Develop Reference

Deserialize/Parse JSON in Java

I am searching a good and dynamic way to parse JSON in Java.
I've seen things such as:
List<String> list = new ArrayList<String>();
JSONArray array = obj.getJSONArray("test");
for(int i = 0 ; i < array.length() ; i++){
list.add(array.getJSONObject(i).getString("testKey"));
}
But that's not what I'm searching. In C# I had something like that:
dynamic results = JsonConvert.DeserializeObject<dynamic>(json);
info.Text = results["test"]["testKey"];
Here's an example of my JSON:
{"date":"07.05.2017 11:44",
"monday":{"1":{"subject":"test","room":"test","status":"test"}}}
So for example I would like to make:
results["monday"]["1"]["subject"];
I hope someone understands my problem and can help me.
Thanks in advance!

The core Java runtime does not offer a JSON parser (edit: technically, it does, see bottom of answer), so you will need a library. See Jackson, Gson, perhaps others.
Even with that, you will not get the dynamic features you want, because Java is statically typed. Example with Jackson:
ObjectMapper mapper = new ObjectMapper();
Map<String, Object> map = mapper.readValue(json, new TypeReference<Map<String, Object>>(){});
map.get("monday").get("1").get("subject");
^^^
This fails because the result of get("monday") is Object, not Map
The "right" approach in Java-land, would be to create a class (or set of classes) that represents your JSON model, and pass it to the JSON parser's "object mapper". But you said "dynamic" so I'm not exploring this here.
So you'll need to cast to Map when you know it's not primitive values:
((Map<String,Map<String,String>>)map.get("monday")).get("1").get("subject");
This works but with a warning about unchecked cast...
All in all, Java is not a dynamic language and I see no way to do exactly what you want (perhaps I'm missing approaches that are still slightly easier than what I have suggested).
Are you limited to Java-the-language or Java-the-platform? In the latter case you can use a dynamic language for the Java platform, such as Groovy, who has excellent features to parse JSON.
EDIT: a funny alternative is to use Java's own JavaScript implementation. This works and is easy and dynamic, but I don't know if it's "good":
String json = "{\"date\":\"07.05.2017 11:44\",\n" +
"\"monday\":{\"1\":{\"subject\":\"test\",\"room\":\"test\",\"status\":\"test\"}}}";
ScriptEngine engine = new ScriptEngineManager().getEngineByName("nashorn");
engine.put("data", json);
System.out.println(engine.eval("JSON.parse(data)['monday']['1']['subject']"));

If you are sure about the value you want to get then you can do following as well :
String str = "{\"date\":\"07.05.2017 11:44\", \"monday\":{\"1\":{\"subject\":\"test\",\"room\":\"test\",\"status\":\"test\"}}}";
JSONObject results= new JSONObject(str);
String str1 = results.getJSONObject("monday").getJSONObject("1").getString("subject");
System.out.println(str1);
For array kind of results, we have to write logic for that. In this case org.json library is used.

You can use GCON library:
https://github.com/google/gson
Very good for parsing JSON objects.

Update data stored in ElasticSearch in Java with TransportClient

I'm trying to update data in Elastic Search in my Java program with TranportClient class. I know I can UPDATE in this way :
XContentBuilder builder = XContentFactory.jsonBuilder().startObject()
.field("source.ivarId", source.ivarId)
.field("source.channel", source.channel)
.field("source.bacId", source.bacId).endObject();
UpdateResponse response = client.prepareUpdate(_index, _type, _id).setDoc(builder.string()).get();
while source is my user-defined class which contains 3 fields : ivarId, channel and bacId.
But I want to know is there any method that could do the same thing, but using another more efficient and easier way, so that I don't need to assign each field inside a class? For example, can I do like this?
XContentBuilder builder = XContentFactory.jsonBuilder().startObject()
.field("source", source).endObject();
UpdateResponse response = client.prepareUpdate(_index, _type, _id).setDoc(builder.string()).get();
I tried the latter method, and I got this exception :
MapperParsingException[object mapping for [source] tried to parse field [source] as object, but found a concrete value]
I'm using Java 1.8.0.121, and both versions of ElasticSearch and TransportClient are 5.1. Thanks!

The answer is much more easier than I thought.
Gson gson = new Gson();
String sourceJsonString = gson.toJson(updateContent);
UpdateResponse response = client
.prepareUpdate(_index, "logs", id).setDoc(sourceJsonString).get();
updateContent is the object that contained new data, just to transform it to Json string, then use that to update, done.

Strategies for iterating through JsonPath array

I just started exploring JsonPath today. I want to explore not just what's possible to do with it, but some effective strategies.
For instance, let's say I have to iterate through an array contained within one element in the json string.
I'm using the "store" example from https://github.com/jayway/JsonPath#path-examples .
To get the list of books itself, I would imagine I could do something like this:
List<?> allBooks = JsonPath.<List<?>>read(context, "$.store.book");
Does it make sense to think about it this way?
It's the logic for iterating through this that I'm uncertain about.
I would have thought I could define a "Book" pojo and then do something like this:
for (int ctr = 0; ctr < allBooks.size(); ++ ctr) {
Book book = JsonPath.<Book>read(context, ".[" + ctr + "]");
System.out.println("book[" + book + "]");
}
However, this doesn't work. The "read" method at this point returns a JSONArray.
The last line in the code sample at https://github.com/jayway/JsonPath#what-is-returned-when is close to what I'm looking at, but this requires parsing the json in every iteration. It seems like the "DocumentContext" class has "read" methods that can take a type parameter, but not "JsonPath".
What are some reasonable strategies for navigating something like this?

JSON path will just return you a list of Maps as you've no doubt already seen. You need a way to tell it how to map these values to an object - for this you will need a custom configuration. There are other providers like Gson etc., but I've only used Jackson.
Configuration configuration = Configuration
.builder()
.jsonProvider(new JacksonJsonProvider())
.mappingProvider(new JacksonMappingProvider())
.build();
The second step is to specify generic type information with a TypeRef and pass it along when reading the tag.
List<Book> allBooks = JsonPath.using(configuration)
.parse(context)
.read("$.store.book", new TypeRef<List<Book>>() {});
As a result you get a nice list of Book objects.

How to parse a jsonObject or Array in Java

I have an API request from my CRM that can either return a jsonObject if there is only one result, or a jsonArray if there are multiple results. Here are what they look like in JSON Viewer
JsonObject:
JsonArray:
Before you answer, this is not my design, it's my CRM's design, I don't have any control over it, and yes, I don't like how it is designed either. The only reason I am not storing the records in my own database and just parsing that, which would be MUCH easier, is because my account is having issues not running some workflows that would allow me to auto add the records. Is there any way to figure out if the result is an object or an array using java? This is for an android app by the way, I need it to display the records on the phone.

You should use OPT command instead of GET
JSONObject potentialObject=response.getJsonObject("resuslt")
.getJsonObject("Potentials");
// here use opt. if the object is null, it means its not the type you specified
JSONObject row=potentialObject.optJsonObject("row");
if(row==null){
// row is json array .
JSONArray rowArray=potentialObject.getJsonArray("row");
// do whatever you like with rowArray
} else {
// row is json object. do whatever you like with it
}

ONE
You can use instanceof keyword to check the instances as in
if(json instanceof JSONObject){
System.out.println("object");
}else
System.out.println("array");
TWO
BUT I think a better way to do this is choose to use only JSONArray so that the format of your results can be predicated and catered for. JSONArrays can contain JSONObjects. That is they can cover the scope of JSONObject.
For example when you get the response (either in a JSONObject or a JSONArray), you need to store that in an instance. What instance are you going to store it in? So to avoid issues use JSONArray to store the response and provide statements to handle that.
THREE
Also you can try method overloading in java or Generics

Simplest way is to use Moshi, so that you dont have to parse, even in the case of the Model changing later, you have to change your pojo and it will work.
Here is the snippet from readme
String json = ...;
Moshi moshi = new Moshi.Builder().build();
JsonAdapter<BlackjackHand> jsonAdapter = moshi.adapter(BlackjackHand.class);
BlackjackHand blackjackHand = jsonAdapter.fromJson(json);
System.out.println(blackjackHand);
https://github.com/square/moshi/blob/master/README.md

Extract just one line from a big JSON in Java

Since my very first days of Java + JSON I tried to extract just some certain parts of a JSON.
But no matter if which of the libraries I used:
Gson
json-simple
javax.json
it never was possible to make it quick and comfortable. Mostly for easy task or even prototyping. It already cost me many hours of different approaches.
Going trough the hierarchy of an JSON
Object jsonObject = gson.fromJson(output, Object.class);
JsonElement jsonTree = gson.toJsonTree(jsonObject);
JsonArray commitList = jsonTree.getAsJsonArray();
JsonElement firstElement = commitList.get(0);
JsonObject firstElementObj = firstElement.getAsJsonObject();
System.out.println(firstElementObj.get("sha"));
JsonElement fileList = firstElementObj.get("files");
This is dirty code for a reason. It shows how many early approaches looks like and how many people cannot achieve it to do it better early.
Deserializing JSON to a Java Object
Your have to analyse the complete JSON to create an complete Java-Object representation just to get access to some single memebers of it. This is a way I never wanted to do for prototyping
JSON is an easy format. But using libraries like that is quite difficult and often an problem for beginner. I've found several different answers via Google and even StackOverflow. But most were quite big larged which required to create a own specific class for the whole JSON-Object.
What is the best approach to make it more beginner-friendly?
or
What is the best beginner-friendly approach?

Using Jackson (which you tagged), you can use JsonPointer expressions to navigate through a tree object:
ObjectMapper mapper = new ObjectMapper();
JsonNode tree = mapper
.readTree("[ { \"sha\": \"foo\", \"files\": [ { \"sha\": \"bar\" }, { \"sha\": \"quux\" } ] } ]");
System.out.println(tree.at("/0/sha").asText());
for (JsonNode file : tree.at("/0/files")) {
System.out.println(file.get("sha").asText());
}
You could also use the ObjectMapper to convert just parts of a tree to your model objects, if you want to start using that:
for (JsonNode fileNode : tree.at("/0/files")) {
FileInfo fileInfo = mapper.convertValue(fileNode, FileInfo.class);
System.out.println(fileInfo.sha);
}
If your target class (FileInfo) specifies to ignore unknown properties (annotate target class with #JsonIgnoreProperties(ignoreUnknown = true) or disable DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES on the mapper), then you can simply declare the properties you are interested in.
"Best" is whatever works to get you going.

Generate Plain Old Java Objects from JSON or JSON-Schema
One little helper I found via my research was an Online-Tool like
http://www.jsonschema2pojo.org/
This is a little help, when you know about that. But the negative side I mentioned at point 2 is still there.

You can use JsonSurfer to selectively extract value or object from big json with streaming JsonPath processor.
JsonSurfer jsonSurfer = JsonSurfer.gson();
System.out.println(jsonSurfer.collectOne(json, "$[0].sha"));
System.out.println(jsonSurfer.collectOne(json, "$[0].files"));