2013-07-24 54 views
0

有人可以帮助我将此代码转换为$ .post,我替换的任何代码都不起作用。谢谢

$.ajax({ 
     type: "POST", 
     url: "http://192.168.254.107/webs/main/ajax/validateLogin.php", 
     data: sendData, 
     success: function(data) { 
      $("#info").html(data); 
     var ReturnMessage = data; 
     var mes = ReturnMessage.toString().trim(); 


     if (mes != "" && mes != "Invalid Login!") { 
       localStorage.setItem("message", ReturnMessage); 
       window.localStorage.setItem("username", username); 
       window.localStorage.setItem("data", data); 
       document.location.href = "trackme.html"; 

      } else { 
       alert(mes); 
      } 

     } 

    }); 
+2

你尝试过什么?你读过这个吗? http://api.jquery.com/jQuery.post/ – vee

回答

1
如何

一下:

$.post('http://192.168.254.107/webs/main/ajax/validateLogin.php', sendData, function(data) { 

    $("#info").html(data); 
    var ReturnMessage = data; 
    var mes = ReturnMessage.toString().trim(); 

    if (mes != "" && mes != "Invalid Login!") { 
     localStorage.setItem("message", ReturnMessage); 
     window.localStorage.setItem("username", username); 
     window.localStorage.setItem("data", data); 
     document.location.href = "trackme.html"; 

    } else { 
     alert(mes); 
    } 

}); 

来源:http://api.jquery.com/jQuery.post/

+0

谢谢@jrthib。有用! – miles

+0

@miles,请将此标记为答案... – Kasma

+0

在这种情况下,sendData是否为对象? – DevlshOne

0

它应该是:

$.post("http://192.168.254.107/webs/main/ajax/validateLogin.php", sendData).done(function(data) { 
      $("#info").html(data); 
     var ReturnMessage = data; 
     var mes = ReturnMessage.toString().trim(); 


     if (mes != "" && mes != "Invalid Login!") { 
       localStorage.setItem("message", ReturnMessage); 
       window.localStorage.setItem("username", username); 
       window.localStorage.setItem("data", data); 
       document.location.href = "trackme.html"; 

      } else { 
       alert(mes); 
      } 

     });