如何在jQuery中使用Ajax

Question

function ajaxFunction(){

    var ajaxRequest;  // The variable that makes Ajax possible!

try{
        // Opera 8.0+, Firefox, Safari
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }

    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4 && ajaxRequest.status==200){
        alert(ajaxRequest.responseText);
        }
    }


    var txt = document.getElementById("data");
    ajaxRequest.open("POST", "hello.php", true);

    ajaxRequest.send("user=" + txt.value); 
        alert("here");
   }

如何在jQuery Ajax中编写完全相同的功能？

Answer 1

var txt = $('#data').val();

$.ajax({
    url: 'hello.php',
    type: 'post',
    data  : { user: txt }, 
    success: function(data) {
        alert(data);
    },
    error : function(err, req) {
        alert("Your browser broke!");
    }
});