I am new to java
my string contains : "{'message': 'hello'}";
I want to convert above string to JSON data that gives 'message' key gives 'hello' as value. can any one please help me out ?.
I would use the json.org library, and do something like this:
import org.json.JSONObject;
public class App
{
public static void main( String[] args )
{
String message = "{'message': 'hello'}";
JSONObject json = new JSONObject(message);
System.out.println(json.get("message"));
}
}
Related
I want to verify that id: 1 belongs to Tiger Nixon
{"status":"success","data":[{"id":"1","employee_name":"Tiger Nixon","employee_salary":"320800","employee_age":"61","profile_image":""}
I have been following this document: https://github.com/rest-assured/rest-assured/wiki/Usage#json-using-jsonpath , this section body("shopping.category.find { it.#type == 'groceries' }.item", hasItems("Chocolate", "Coffee"));
Currently I am getting a null pointer exception. I could use some help to come up with a solution. Thanks in advance for your time.
import com.google.gson.JsonObject;
import io.restassured.RestAssured;
import org.testng.Assert;
public class RestAssuredExample_2 {
public void getResponse() {
JsonObject responseObject = RestAssured.get("http://dummy.restapiexample.com/api/v1/employees")
.then()
.extract().response().as(JsonObject.class);
String emp1Name = responseObject.get("data.find{it.#id=='1'}.employee_name").toString();
Assert.assertEquals(emp1Name, "Tiger Nixon");
}
public static void main(String[] args) {
RestAssuredExample_2 rest2 = new RestAssuredExample_2();
rest2.getResponse();
}
}
Error:
Exception in thread "main" java.lang.NullPointerException
at HTTPz.RestAssuredExample_2.getResponse(RestAssuredExample_2.java:16)
Tried to replicate this scenario
String responseObject = RestAssured.get("http://dummy.restapiexample.com/api/v1/employees").then().extract().asString();
JsonPath js = new JsonPath(responseObject);
String emp1Name = js.get("data.find {it.id =='1'}.employee_name").toString();
System.out.println(emp1Name);
And I get the value for emp1Name as "Tiger Nixon"
Difference :
Original : it.#id=='1'
Mine : it.id =='1'
It seems like # is for XMLPath and not for JSONPath and honestly I wouldn't know about it in detail as well cause I have been using JSONPath only for a very long time :)
I have the following xml:
<root><field>test </field></root>
When I try to convert it to json,
String xmlString = "<root><field>test </field></root>";
String jsonString = XML.toJSONObject(xmlString, false).toString();
System.out.println(jsonString);
the result is like this:
{"root":{"field":"test"}}
How can I keep spaces in strings when converting?
Underscore-java library can keep spaces while conversion from XML to JSON. I am the maintainer of the project.
import com.github.underscore.U;
public class MyClass {
public static void main(String args[]) {
System.out.println(U.xmlToJson("<root><field>test </field></root>"));
}
}
Output:
{
"root": {
"field": "test "
},
"#omit-xml-declaration": "yes"
}
I have a very large JSON file in the following format:
[{"fullname": "name1", "id": "123"}, {"fullname": "name2", "id": "245"}, {"fullname": "name3", "id": "256"}]
It looks like a JSONArray. All the records are written in the same line.
Can you help me how can I parse this file using Java. I want to read each JSON object and display all the fullname and ids. Below is my attempt, but my code is not working:
import org.apache.commons.lang3.StringEscapeUtils;
import org.json.simple.JSONArray;
import org.json.simple.JSONObject;
import org.json.simple.parser.JSONParser;
import org.json.simple.parser.ParseException;
public class JSONFileReaderDriver {
public static void main(String[] args) throws FileNotFoundException,
IOException, ParseException
{
String filename="Aarau";
String src="C:\\"+filename+".json";
JSONParser parser = new JSONParser();
JSONObject obj;
try
{
BufferedReader br=new BufferedReader (new FileReader(src));
obj = (JSONObject) new JSONParser().parse(row);
String fullname=obj.get("fullname");
String id=obj.get("id");
System.out.println ("fullname: "+fullname+" id: "+id);
}catch(Exception e)
{e.printStackTrace();}
br.close();
}
}
Make your life easy and use an ObjectMapper.
This way you simply define a Pojo with the same properties as you json object.
In you case you need a Pojo that looks like this:
public class Person{
private String fullname;
private int id;
public Person(String fullname, int id) {
this.fullname = fullname;
this.id = id;
}
public String getFullname() {
return fullname;
}
public int getId() {
return id;
}
}
With that you only need to do:
ObjectMapper objectMapper = new ObjectMapper();
List<Person> persons = objectMapper.readValue(myInputStream, TypeFactory.defaultInstance().constructCollectionType(List.class, Person.class));
This is a hassle free and type safe approach.
Dependencies needed:
https://github.com/FasterXML/jackson-databind
<dependency>
<groupId>com.fasterxml.jackson.core</groupId>
<artifactId>jackson-databind</artifactId>
</dependency>
you can use Json.simple java api , below is code that can helpful to you
byte[] bFile = Files.readAllBytes(new File("C:/xyz.json").toPath());
JSONArray root = (JSONArray) JSONValue.parseWithException(bFile);
JSONObject rootObj = (JSONObject) root.get(0);
You can get values from JSONObject based on key , it also depends on format of your json as there could be nested json data. So you have to extract data accordingly . Apart from this you can use jackson parser api or GSON as well.
Okay folks...just solved my problem. I am posting the solution in case someone runs into the same issue again, can use my solution. My solution is partly motivated by Rahul Rabhadiya. Thanks, dude.
try{
row=br.readLine();
JSONArray root = (JSONArray) JSONValue.parseWithException(row);
for (int i=0;i<root.size();i++)
{
JSONObject rootObj = (JSONObject) root.get(i);
String fullname=(String) rootObj.get("fullname");
System.out.println (fullname);
}
}catch(Exception e)
{
e.printStackTrace();
}
Your json is a JSONArray, so when you are parsing it, you need to parse it as a JSONArray.
JSONParser jsonParser = new JSONParser();
JSONArray a = (JSONArray) jsonParser.parse(new FileReader(src));
for (Object o : a) {
// access your object here.
}
I would suggest going with Jackson, in particular the Jackson Streaming API which is perfect for parsing large arrays like this.
See this answer: https://stackoverflow.com/a/24838392/3765428
I've the following code, but when I am saving below JSON in database its giving me wrong url like {"#url#":"https:\/\/www.test.com\/test"}
import org.json.simple.JSONObject;
public class DemoURL {
private static String url = "https://www.test.com/test";
public static void main(String[] args) {
JSONObject msgJson = new JSONObject();
msgJson.put("#url#", url);
System.out.println(msgJson.toString());
}
}
I want url like {"#url#":"https://www.test.com/test"}
Please suggest how to fix it?
Here is the solution:
public class App{
private static String url = "https://www.test.com/test";
public static void main(String[] args) {
JSONObject msgJson = new JSONObject();
msgJson.put("#url#", url);
System.out.println(getCleanURL(msgJson.toString()));
}
private static String getCleanURL(String url){
return url.replaceAll("\\\\", "").trim();
}
}
This gives correct output, simply run this code. This will store exact value in the database.
{"#url#":"https://www.test.com/test"}
You are using org.json.simple JSON library. JSON-Simple escapes char from String.
You can't change this thing, as its not configurable.
But you can use org.json JSON library, this will not escape String, and good part is, you don't have to change your code, existing syntax will work fine.
e.g.
import org.json.JSONObject;
public class DemoURL {
private static String url = "https://www.test.com/test";
public static void main(String[] args) {
JSONObject msgJson = new JSONObject();
msgJson.put("#url#", url);
System.out.println(msgJson.toString());
}
}
output : {"#url#":"https://www.test.com/test"}
Try replacing slash(/) with unicode character \u2215 before passing it to JSON object.
try to quote the string with single quotes, something like this
JSONObject msgJson = new JSONObject();
msgJson.put("#url#", "\'"+url+"\'");
System.out.println(msgJson.toString());
I have a json string like this:
{"a":"vala", "b":"valb", "c":"valc"}
I want to convert the above string to a JSONObject so that I can do something like:
testObject.remove("b");
testObject.remove("c");
So that I can easily print out the json string of:
{"a":"vala"}
What is the simplest way for me to do this?
org-json-java can do the things you want
import org.json.JSONException;
import org.json.JSONObject;
public class JsonTest {
public static void main(String[] args) {
try {
JSONObject testObject = new JSONObject("{\"a\":\"vala\", \"b\":\"valb\", \"c\":\"valc\"}");
testObject.remove("b");
testObject.remove("c");
System.out.println(testObject.toString()); // This prints out the json string
} catch (JSONException e) {
e.printStackTrace();
}
}
}
The execution result is
{"a":"vala"}
Make sure you've downloaded and imported org.json-YYYYMMDD.jar from here before you run the above code.
Firstly I think you need change title of question.
Secondly, If you want change from String to Json just using org.json library
JSONObject jsonObj = new JSONObject("{\"a\":\"valuea\",\"b\":\"valueb\"}");
Are you tried remove. http://www.json.org/javadoc/org/json/JSONObject.html#remove(java.lang.String)
Json.remove(key)