javascript - 这个post提交方式哪里写的不对呢？

提交后插入不了数据 是我ajax写的不对吧？
tt.php

<code>


    <title></title>
    <script type="text/javascript">
        function ajax(url,data,funsucc){
            var oAjax=new XMLHttpRequest();
            oAjax.open('POST',url,true);                   
            oAjax.setRequestHeader("Content-Type","application/x-www-form-urlencoded;charset=UTF-8");
            oAjax.send("aa="+data);    
            oAjax.onreadystatechange=function(){
              if(oAjax.readyState==4){
                if(oAjax.status==200){
                  funsucc(oAjax.responseText);
                }
             }
           }
        }
    </script>
    <script type="text/javascript">
        window.onload=function(){
            var oTxt=document.getElementById('txt1');
            var oBtn=document.getElementById('btn1');
            oBtn.onclick=function(){
                ajax("ajax.php",oTxt.value,function(){
                    window.location.reload();
                });
            }
        }
    </script>


<form method="post">
    <input type="text" id="txt1">
    <button id="btn1" type="submit">提交</button>
</form>

</code>

ajax.php

<code><?php $pdo=new PDO("mysql:host=localhost;dbname=t1","root","");
$txt=$_POST["aa"];      
$stmt=$pdo->prepare("insert into ajax(txt)values(?)");
$stmt->execute(array($txt));
?></code>

回复内容：

提交后插入不了数据 是我ajax写的不对吧？
tt.php

<code>


    <title></title>
    <script type="text/javascript">
        function ajax(url,data,funsucc){
            var oAjax=new XMLHttpRequest();
            oAjax.open('POST',url,true);                   
            oAjax.setRequestHeader("Content-Type","application/x-www-form-urlencoded;charset=UTF-8");
            oAjax.send("aa="+data);    
            oAjax.onreadystatechange=function(){
              if(oAjax.readyState==4){
                if(oAjax.status==200){
                  funsucc(oAjax.responseText);
                }
             }
           }
        }
    </script>
    <script type="text/javascript">
        window.onload=function(){
            var oTxt=document.getElementById('txt1');
            var oBtn=document.getElementById('btn1');
            oBtn.onclick=function(){
                ajax("ajax.php",oTxt.value,function(){
                    window.location.reload();
                });
            }
        }
    </script>


<form method="post">
    <input type="text" id="txt1">
    <button id="btn1" type="submit">提交</button>
</form>

</code>

ajax.php

<code><?php $pdo=new PDO("mysql:host=localhost;dbname=t1","root","");
$txt=$_POST["aa"];      
$stmt=$pdo->prepare("insert into ajax(txt)values(?)");
$stmt->execute(array($txt));
?></code>

1、提交按钮点击默认会触发onsubmit事件，而你给它绑定的onclick事件里没有取消默认事件；

<code>oBtn.onclick=function(e){
    var e=window.event||e;
    e.preventDefault&&e.preventDefault();
    e.returnValue&&e.returnValue=false;
}</code>

2、采用默认onsubmit，无视ajax，txt1加上name="aa"；

<code>function ajax(url,data,funsucc){
            var oAjax=new XMLHttpRequest();
            oAjax.open('POST',url,true);                   
            oAjax.setRequestHeader("Content-Type","application/x-www-form-urlencoded;charset=UTF-8");
            oAjax.send("aa="+data);//在这里打个断点看看
            oAjax.onreadystatechange=function(){
              if(oAjax.readyState==4){
                if(oAjax.status==200){
                  funsucc(oAjax.responseText);
                }
             }
           }
        }</code>

ajax这个函数这样改：

<code>function ajax(url,data,funsucc){
            var oAjax=new XMLHttpRequest();
            oAjax.open('POST',url,true);                   
            oAjax.setRequestHeader("Content-Type","application/x-www-form-urlencoded;charset=UTF-8");          
            oAjax.onreadystatechange=function(){
              if(oAjax.readyState==4){
                if(oAjax.status==200){
                  funsucc(oAjax.responseText);
                }
             }
           }
              oAjax.send("aa="+data);   
        }</code>

异步调用，不然数据发送不出去