javascript - 這個post提交方式哪裡寫的不對呢？-php教程-PHP中文網

javascript - 這個post提交方式哪裡寫的不對呢？

WBOY

發布： 2016-08-18 09:15:49

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1250 人瀏覽過

提交後插入不了資料是我ajax寫的不對吧？
tt.php

<code><!DOCTYPE html>
<html>
<head>
    <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>
</head>
<body>
<form method="post">
    <input type="text" id="txt1">
    <button id="btn1" type="submit">提交</button>
</form>
</body>
</html></code>

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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>

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回覆內容：

提交後插入不了資料是我ajax寫的不對吧？
tt.php

<code><!DOCTYPE html>
<html>
<head>
    <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>
</head>
<body>
<form method="post">
    <input type="text" id="txt1">
    <button id="btn1" type="submit">提交</button>
</form>
</body>
</html></code>

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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>

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1、提交按鈕點擊預設會觸發onsubmit事件，而你給它綁定的onclick事件裡沒有取消預設事件；

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

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2、採用預設onsubmit，無視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>

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ajax這個函數這樣改：

<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>

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異步調用，不然資料發送不出去