fis);
//the error is thrown on below line while reading the above json.
JsonObject jsonObject = jsonReader.readObject();
jsonReader.close();
fis.close();
System.out.println(jsonObject.getInt("issueCounter"));
//reading arrays from json
JsonArray jsonArrayData = jsonObject.getJsonArray("data");
Project [] prj = new Project[jsonArrayData.size()];
int index = 0;
for(JsonValue value : jsonArrayData){
JSONObject jsonObj = new JSONObject(value.toString());
System.out.println(jsonObj.getString("name"));
System.out.println(jsonObj.getInt("prjId"));
//this is also the place where I am stuck, I know I need to construct an array out of it by obtaining issue attribute. Below is very very wrong.
/*
JsonArray jsonArrayIssue = jsonObj.getJsonArray("issue");
for(JsonValue issue : jsonArrayIssue){
JSONObject jsonIssueObj = new JSONObject(issue.toString());
System.out.println(jsonIssueObj.getString("status"));
System.out.println(jsonIssueObj.getInt("id"));
}
*/
}
Any help or pointers is deeply appreciated. I can tweak the json if its required ultimately I need to maintain an array of issues.
任何帮助或指示深表感谢。如果需要，我可以调整 json 最终我需要维护一系列问题。
回答by isah
The problem as others said is the JSON. "id": 00001 <-- this is a number, numbers cannot start with a leading zero as per JSON stadard. If you control the JSON you should tweak it.
正如其他人所说，问题是 JSON。"id": 00001 <-- 这是一个数字，根据 JSON 标准，数字不能以前导零开头。如果您控制 JSON，则应该对其进行调整。
Alternatively ff you don't, you can use a less strict parser like org.json.simplehttps://mvnrepository.com/artifact/com.googlecode.json-simple/json-simpleThe code will be the same as yours, just adjusted to org.json.simple
或者，如果您不这样做，您可以使用不太严格的解析器，例如org.json.simplehttps://mvnrepository.com/artifact/com.googlecode.json-simple/json-simple代码将与您的相同，只是调整为org.json.simple
try { ...
JSONObject rootJSON = (JSONObject) new JSONParser().parse(jsonString);
JSONArray dataList = (JSONArray) rootJSON.get("data");
for(Object projectObj: dataList.toArray()){
JSONObject project = (JSONObject)projectObj;
JSONArray issueList = (JSONArray) project.get("issue");
for(Object issueObj: issueList.toArray()){
JSONObject issue = (JSONObject) issueObj;
//do something with the issue
}
}
} catch (ParseException e) {
//do smth
e.printStackTrace();
}
回答by Javasamurai
Your json data is invalid.You can check here.
http://jsonlint.com...issue": [{ "id": 00001,Your id must be string number,string,boolean.Send 1,2,3,.... as return values and check if it works.
"status": ----------------------^
您的 json 数据无效。您可以在这里查看。
http://jsonlint.com您的 id 必须是字符串 number,string,boolean.Send 1,2,3,.... 作为返回值并检查它是否有效。...issue": [{ "id": 00001,
"status": ----------------------^
回答by David
Your code looks okay the problem is the JSON formatting. Specifically the following lines:
您的代码看起来没问题，问题在于 JSON 格式。具体如下几行：
"id": 00001, "id": 00002, "id": 00003, "id": 00004,
“id”：00001，“id”：00002，“id”：00003，“id”：00004，
Basically if you want it in that format you will need to set them as strings by wrapping the values in quotations i.e. "id": "00001" or you can use a valid number i.e. "id": 1
基本上，如果您希望采用这种格式，则需要通过将值括在引号中来将它们设置为字符串，即“id”：“00001”，或者您可以使用有效数字，即“id”：1
