Question

今天我有点问题。我有一个我一直用于我的网站的ajax脚本，但出于效率原因，我需要将其转换为jQuery。我一直在研究如何做到这一点，并且没有这样的运气。

这是我的剧本：

<script type='text/javascript'>
//<![CDATA[
function ajaxFunction2(){
var ajaxRequest;  

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){
        var ajaxDisplay = document.getElementById('editpagecolors');
        ajaxDisplay.innerHTML = ajaxRequest.responseText;

    }
   }
  var bandname = document.getElementById('bandname').value;
  var musicstyle = document.getElementById('musicstyle').value;
  var websiteurl = document.getElementById('websiteurl').value;
  var aboutme = document.getElementById('aboutme').value;
  var chooser = document.getElementById('chooser').value;
  var chooser2 = document.getElementById('chooser2').value;
  var chooser3 = document.getElementById('chooser3').value;
  var chooser5 = document.getElementById('chooser5').value;
    var chooser6 = document.getElementById('chooser6').value;
  var chooser7 = document.getElementById('chooser7').value;
  var params = 'bandname=' + bandname + '&musicstyle=' + musicstyle +   '&websiteurl=' + websiteurl + '&aboutme=' + aboutme + '&chooser=' + chooser + '&chooser2='   + chooser2 + '&chooser3=' + chooser3 + '&chooser5=' + chooser5 + '&chooser6=' + chooser6 + '&chooser7=' + chooser7;
  ajaxRequest.open("POST", 'ajaxeditprofile.php', true);  
  ajaxRequest.setRequestHeader('Content-Type','application/x-www-form-urlencoded');
  ajaxRequest.send(params);

}
//]]>
</script>

甚至可以将其转换为jQuery，如果是这样，我将如何做到这一点。非常感谢你们的帮助！

Answer 1

    function ajaxFunction2(){

      var bandname = $('#bandname').val();
      var musicstyle = $('#musicstyle').val();
      var websiteurl = $('#websiteurl').val();
      var aboutme = $('#aboutme').val();
      var chooser = $('#chooser').val();
      var chooser2 = $('#chooser2').val();
      var chooser3 = $('#chooser3').val();
      var chooser5 = $('#chooser5').val();
        var chooser6 = $('#chooser6').val();
      var chooser7 = $('#chooser7').val();
      var params = 'bandname=' + bandname + '&musicstyle=' + musicstyle +   '&websiteurl=' + websiteurl + '&aboutme=' + aboutme + '&chooser=' + chooser + '&chooser2='   + chooser2 + '&chooser3=' + chooser3 + '&chooser5=' + chooser5 + '&chooser6=' + chooser6 + '&chooser7=' + chooser7; 

        $.post('ajaxeditprofile.php', params, function( data){                                             
            $('#editpagecolors').html( data);              
        });

    }

Answer 2

$.ajax({
   url: 'yourUrl',
   data: 'yourserializeddata',
   type: 'POST'
}).success(function(responseHtml) {
   $('#editpagecolors').html(responseHtml);
});

http://api.jquery.com/jQuery.ajax/

Answer 3

查看这里的指南at netplus

以上不起作用吗？它不是非常有效但是将它包装在jquery中并且加载jquery库也不是那么高效，我也怀疑将其移植到jquery将会带来很多节省（如jquery最终会在内部调用上面的内容）。

将ajax脚本转换为jQuery

3 个答案: