How can I convert JSON format string into a real object in JS?
-
I am using AngularJS. I get JSON format string from Server using RESTful, but now I want to work with my json response as a real js object I can transfer to other pages and do even more. How to do that? Is there a good framework that does it for me? My object is complicated, not a simple one.
-
Answer:
Angularjs web site do...
Jim Hatton at Quora Visit the source
Other answers
These are available methods AFAIK. Choose one from these Use JSON.parse JSON.parse(stringifiedObject) Use new function, var parsed = new Function('return ' + stringifiedObject)(); use eval, eval(stringifiedObject) I like the beauty of new Function though it has some minor side effects :).
Prathap Reddy
you can use javascript's parse function. var jsObject = JSON.parse(jsonString);
Levan Gulisashvili
If you are looking for an application to do it, there is https://itunes.apple.com/us/app/textlab/id1024903185?ls=1&mt=12 for Mac OS which has a feature to convert JSON into JavaScript object.
Ondrej Kvasnovsky
I will give the answer assuming you are using jQuery. var result = null; $.ajax({ type: 'post', url: '/result.php', data: "gimme some json dude", dataType: 'json', success: function(res){ result = res; // Now this $result is a member of the window object and can accessed globally. You can test it for null. If null, it doesn't hold value else it does. } }); //Suppose you want to parse it var parsed_result = JSON.parse(result); Hope this helps. Good Luck!
Christian M Raymonds
Nothing if you are using the $http service. The default response transform function of the service can automatically detect JSON and parse it into an object in the data variable of the success handler.
Arjun Mathai
You should return a json response from your API instead of a string. You should be able to set the content type of your response to JSON and then you will not have to parse it to JSON object.
Imad Hashmi
You could just use AngularJS: var jsObj = angular.fromJSON(<your_json_string>); Reference: https://docs.angularjs.org/api/ng/function/angular.fromJson or use proper browser json parse: var jsObj = JSON.parse(<you_json_string>); Reference: http://www.w3schools.com/json/json_eval.asp
Diego Moreira
You can use the gson api to do this. in java: for converting json to java object use: fromJson(jsonObject,javaclassname.class) for converting java object to json object use: toJson(javaObject) in javascript - use stringify method to convert the json to string, and parse(jsonstring) to convert the string to json
Mohammed Nazaar
You can just use JSON.parse() as other answers suggest here but if you are using AngularJS take a look at https://github.com/mgonto/restangular a very good rest service that simplifies everything. and will give you restangular object as the result which contains more than just a parsed JSON object.
S K Amarnath
Related Q & A:
- How can I echo characters before and after a string?Best solution by stackoverflow.com
- How can I convert a string number to a number in Perl?Best solution by Stack Overflow
- How can I convert real player audio files into mp3?Best solution by Yahoo! Answers
- How can I convert a video to mp3 format?Best solution by Yahoo! Answers
- How can I get my regular avatar of a real picture back as my avatar instead of the one now?Best solution by Yahoo! Answers
Just Added Q & A:
- How many active mobile subscribers are there in China?Best solution by Quora
- How to find the right vacation?Best solution by bookit.com
- How To Make Your Own Primer?Best solution by thekrazycouponlady.com
- How do you get the domain & range?Best solution by ChaCha
- How do you open pop up blockers?Best solution by Yahoo! Answers
For every problem there is a solution! Proved by Solucija.
-
Got an issue and looking for advice?
-
Ask Solucija to search every corner of the Web for help.
-
Get workable solutions and helpful tips in a moment.
Just ask Solucija about an issue you face and immediately get a list of ready solutions, answers and tips from other Internet users. We always provide the most suitable and complete answer to your question at the top, along with a few good alternatives below.