I am getting an inputstream and converting it to String and then to JSONObject
Below is snippet for converting inputstream to String and then JSONObject.
But after converting to json object(line 6) I am getting only the first object instead of all the objects
Can you please let me know why I am getting only one object instead of all the n objects
InputStream in = new BufferedInputStream(conn.getInputStream());
String result = org.apache.commons.io.IOUtils.toString(in, "UTF-8");
int i =result.indexOf("{");
String forResult=result.substring(i);
System.out.println(forResult); // Result 1
JSONObject jsonObject = new JSONObject(forResult); // Line 6
System.out.println(jsonObject); // Result 2
After converting it to String it look like this
Result -1
{
"test_expr":"",
"val_expr":"someVale",
"val_advanced":true,
"machine_val":null
}, {...// n times}
Result-2 -only first object
{
"test_expr":"",
"val_expr":"someVale",
"val_advanced":true,
"machine_val":null
}
Thanks and please bear my ignorance as I am completly new in java
Because you json is not valid .There is a comma between JSONObject .
Change to this .
{
{
"test_expr":"",
"val_expr":"someVale",
"val_advanced":true,
"machine_val":null
},
{
"test_expr":"",
"val_expr":"someVale",
"val_advanced":true,
"machine_val":null
}
...
}
or
[
{
"test_expr":"",
"val_expr":"someVale",
"val_advanced":true,
"machine_val":null
},
{
"test_expr":"",
"val_expr":"someVale",
"val_advanced":true,
"machine_val":null
}
...
]
The source of JSONObject
/**
* Creates a new {#code JSONObject} with name/value mappings from the JSON
* string.
*
* #param json a JSON-encoded string containing an object.
* #throws JSONException if the parse fails or doesn't yield a {#code
* JSONObject}.
*/
public JSONObject(String json) throws JSONException {
this(new JSONTokener(json));
}
So a JSON-encoded string containing an object(like {}).
// make sure that you result contain {}
result = "{your data here}"
JSONObject json_data = new JSONObject(result);
And if you use JSONArray ,you should contain [] in your JSON
result = "[json data]";
JSONArray jsonArray = new JSONArray(result);
Well, concatenation of multiple jsons is not a valid json. Your parsing library should have rejected such an input, but it seems that it just stopped at the end of the valid object.
You can wrap the list it into an array: [{...},{...},{...}]. Then the parser will be able to correctly interpret it as an array.
I guess you are getting JSONArray object instead of JSONObject. Why do you need to get sub string of the result? Json array can start with [. See the difference between JSONObject and JSONArray. Difference between JSONObject and JSONArray
Thanks to Sotirios Delimanolis.
I am able to resolve the problem by using JSONParser
InputStream in = new BufferedInputStream(conn.getInputStream());
String result = org.apache.commons.io.IOUtils.toString(in, "UTF-8");
JSONParser parser = new JSONParser();
org.json.simple.JSONArray modelArrary =(org.json.simple.JSONArray) parser.parse(result) ;
System.out.println(modelArrary);
Related
I have a goal to verify that certain JSON that I've got from RabbitMQ corresponds to one of expected JSONs in an array in a single file.
In other words, I need to verify that this JSON:
{
"networkCode":"network",
"programId":"92000"
}
is present in this JSON array:
[
{
"networkCode":"network",
"programId":"92000"
},
{
"networkCode":"network",
"programId":"92666"
}
]
Thank you very much for help!
Some part of my code
//GET DESIRABLE JSON
String message = new String(delivery.getBody(), StandardCharsets.UTF_8);
JSONObject myJSON= new JSONObject(message);
//GET THE JSON ARRAYS FROM FILE
JSONParser parser = new JSONParser();
Object expectedJSONs= parser.parse(new FileReader("C:\\amqpclient\\src\\test\\java\\tradeDoubler\\ExpectedDTO.json"));
JSONArray expectedArray = (JSONArray) expectedJSONs;
JSONAssert.assertEquals(
myJSON, expectedArray , JSONCompareMode.LENIENT);
Compilation says that cannot resolve this
Exception in thread "main" java.lang.AssertionError: Expecting a JSON array, but passing in a JSON object
Org.json library is quite easy to use.
Example code below:
import org.json.*;
JSONObject obj = new JSONObject(" yourJSONObjectHere ");
JSONArray arr = obj.getJSONArray("networkArray");
for (int i = 0; i < arr.length(); i++)
{
String networkCode = arr.getJSONObject(i).getString("networkCode");
......
}
By iterating on your JSONArray, you can check if each object is equal to your search.
You may find more examples from: Parse JSON in Java
May I suggest you to use the Gson Library?
You can use something like this. But It will throw an exception if the json doesn't match/contains the fields.
Type listType = new TypeToken<ArrayList<YourJavaClassJsonModel>>() {
}.getType();
List<YourJavaClassJsonModel> resultList = gson.fromJson(JsonString, listType);
Hope it may help
You could use a JSON parser to convert the JSON to a Java object (Jackson and GSON are good options), and then check that object.
This is the JSON file I am working with
{"sentiment":
{"document":
{
"label": "positive",
"score": 0.53777
}
}
}
I need to access the value in label and score. using java. How can I do that?
Find below the code I am using right now:
JSONParser parser = new JSONParser();
try
{
Object object = parser
.parse(new FileReader("output_nlu_sentiment.json"));
//convert Object to JSONObject
JSONObject jsonObject = new JSONObject();
JSONObject sentimentobject= new JSONObject();
JSONObject documentobject = new JSONObject();
sentimentobject= (JSONObject) jsonObject.get("sentiment");
documentobject= (JSONObject) sentimentobject.get("document");
String label = (String) documentobject.get("label");
//float score = (float) jsonObject.get("score");
System.out.println(label);
String test = (String) sentimentobject.get("label");
System.out.println(test);
} catch(FileNotFoundException fe)
{
fe.printStackTrace();
}
catch(Exception e)
{
e.printStackTrace();
}
Why is it printing the value as null.
You might want to have a look at JacksonXml for json parsing.
Right now the problem is that you're not using the JsonObject returned by parser.parse(...).
Instead you use the get method on objects you just created. This of course means that you don't getthe valie you want to.
Try to use following code (JSONObject jsonObject = (JSONObject) object instead of JSONObject jsonObject = new JSONObject();), because you didn't use object at all, just create new empty JSONObject.
JSONParser parser = new JSONParser();
try
{
Object object = parser
.parse(new FileReader("output_nlu_sentiment.json"));
//convert Object to JSONObject
JSONObject jsonObject = (JSONObject) object;
JSONObject sentimentobject = (JSONObject) jsonObject.get("sentiment");
JSONObject documentobject= (JSONObject) sentimentobject.get("document");
String label = (String) documentobject.get("label");
System.out.println(label);
float score = (float) documentobject.get("score");
System.out.println(score );
}catch(FileNotFoundException fe)
{
fe.printStackTrace();
}
catch(Exception e)
{
e.printStackTrace();
}
You have to make use of object created in Object object = parser.parse(new FileReader("output_nlu_sentiment.json")); while creating the jsonObject
For that you can look at the code below:
Object object = parser
.parse(new FileReader("file2.json"));
//convert Object to JSONObject
JSONObject jsonObject = (JSONObject) object;
JSONObject sentimentobject= new JSONObject();
JSONObject documentobject = new JSONObject();
sentimentobject= (JSONObject) jsonObject.get("sentiment");
documentobject= (JSONObject) sentimentobject.get("document");
String label = (String) documentobject.get("label");
//float score = (float) jsonObject.get("score");
System.out.println(label);
String test = (String) sentimentobject.get("label");
You will get the positive printed on console.
you should see the content in para 'sentimentobject',force convert into class JSONObject can not get the value you want.
I prefer the FasterXML Jackson support to parse JSON into plain old Java objects (POJOs). These POJOs are often called Data Transfer Objects (DTOs) and give you a way to turn your JSON fields into properly typed members of the corresponding DTO.
Here is an example method to do that. The ObjectMapper(s) are generally maintained as statics somewhere else because FasterXML's implementation caches information to improve efficiency of object mapping operations.
static final ObjectMapper mapper = new ObjectMapper();
This is the JSON deserialization method:
public static <T> T deserializeJSON(
final ObjectMapper mapper, final InputStream json,
final Class<T> clazz)
throws JsonParseException, UnrecognizedPropertyException,
JsonMappingException, IOException
{
final String sourceMethod = "deserializeJSON";
logger.entering(sourceClass, sourceMethod);
/*
* Use Jackson support to map the JSON into a POJO for us to process.
*/
T pojoClazz;
pojoClazz = mapper.readValue(json, clazz);
logger.exiting(sourceClass, sourceMethod);
return pojoClazz;
}
Assuming I have a class called FooDTO, which has the appropriate Jackson annotations/getters/setters (note you must always provide a default empty public constructor), you can do this:
FooDTO foo = deserializeJSON(mapper, inputstream, FooDTO.class);
The deserialization throws a few different exceptions (all of which have IOException as their parent class) that you will need to handle or throw back to the caller.
Here besides of the correction alreay addressed in comments and other answers, I include some other changes you can benefit of:
It is not necessary to initialize the JSONObjects with a new instance that is going to be ovewritten in the next line.
You can use getJSONObject(), getString(), getFloat() instead of get(), in this way you don't need to cast the result.
public void parseJson() {
JSONParser parser = new JSONParser();
try
{
JSONObject jsonObject = new JSONParser().parse(new FileReader("output_nlu_sentiment.json"));
JSONObject sentimentobject= null;
JSONObject documentobject = null;
sentimentobject= jsonObject.getJSONObject("sentiment");
documentobject= sentimentobject.getJSONObject("document");
String label = documentobject.getString("label");
float score = documentobject.getFloat("score");
String output = String.format("Label: %s Score: %f", label, score);
System.out.println(output);
}catch(FileNotFoundException fe){
fe.printStackTrace();
}catch(Exception e){
e.printStackTrace();
}
}
Also for this kind of objects, where the attribute names could act as object properties, I suggest you take a look at Gson library. After modeling the json as a composition of POJOs, the parsing takes 1 line of code.
My JSON Structure will vary depend on the request. But the content inside each element remain same. For Example:
JSON1:
{
"h1": {
"s1":"s2"
},
"c1": {
"t1:""t2"
}
}
JSON2:
{
"h1": {
"s1":"s2"
},
"c2": {
"x1:""x2"
}
}
In the above example, elements inside h1,c1 and c2 are constant. Please let me know how to convert JSON to JAVA Object
Regards
Udhaya
First of all You need to understand Json Structure cause above format is incorrect visit this
and this
And you can use Google Gson or Json for parsing the result json String .
"t1:""t2" json format incorrect
Used
"t1":"t2"
Instead of
"t1:""t2"
and also used
"x1": "x2"
Instead of
"x1:""X2"
Code to take in java
try {
JSONObject jsonObject = new JSONObject(response);
JSONObject jsonsubObject = jsonObject.getJSONObject("h1");
String s1 = jsonsubObject.getString("s2");
JSONObject jsonsubObject1 = jsonObject.getJSONObject("c1");
String t1 = jsonsubObject1 .getString("t2");
}
catch (JSONException e) {
e.printStackTrace();
}
Use Google Gson:
Gson gson = new Gson();
ClassName object;
try {
object = gson.fromJson(json, ClassName.class);
} catch (com.google.gson.JsonSyntaxException ex) {
//the json wasn't valid json
}
String validJson = gson.toJson(obj); //obj is an instance of any class
json must be a valid JSON String
import org.json.JSONObject;
you can simple pass your data in constructor of JSONObject it automatically handle, you need to throws JSONException which may occur during conversion id format of data is not correct
String data = "{'h1':{'s1':'s2'},'c1':{'t1:''t2'}}";
JSONObject jsnobject = new JSONObject(data);
I'm trying to "create" a JSONObject. Right now I'm using JSON-Simple and I'm trying to do something along the lines of this (sorry if any typo's are made in this example JSON file)
{
"valuedata": {
"period": 1,
"icon": "pretty"
}
}
Right now I'm having issues finding on how to write valuedata into a JSON file through Java, what I did try was:
Map<String, String> t = new HashMap<String, String>();
t.put("Testing", "testing");
JSONObject jsonObject = new JSONObject(t);
but that just did
{
"Testing": "testing"
}
Whatr you want to do is put another JSONObject inside your JSONObject "jsonObject", in the field "valuedata" to be more exact. You can do this like that...
// Create empty JSONObect here: "{}"
JSONObject jsonObject = new JSONObject();
// Create another empty JSONObect here: "{}"
JSONObject myValueData = new JSONObject();
// Now put the 2nd JSONObject into the field "valuedata" of the first:
// { "valuedata" : {} }
jsonObject.put("valuedata", myValueData);
// And now add all your fields for your 2nd JSONObject, for example period:
// { "valuedata" : { "period" : 1} }
myValueData.put("period", 1);
// etc.
Following is example which shows JSON object streaming using Java JSONObject:
import org.json.simple.JSONObject;
class JsonEncodeDemo
{
public static void main(String[] args)
{
JSONObject obj = new JSONObject();
obj.put("name","foo");
obj.put("num",new Integer(100));
obj.put("balance",new Double(1000.21));
obj.put("is_vip",new Boolean(true));
StringWriter out = new StringWriter();
obj.writeJSONString(out);
String jsonText = out.toString();
System.out.print(jsonText);
}
}
While compile and executing above program, this will produce following result:
{"balance": 1000.21, "num":100, "is_vip":true, "name":"foo"}
Here is my json Object.
{"id":"mrbbt6f3fa99gld0m6n52osge0",
"name_value_list":
{"user_default_dateformat":{"name":"user_default_dateformat","value":"m/d/Y"}},
"module_name":"Users"}
I got id,and module_name through following code.How can i get user_default_dateformat?.
I know it may so simple but I am a newbie in json.
String jsonResponse;
while ((jsonResponse = br.readLine()) != null) {
jsonOutput = jsonResponse;
}
JSONObject job = new JSONObject(jsonOutput);
System.out.println(job);// i can see the same json object
that i showen above.
sessionID = job.get("id").toString();
Exception generating coge
JSONObject job2=new JSONObject(job);
dateFormat = job2.get("user_default_dateformat").toString();
The Eexception is
org.json.JSONException: JSONObject["user_default_dateformat"] not found.
Thanks,
name_value_list is also an Object.
JSONObject job2 = new JSONObject(job.get("name_value_list"));
So there you get
job2.get("user_default_dateformat");
Every {} in your JSON is an object. So for every String you get which is something like {"xy":"za","ab":"cd"} you have to cast it to the JSONObject
Edit for your error:
As you can see in your code the line:
JSONObject job2=new JSONObject(job);
will try to generate a JSONObject out of your JSONObject.
You have to get the JSONObject in your JSONObject.
You want to get the user_default_dateformat which is in your JSONObject:
String name_value_list_string = job.get("name_value_list").toString();
//this string is another json-string which contains the user_default_dateformat
JSONObject name_value_list_object = new JSONObject(name_value_list_string);
//This JSONObject contains the user_default_dateformat but this is also a JSONObject
String user_default_dateformat_string = name_value_list_object.get("user_default_dateformat").toString();
//this String contains the user_default_dateformat JSONString
JSONObject user_default_dateformat_object = new JSONObject(user_default_dateformat_string);
//This JSONObject contains the String values of your user_default_dateformat
if you are using JSONSimple library you can use this:
jsonObject = (JSONObject) new JSONParser().parse(jsonstr);
System.out.println((JSONObject)jsonObject.get("name_value_list"))).get("user_default_dateformat"));
This should give you the required result.