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Home Backend Development PHP Tutorial ThinkPhp ajax 提交

ThinkPhp ajax 提交

Jun 23, 2016 pm 01:48 PM
ajax thinkphp submit

本人新手 弱弱的问个问题 
function aa(){
        //ThinkAjax.sendForm(表单ID,URL,回调函数,信息显示的地方);
        ThinkAjax.sendForm('frm','app/Lib/modules/dealModule/add',wc);   
    }
    function wc(data,status){
        if(status!=1){
            alert('发送失败');
        }else{
            alert('2')
        }   
    }

-------------------------------------------------------------------------------------------------------------------
  
function TiJiao()

var name = document.getElementById("UserNaem").value;
var Eamil=document.getElementById("Eamil").value;
var Phone=document.getElementById("Phone").value;
var JinEr=document.getElementById("JinEr").value;

jQuery.ajax({ 
url:'app/Lib/modules/dealModule/add',
        type:'POST',
        data: {'UserNaem':name,'Eamil':Eamil,'Phone':Phone,'JinEr':JinEr},
        success:function(result){
           alert(result)
        },
        error:function(msg){
            alert('Error:');
        }
    });  
 } 



两个js 上面的两个引用的js一直prototype.js 报错Uncaught TypeError: undefined is not a function 这个错误  不知道怎么提交到后台


回复讨论(解决方案)

在线等。。。急  不知道是哪里错了

分比较少  感谢各位了。。。。

难道都下班了?

jQuery.ajax({ ......
这分明是 jQuery.js
你怎么说是 prototype.js 呢?

这个是Jquery吧。

没有引用jquery.min.js吧

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