Question

thinkPHP的Ajax问题,为什么我会找不到我的Form的ID

<script type="text/javascript">
  function show(){
    ThinkAjax.sendForm('frm','__URL__/getID',wc); 
  }
  function wc(data,status){
    if(status!=1){
      console.log('发送失败');
    }else{
      console.log(data.questionID);
    } 
  }
</script>
<script type="text/javascript" src="__PUBLIC__/Js/ThinkAjax.js"></script>

这是form代码
<form action="" method="POST" id="frm" name="frm">
<button type="button" data-toggle="modal" data-target="#editques" class="templatemo-blue-button editfold" onClick="show()">修改
</button>
</form>
提交的时候提示两个问题
Uncaught ReferenceError: Form is not defined
Uncaught TypeError: Cannot read property 'find' of undefined

Answer 1

先加载 ThinkAjax 文件 再加载js代码