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 .

Dipika · 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 successvalue :' + 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 (xhrerrorTypeexception) { 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 successvalue :' + 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 T P · Answer

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

Dipika · Answer

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