JavaScript Ajax Call

Question

1.How to Give the url of JavaScript ajax call???? 2.Any tutorial which teach me step by step how to call JavaScript ajax call in asp.net??????

Anubhav Chaudhary · Answer

You can use jQuery Ajax call like this one:- jQuery.ajax({ type: 'POST'  contentType: application/json; charset=utf-8  url: 'WebService1.asmx/getCityData'  data: {'stateName':' + stateName + '}  dataType: 'JSON'  success: function (response) { StateCityChart(response); return false ; } error: function (er sw sq) { //alert(Error occured in application); } }); 1- In url you need to pass your pagepath /methodname. 2- If you want to send some data then pass it in data:  otherwise pass it blank. If you are passing the data then remember to pass it with the name of method otherwise call will not be made. You can check my this article for any other help. http://www.c-sharpcorner.com/UploadFile/cd7c2e/dynamic-chart-creation-using-jquerygoogle-charts/ If you want any other help you can ask me.

Saroj Kumar Sahu · Answer

Hi Please visit the below link for Ajax call: http://encosia.com/using-jquery-to-directly-call-aspnet-ajax-page-methods/ http://www.c-sharpcorner.com/UploadFile/dacca2/understand-jquery-ajax-function-call-code-behind-C-Sharp-method/ http://www.aspsnippets.com/Articles/Calling-ASPNet-WebMethod-using-jQuery-AJAX.aspx http://www.aspsnippets.com/Articles/Call-ASPNet-Page-Method-using-jQuery-AJAX-Example.aspx http://stackoverflow.com/questions/14819607/ajax-call-to-asp-net-web-method-using-jquery

Shuvojit Halder · Answer

You can use function showCustomer(str) { var xmlhttp; if (str==) { document.getElementById(txtHint).innerHTML=; return; } if (window.XMLHttpRequest) {// code for IE7+ Firefox Chrome Opera Safari xmlhttp=new XMLHttpRequest(); } else {// code for IE6 IE5 xmlhttp=new ActiveXObject(Microsoft.XMLHTTP); } xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById(txtHint).innerHTML=xmlhttp.responseText; } } xmlhttp.open(GETgetcustomer.aspx?q=+strtrue); xmlhttp.send(); }