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Answers for "how to make an ajax call"

Javascript
5

ajax call

$.ajax({
        url: "Url",
        dataType: "json",
        type: "Post",
        async: true,
        data: {"Key":value,"Key2":value2},
        success: function (data) {
                
        },
        error: function (xhr, exception, thrownError) {
            var msg = "";
            if (xhr.status === 0) {
                msg = "Not connect.\n Verify Network.";
            } else if (xhr.status == 404) {
                msg = "Requested page not found. [404]";
            } else if (xhr.status == 500) {
                msg = "Internal Server Error [500].";
            } else if (exception === "parsererror") {
                msg = "Requested JSON parse failed.";
            } else if (exception === "timeout") {
                msg = "Time out error.";
            } else if (exception === "abort") {
                msg = "Ajax request aborted.";
            } else {
                msg = "Error:" + xhr.status + " " + xhr.responseText;
            }
            if (callbackError) {
                callbackError(msg);
            }
           
        }
    });
Posted by: Guest on January-06-2021
3

make ajax calls with jQuery

// GET Request
    $.ajax({
        url: "example.php?firstParam=Hello&secondParam=World", //you can also pass get parameters
        dataType: 'json',	//dataType you expect in the response from the server
        timeout: 2000
    }).done(function (data, textStatus, jqXHR) {
        //your code here
    }).fail(function (jqXHR, textStatus, errorThrown) {
        console.log("jqXHR:" + jqXHR);
        console.log("TestStatus: " + textStatus);
        console.log("ErrorThrown: " + errorThrown);
    });

//POST Request
    var formData = {name: "John", surname: "Doe", age: "31"}; //Array 
    $.ajax({
        url: "example.php",
        type: "POST", // data type (can be get, post, put, delete)
        data: formData, // data in json format
       	timeout: 2000,	//Is useful ONLY if async=true. If async=false it is useless
        async: false, // enable or disable async (optional, but suggested as false if you need to populate data afterwards)
        success: function (data, textStatus, jqXHR) {
            //your code here
        },
        error: function (jqXHR, textStatus, errorThrown) {
            console.log("jqXHR:" + jqXHR);
            console.log("TestStatus: " + textStatus);
            console.log("ErrorThrown: " + errorThrown);
        }
    });


//Alternatively, the old aproach is
    $.ajax({
        url: "api.php?action=getCategories",
        dataType: 'json',
        timeout: 2000,
        success: function (result, textStatus, jqXHR) {   //jqXHR = jQuery XMLHttpRequest
            /*You could put your code here but this way of doing it is obsolete. Better to use .done()*/
        },
        error: function (jqXHR, textStatus, errorThrown) {
            console.log("jqXHR:" + jqXHR);
            console.log("TestStatus: " + textStatus);
            console.log("ErrorThrown: " + errorThrown);
        }
    });
Posted by: Guest on April-06-2021
6

ajax syntax in javascript

var id = empid;

$.ajax({
    type: "POST",
    url: "../Webservices/EmployeeService.asmx/GetEmployeeOrders",
    data: "{empid: " + empid + "}",
    contentType: "application/json; charset=utf-8",
    dataType: "json",
    success: function(result){
        alert(result.d);
        console.log(result);
    }
});
Posted by: Guest on September-30-2020
3

how to make ajax request javascript

//Change the text of a <div> element using an AJAX //request:
//using JQuery


$("button").click(function(){
  $.ajax({url: "demo_test.txt", success: function(result){
    $("#div1").html(result);
  }});
});



//To send a request to a server, we use the open() //and send() methods of the XMLHttpRequest object:
// Javascript


xhttp.open("GET", "ajax_info.txt", true);
xhttp.send();

//example below
<html>
<body>

<h1>The XMLHttpRequest Object</h1>

<button type="button" onclick="loadDoc()">Request data</button>

<p id="demo"></p>


<script>
function loadDoc() {
  var xhttp = new XMLHttpRequest();
  xhttp.onreadystatechange = function() {
    if (this.readyState == 4 && this.status == 200) {
      document.getElementById("demo").innerHTML = this.responseText;
    }
  };
  xhttp.open("GET", "demo_get.asp", true);
  xhttp.send();
}
</script>

</body>
</html>
Posted by: Guest on January-17-2020

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