How to combine two or more JSON objects into one JSONObject

I've two JSON Objects as below

firstJSON: {"200":"success"} 

secondJSON: {"401":"not found"}

I need to combine them as {"codes":{"200":"success"},{"401":"not found"}} using java.

I tried in 3 ways but couldn't achieve the desired output. Could you please help with code snippet?

code which I've tried as below

JSONObject firstJSON = new JSONObject();

firstJSON.put("200", "success");

String firstJSONStr = firstJSON.toString();

System.out.println("firstJSONStr--> " + firstJSONStr);



JSONObject secondJSON = new JSONObject();

secondJSON.put("401", "not found");

String secondJSONStr = secondJSON.toString();

System.out.println("secondJSONStr--> "+secondJSONStr);



String finalJSONStr = firstJSONStr + "," + secondJSONStr;

JSONObject finalJSON1 = new JSONObject();

finalJSON1.put("codes", new JSONObject(finalJSONStr));

System.out.println("finalJSON1--> " + finalJSON1.toString());



JSONObject finalJSON2 = new JSONObject();

finalJSON2.put("codes", finalJSONStr);

System.out.println("finalJSON2--> " + finalJSON2.toString());



JSONObject finalJSON3 = new JSONObject();

ArrayList<JSONObject> jsonArray = new ArrayList<JSONObject>();

jsonArray.add(firstJSON);

jsonArray.add(secondJSON);

finalJSON3.put("codes", jsonArray);

System.out.println("finalJSON3--> " + finalJSON3.toString());

Output:

firstJSONStr--> {"200":"success"}

secondJSONStr--> {"401":"not found"}

finalJSON1--> {"codes":{"200":"success"}}

finalJSON2--> {"codes":"{"200":"success"},{"401":"not found"}"}

finalJSON3--> {"codes":[{"200":"success"},{"401":"not found"}]}

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

asked Jan 2 at 10:20

Narasimha

1615

this might be helpful stackoverflow.com/questions/20346790/…

– Supun Dharmarathne
Jan 2 at 10:35

Note that your desired output is not valid JSON

– Hulk
Jan 2 at 10:35

1

I think the reason you are finding it difficult to get the json you require is because it simply isn't valid json: {"codes":{"200":"success"},{"401":"not found"}}. Here you have codes object {"200":"success"} and then an anonymous object {"401":"not found"}

– karen
Jan 2 at 10:35

hi folks, desired json which I'm trying to generate will be valid one

– Narasimha
Jan 2 at 11:09

@Narasimha {"codes":{"200":"success"},{"401":"not found"}} Is it valid JSON ?

– varatharajan
Jan 2 at 11:25

|
show 4 more comments

I've two JSON Objects as below

firstJSON: {"200":"success"} 

secondJSON: {"401":"not found"}

I need to combine them as {"codes":{"200":"success"},{"401":"not found"}} using java.

I tried in 3 ways but couldn't achieve the desired output. Could you please help with code snippet?

code which I've tried as below

JSONObject firstJSON = new JSONObject();

firstJSON.put("200", "success");

String firstJSONStr = firstJSON.toString();

System.out.println("firstJSONStr--> " + firstJSONStr);



JSONObject secondJSON = new JSONObject();

secondJSON.put("401", "not found");

String secondJSONStr = secondJSON.toString();

System.out.println("secondJSONStr--> "+secondJSONStr);



String finalJSONStr = firstJSONStr + "," + secondJSONStr;

JSONObject finalJSON1 = new JSONObject();

finalJSON1.put("codes", new JSONObject(finalJSONStr));

System.out.println("finalJSON1--> " + finalJSON1.toString());



JSONObject finalJSON2 = new JSONObject();

finalJSON2.put("codes", finalJSONStr);

System.out.println("finalJSON2--> " + finalJSON2.toString());



JSONObject finalJSON3 = new JSONObject();

ArrayList<JSONObject> jsonArray = new ArrayList<JSONObject>();

jsonArray.add(firstJSON);

jsonArray.add(secondJSON);

finalJSON3.put("codes", jsonArray);

System.out.println("finalJSON3--> " + finalJSON3.toString());

Output:

firstJSONStr--> {"200":"success"}

secondJSONStr--> {"401":"not found"}

finalJSON1--> {"codes":{"200":"success"}}

finalJSON2--> {"codes":"{"200":"success"},{"401":"not found"}"}

finalJSON3--> {"codes":[{"200":"success"},{"401":"not found"}]}

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

asked Jan 2 at 10:20

Narasimha

1615

this might be helpful stackoverflow.com/questions/20346790/…

– Supun Dharmarathne
Jan 2 at 10:35

Note that your desired output is not valid JSON

– Hulk
Jan 2 at 10:35

1

I think the reason you are finding it difficult to get the json you require is because it simply isn't valid json: {"codes":{"200":"success"},{"401":"not found"}}. Here you have codes object {"200":"success"} and then an anonymous object {"401":"not found"}

– karen
Jan 2 at 10:35

hi folks, desired json which I'm trying to generate will be valid one

– Narasimha
Jan 2 at 11:09

@Narasimha {"codes":{"200":"success"},{"401":"not found"}} Is it valid JSON ?

– varatharajan
Jan 2 at 11:25

|
show 4 more comments

I've two JSON Objects as below

firstJSON: {"200":"success"} 

secondJSON: {"401":"not found"}

I need to combine them as {"codes":{"200":"success"},{"401":"not found"}} using java.

I tried in 3 ways but couldn't achieve the desired output. Could you please help with code snippet?

code which I've tried as below

JSONObject firstJSON = new JSONObject();

firstJSON.put("200", "success");

String firstJSONStr = firstJSON.toString();

System.out.println("firstJSONStr--> " + firstJSONStr);



JSONObject secondJSON = new JSONObject();

secondJSON.put("401", "not found");

String secondJSONStr = secondJSON.toString();

System.out.println("secondJSONStr--> "+secondJSONStr);



String finalJSONStr = firstJSONStr + "," + secondJSONStr;

JSONObject finalJSON1 = new JSONObject();

finalJSON1.put("codes", new JSONObject(finalJSONStr));

System.out.println("finalJSON1--> " + finalJSON1.toString());



JSONObject finalJSON2 = new JSONObject();

finalJSON2.put("codes", finalJSONStr);

System.out.println("finalJSON2--> " + finalJSON2.toString());



JSONObject finalJSON3 = new JSONObject();

ArrayList<JSONObject> jsonArray = new ArrayList<JSONObject>();

jsonArray.add(firstJSON);

jsonArray.add(secondJSON);

finalJSON3.put("codes", jsonArray);

System.out.println("finalJSON3--> " + finalJSON3.toString());

Output:

firstJSONStr--> {"200":"success"}

secondJSONStr--> {"401":"not found"}

finalJSON1--> {"codes":{"200":"success"}}

finalJSON2--> {"codes":"{"200":"success"},{"401":"not found"}"}

finalJSON3--> {"codes":[{"200":"success"},{"401":"not found"}]}

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

asked Jan 2 at 10:20

Narasimha

1615

I've two JSON Objects as below

firstJSON: {"200":"success"} 

secondJSON: {"401":"not found"}

I need to combine them as {"codes":{"200":"success"},{"401":"not found"}} using java.

I tried in 3 ways but couldn't achieve the desired output. Could you please help with code snippet?

code which I've tried as below

JSONObject firstJSON = new JSONObject();

firstJSON.put("200", "success");

String firstJSONStr = firstJSON.toString();

System.out.println("firstJSONStr--> " + firstJSONStr);



JSONObject secondJSON = new JSONObject();

secondJSON.put("401", "not found");

String secondJSONStr = secondJSON.toString();

System.out.println("secondJSONStr--> "+secondJSONStr);



String finalJSONStr = firstJSONStr + "," + secondJSONStr;

JSONObject finalJSON1 = new JSONObject();

finalJSON1.put("codes", new JSONObject(finalJSONStr));

System.out.println("finalJSON1--> " + finalJSON1.toString());



JSONObject finalJSON2 = new JSONObject();

finalJSON2.put("codes", finalJSONStr);

System.out.println("finalJSON2--> " + finalJSON2.toString());



JSONObject finalJSON3 = new JSONObject();

ArrayList<JSONObject> jsonArray = new ArrayList<JSONObject>();

jsonArray.add(firstJSON);

jsonArray.add(secondJSON);

finalJSON3.put("codes", jsonArray);

System.out.println("finalJSON3--> " + finalJSON3.toString());

Output:

firstJSONStr--> {"200":"success"}

secondJSONStr--> {"401":"not found"}

finalJSON1--> {"codes":{"200":"success"}}

finalJSON2--> {"codes":"{"200":"success"},{"401":"not found"}"}

finalJSON3--> {"codes":[{"200":"success"},{"401":"not found"}]}

java json

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

asked Jan 2 at 10:20

Narasimha

1615

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

asked Jan 2 at 10:20

Narasimha

1615

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

edited Jan 2 at 11:13

Karol Dowbecki

24.8k93759

asked Jan 2 at 10:20

Narasimha

1615

asked Jan 2 at 10:20

Narasimha

1615

asked Jan 2 at 10:20

Narasimha

1615

this might be helpful stackoverflow.com/questions/20346790/…

– Supun Dharmarathne
Jan 2 at 10:35

Note that your desired output is not valid JSON

– Hulk
Jan 2 at 10:35

1

I think the reason you are finding it difficult to get the json you require is because it simply isn't valid json: {"codes":{"200":"success"},{"401":"not found"}}. Here you have codes object {"200":"success"} and then an anonymous object {"401":"not found"}

– karen
Jan 2 at 10:35

hi folks, desired json which I'm trying to generate will be valid one

– Narasimha
Jan 2 at 11:09

@Narasimha {"codes":{"200":"success"},{"401":"not found"}} Is it valid JSON ?

– varatharajan
Jan 2 at 11:25

|
show 4 more comments

this might be helpful stackoverflow.com/questions/20346790/…

– Supun Dharmarathne
Jan 2 at 10:35

Note that your desired output is not valid JSON

– Hulk
Jan 2 at 10:35

1

I think the reason you are finding it difficult to get the json you require is because it simply isn't valid json: {"codes":{"200":"success"},{"401":"not found"}}. Here you have codes object {"200":"success"} and then an anonymous object {"401":"not found"}

– karen
Jan 2 at 10:35

hi folks, desired json which I'm trying to generate will be valid one

– Narasimha
Jan 2 at 11:09

@Narasimha {"codes":{"200":"success"},{"401":"not found"}} Is it valid JSON ?

– varatharajan
Jan 2 at 11:25

this might be helpful stackoverflow.com/questions/20346790/…

– Supun Dharmarathne
Jan 2 at 10:35

Note that your desired output is not valid JSON

– Hulk
Jan 2 at 10:35

I think the reason you are finding it difficult to get the json you require is because it simply isn't valid json: {"codes":{"200":"success"},{"401":"not found"}}. Here you have codes object {"200":"success"} and then an anonymous object {"401":"not found"}

– karen
Jan 2 at 10:35

hi folks, desired json which I'm trying to generate will be valid one

– Narasimha
Jan 2 at 11:09

@Narasimha {"codes":{"200":"success"},{"401":"not found"}} Is it valid JSON ?

– varatharajan
Jan 2 at 11:25

|
show 4 more comments

2 Answers
2

active

oldest

votes

Your expected JSON {"codes":{"200":"success"},{"401":"not found"}} is invalid. You can verify it with https://jsonlint.com/ which will produce an error:

Error: Parse error on line 4:

...00": "success"   },  {       "401": "not foun

---------------------^

Expecting 'STRING', got '{'

You most likely want an array to group the first and second object which results in below JSON (do notice the square brackets [ and ]):

{"codes":[{"200":"success"},{"401":"not found"}]}

This can be achieved with:

JSONObject first = new JSONObject();

first.put("200", "success");



JSONObject second = new JSONObject();

second.put("401", "not found");



JSONArray codes = new JSONArray();

codes.put(first);

codes.put(second);



JSONObject root = new JSONObject();

root.put("codes", codes);



System.out.println(root);

edited Jan 2 at 11:10

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

hi Karol, It's same as finalJSON3 code snippet

– Narasimha
Jan 2 at 11:08

@Narasimha your desired output is not a valid JSON. You can craft it manually using StringBuilder but no sane JSON library will let you create {"codes":{"200":"success"},{"401":"not found"}}

– Karol Dowbecki
Jan 2 at 11:09

add a comment |

Got the logic finally.

            JSONObject successJSON = new JSONObject();

            successJSON.put("description", "success");



            JSONObject scJSON = new JSONObject();

            scJSON.put("200", successJSON);







            JSONObject failJSON = new JSONObject();

            failJSON.put("description","failure");





            scJSON.put("401", failJSON);





            JSONObject finalJSON = new JSONObject();

            finalJSON.put("codes", scJSON);



            System.out.println("finalJSON --> "+finalJSON.toString());

answered Jan 3 at 8:15

Narasimha

1615

output: finalJSON --> {"codes":{"200":{"description":"success"},"401":{"description":"failure"}}}

– Narasimha
Jan 3 at 8:15

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Your expected JSON {"codes":{"200":"success"},{"401":"not found"}} is invalid. You can verify it with https://jsonlint.com/ which will produce an error:

Error: Parse error on line 4:

...00": "success"   },  {       "401": "not foun

---------------------^

Expecting 'STRING', got '{'

You most likely want an array to group the first and second object which results in below JSON (do notice the square brackets [ and ]):

{"codes":[{"200":"success"},{"401":"not found"}]}

This can be achieved with:

JSONObject first = new JSONObject();

first.put("200", "success");



JSONObject second = new JSONObject();

second.put("401", "not found");



JSONArray codes = new JSONArray();

codes.put(first);

codes.put(second);



JSONObject root = new JSONObject();

root.put("codes", codes);



System.out.println(root);

edited Jan 2 at 11:10

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

hi Karol, It's same as finalJSON3 code snippet

– Narasimha
Jan 2 at 11:08

@Narasimha your desired output is not a valid JSON. You can craft it manually using StringBuilder but no sane JSON library will let you create {"codes":{"200":"success"},{"401":"not found"}}

– Karol Dowbecki
Jan 2 at 11:09

add a comment |

Your expected JSON {"codes":{"200":"success"},{"401":"not found"}} is invalid. You can verify it with https://jsonlint.com/ which will produce an error:

Error: Parse error on line 4:

...00": "success"   },  {       "401": "not foun

---------------------^

Expecting 'STRING', got '{'

You most likely want an array to group the first and second object which results in below JSON (do notice the square brackets [ and ]):

{"codes":[{"200":"success"},{"401":"not found"}]}

This can be achieved with:

JSONObject first = new JSONObject();

first.put("200", "success");



JSONObject second = new JSONObject();

second.put("401", "not found");



JSONArray codes = new JSONArray();

codes.put(first);

codes.put(second);



JSONObject root = new JSONObject();

root.put("codes", codes);



System.out.println(root);

edited Jan 2 at 11:10

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

hi Karol, It's same as finalJSON3 code snippet

– Narasimha
Jan 2 at 11:08

@Narasimha your desired output is not a valid JSON. You can craft it manually using StringBuilder but no sane JSON library will let you create {"codes":{"200":"success"},{"401":"not found"}}

– Karol Dowbecki
Jan 2 at 11:09

add a comment |

Your expected JSON {"codes":{"200":"success"},{"401":"not found"}} is invalid. You can verify it with https://jsonlint.com/ which will produce an error:

Error: Parse error on line 4:

...00": "success"   },  {       "401": "not foun

---------------------^

Expecting 'STRING', got '{'

You most likely want an array to group the first and second object which results in below JSON (do notice the square brackets [ and ]):

{"codes":[{"200":"success"},{"401":"not found"}]}

This can be achieved with:

JSONObject first = new JSONObject();

first.put("200", "success");



JSONObject second = new JSONObject();

second.put("401", "not found");



JSONArray codes = new JSONArray();

codes.put(first);

codes.put(second);



JSONObject root = new JSONObject();

root.put("codes", codes);



System.out.println(root);

edited Jan 2 at 11:10

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

Your expected JSON {"codes":{"200":"success"},{"401":"not found"}} is invalid. You can verify it with https://jsonlint.com/ which will produce an error:

Error: Parse error on line 4:

...00": "success"   },  {       "401": "not foun

---------------------^

Expecting 'STRING', got '{'

You most likely want an array to group the first and second object which results in below JSON (do notice the square brackets [ and ]):

{"codes":[{"200":"success"},{"401":"not found"}]}

This can be achieved with:

JSONObject first = new JSONObject();

first.put("200", "success");



JSONObject second = new JSONObject();

second.put("401", "not found");



JSONArray codes = new JSONArray();

codes.put(first);

codes.put(second);



JSONObject root = new JSONObject();

root.put("codes", codes);



System.out.println(root);

edited Jan 2 at 11:10

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

edited Jan 2 at 11:10

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

answered Jan 2 at 10:41

Karol Dowbecki

24.8k93759

hi Karol, It's same as finalJSON3 code snippet

– Narasimha
Jan 2 at 11:08

@Narasimha your desired output is not a valid JSON. You can craft it manually using StringBuilder but no sane JSON library will let you create {"codes":{"200":"success"},{"401":"not found"}}

– Karol Dowbecki
Jan 2 at 11:09

add a comment |

hi Karol, It's same as finalJSON3 code snippet

– Narasimha
Jan 2 at 11:08

@Narasimha your desired output is not a valid JSON. You can craft it manually using StringBuilder but no sane JSON library will let you create {"codes":{"200":"success"},{"401":"not found"}}

– Karol Dowbecki
Jan 2 at 11:09

hi Karol, It's same as finalJSON3 code snippet

– Narasimha
Jan 2 at 11:08

@Narasimha your desired output is not a valid JSON. You can craft it manually using StringBuilder but no sane JSON library will let you create {"codes":{"200":"success"},{"401":"not found"}}

– Karol Dowbecki
Jan 2 at 11:09

add a comment |

Got the logic finally.

            JSONObject successJSON = new JSONObject();

            successJSON.put("description", "success");



            JSONObject scJSON = new JSONObject();

            scJSON.put("200", successJSON);







            JSONObject failJSON = new JSONObject();

            failJSON.put("description","failure");





            scJSON.put("401", failJSON);





            JSONObject finalJSON = new JSONObject();

            finalJSON.put("codes", scJSON);



            System.out.println("finalJSON --> "+finalJSON.toString());

answered Jan 3 at 8:15

Narasimha

1615

output: finalJSON --> {"codes":{"200":{"description":"success"},"401":{"description":"failure"}}}

– Narasimha
Jan 3 at 8:15

add a comment |

Got the logic finally.

            JSONObject successJSON = new JSONObject();

            successJSON.put("description", "success");



            JSONObject scJSON = new JSONObject();

            scJSON.put("200", successJSON);







            JSONObject failJSON = new JSONObject();

            failJSON.put("description","failure");





            scJSON.put("401", failJSON);





            JSONObject finalJSON = new JSONObject();

            finalJSON.put("codes", scJSON);



            System.out.println("finalJSON --> "+finalJSON.toString());

answered Jan 3 at 8:15

Narasimha

1615

output: finalJSON --> {"codes":{"200":{"description":"success"},"401":{"description":"failure"}}}

– Narasimha
Jan 3 at 8:15

add a comment |

Got the logic finally.

            JSONObject successJSON = new JSONObject();

            successJSON.put("description", "success");



            JSONObject scJSON = new JSONObject();

            scJSON.put("200", successJSON);







            JSONObject failJSON = new JSONObject();

            failJSON.put("description","failure");





            scJSON.put("401", failJSON);





            JSONObject finalJSON = new JSONObject();

            finalJSON.put("codes", scJSON);



            System.out.println("finalJSON --> "+finalJSON.toString());

answered Jan 3 at 8:15

Narasimha

1615

Got the logic finally.

            JSONObject successJSON = new JSONObject();

            successJSON.put("description", "success");



            JSONObject scJSON = new JSONObject();

            scJSON.put("200", successJSON);







            JSONObject failJSON = new JSONObject();

            failJSON.put("description","failure");





            scJSON.put("401", failJSON);





            JSONObject finalJSON = new JSONObject();

            finalJSON.put("codes", scJSON);



            System.out.println("finalJSON --> "+finalJSON.toString());

answered Jan 3 at 8:15

Narasimha

1615

answered Jan 3 at 8:15

Narasimha

1615

answered Jan 3 at 8:15

Narasimha

1615

answered Jan 3 at 8:15

Narasimha

1615

output: finalJSON --> {"codes":{"200":{"description":"success"},"401":{"description":"failure"}}}

– Narasimha
Jan 3 at 8:15

add a comment |

output: finalJSON --> {"codes":{"200":{"description":"success"},"401":{"description":"failure"}}}

– Narasimha
Jan 3 at 8:15

output: finalJSON --> {"codes":{"200":{"description":"success"},"401":{"description":"failure"}}}

– Narasimha
Jan 3 at 8:15

add a comment |

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