Error $.parsejson Is Not A Function
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Parse Json In Jquery
a minute: Sign up which function can i use instead of parseJSON jquery version 1.3.2 up vote 1 down vote favorite hi i am using jquery version 1.3.2 , i use it because my jquery validation plugin does not work properly with jquery 1.6.2 , but when using version 1.3.2 parseJSON is not a function appears in firebug . i searched and unexpected token o in json at position 1 found the answer that is because parseJSON was added in version 1.4.1 so what function can i use instead of parseJSON please help................... javascript jquery share|improve this question edited Oct 7 '11 at 7:26 VolkerK 72.9k11121182 asked Oct 7 '11 at 7:22 Kanishka Panamaldeniya 8,6581474131 2 github.com/douglascrockford/JSON-js maybe? –Yoshi Oct 7 '11 at 7:24 add a comment| 3 Answers 3 active oldest votes up vote 4 down vote accepted You may use JSON.Parse(). var jsontext = '{"firstname":"Jesper","surname":"Aaberg","phone":["555-0100","555-0120"]}'; var contact = JSON.parse(jsontext); var fullname = contact.surname + ", " + contact.firstname; share|improve this answer edited Feb 12 '15 at 11:39 Marcel 5,279134797 answered Oct 7 '11 at 7:25 Kai 2,11011025 thank you very much............... :D –Kanishka Panamaldeniya Nov 3 '11 at 13:07 1 Keep in mind that this isn't available <=IE7 –Aaron Apr 3 '13 at 2:56 It should be JSON.parse(). –RN Kushwaha Sep 30 '15 at 13:01 add a comment| up vote 1 down vote You can always eval() JSON http://www.json.org/js.html Or use one of the existing js libraries to parse JSON. This, for example: https://github.co
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policies of this site About Us Learn more about Stack Overflow the company Business Learn more about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community http://stackoverflow.com/questions/7684189/which-function-can-i-use-instead-of-parsejson-jquery-version-1-3-2 Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up jQuery - parseJSON not a function? up vote 0 down vote favorite I'm using Wordpress 3 and jQuery 1.4.2. Firebug tells me $.parseJSON is not a function http://stackoverflow.com/questions/3709958/jquery-parsejson-not-a-function and I'm stumped as to why. Any suggestions are appreciated. $(document).ready(function(){ $('#subscribe_form_submit').click(function(){ function updatePage(theResponse, textStatus, XMLHttpRequest){ theResponse = $.parseJSON(theResponse); console.log(theResponse); if(theResponse == "OK!"){ $('#subscribe').fadeOut('fast', function(){ $('#subscribe').html("
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"); $('#subscribe').fadeIn('slow'); }); } else{ theResponse = $.parseJSON(theResponse); console.log(theResponse); } } var theData = $('#subscribe').serialize(); $.ajax({ type: 'GET', url: 'http://www.foo.com/wp-content/themes/thesis_17/custom/subscribe.php?' + theData, dataType: 'json', success: updatePage(), error: function(xhr, textStatus, errorThrown){ console.log((errorThrown ? errorThrown : xhr.status)); } }) }); }); jquery share|improve this question asked Sep 14 '10 at 14:41 Tim76 4092827 Look the API documentation api.jquery.com/jQuery.parseJSON, –Felix G Sep 14 '10 at 14:46 1 In the second last line there's missing a semicolon. Adding that maybe that solves the problem... –davehauser Sep 14 '10 at 14:46 add a comment| 3 Answers 3 active oldest votes up vote 2 down vote accepted I'm not sure why you get that error message, but I don't think that's the real
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a GitHub account Sign in Create a gist now Instantly share code, notes, and snippets. Star 27 Fork 11 zuch/parse.js Created Sep 14, 2012 Embed What would you like to do? Embed Embed this gist in your website. Embed Share Copy sharable URL for this gist. Share Clone via HTTPS Clone with Git or checkout with SVN using the repository's web address. HTTPS Learn more about clone URLs Download ZIP Code Revisions 3 Stars 27 Forks 11 jquery - Parse JSON with jQuery Example Raw parse.js /********************************** example **************************************/ var base_url = 'http://people.cs.uct.ac.za/~swatermeyer/VulaMobi/'; function example() { var response = ""; var form_data = { username: username, password: password }; $.ajax({ type: "POST", url: base_url + "ajax.php?test/json", data: form_data, success: function(response) { /*response = '[{"Language":"jQuery","ID":"1"},{"Language":"C#","ID":"2"}, {"Language":"PHP","ID":"3"},{"Language":"Java","ID":"4"}, {"Language":"Python","ID":"5"},{"Language":"Perl","ID":"6"}, {"Language":"C++","ID":"7"},{"Language":"ASP","ID":"8"}, {"Language":"Ruby","ID":"9"}]'*/ console.log(response); var json_obj = $.parseJSON(response);//parse JSON var output="
- "; for (var i in json_obj) { output+="
- " + json_obj[i].Language + ", " + json_obj[i].ID + ""; } output+=""; $('span').html(output); }, dataType: "json"//set to JSON }) } Owner zuch commented Sep 14, 2012 Helpful Tutorials on Parsing with jQuery http://jquerybyexample.blogspot.com/2012/05/how-to-read-and-parse-json-using-jquery.html http://webhole.net/2009/11/28/how-to-read-json-with-javascript/ tatthien commented Jul 24, 2013 Thanks. It's solve my problem :) dianneanes commented Jul 25, 2013 hi, i would like to know how can i go about getting this result: (Main Category: Shoes, Sub_Category1:Women, Sub_Category2: Sandals, Sub_Category3: Slide) from this json data example "Shoes>Women>Sandals>Slide". I need to separate each category and show it in the menu bar. thanks. Owner zuch commented Aug 14, 2013 @dianneanes Check out this gist: Populate List with JSON using jQuery Hope it helps