button should not postback if jquery.ajax return false

Question

i have a button i want that it should not be postback to serverside if jquery ajax featching value from database return false.if jquery ajax return true.button will postback to serverside code

Susanta Rout · Answer

Vasanti sorry to say i dont want to use async:false. bcz its some perfomace problem

ravi verma · Answer

Try to use input type button not input type Submit .

John · Answer

this works for me $(document).ready(function () { $("# ").click(function () { var test; $.ajax({ type: "POST", data: {}, async: false, url: "WebService.asmx/HelloWorld", contentType: 'application/json; charset=utf-8', dataType: 'json', error: function (xhr, errorType, exception) { try { alert(xhr.StackTrace); var responseText; responseText = $.parseJSON(xhr.responseText); var v = JSON.stringify(xhr.responseText); $("#msg").html(v); alert('error :' + v); var d = xhr.stackTrace; test = false; } catch (e) { } }, success: function (response) { alert('sucess :' + response.d); if (response.d == "Hello World") { test = true; $('# ').submit(); alert('inside success,value :' + test); test= true; } else { test= false; } //alert("t- " + test); // here we can call server side code } }); //alert('outside, value:' + test); return test; }); });

Susanta Rout · Answer

button id.submit() not work if it is sucess...its urent plz some buddy help me

Susanta Rout · Answer

var test; $.ajax({ type: "POST", url: "WebService1.asmx/HelloWorld", contentType: 'application/json; charset=utf-8', dataType: 'json', error: function (xhr,errorType,exception) { try { alert(xhr.StackTrace); var responseText; responseText = $.parseJSON(xhr.responseText); var v = JSON.stringify(xhr.responseText); $("#msg").html(v); alert('error :' + v); var d = xhr.stackTrace; } catch(e) { } return false; }, success:function(response) { alert('sucess :' + response.d); if(response.d=="Hello World") { test = true; $('# ').submit(); alert('inside success,value :' + test); } else { test = false; //return test; } // alert(test);// here we can call server side code } }); alert('outside, value:' + test); return false; });

Susanta Rout · Answer

Vasanti i already applied what you are suggesting about.

Midhun Tp · Answer

Hi, Try below code - $.ajax({ url : 'yoururl', success : function(data) { if (data == "true") $('#yourbuttonid').submit(); else return false; } } }); return false;

John · Answer

write the jquery.ajax code on client side button click and then return true false accordingly

button should not postback if jquery.ajax return false

Answers (8)