2012-08-04 130 views
1

我是symfony的新手。我創建了一個名爲index.html.twig的樹枝文件。在這裏我寫了一個代碼如下:無法獲得ajax響應

<script language="javascript"> 
    function onsub() 
    { 
     alert(document.getElementById('source').value); 
     var http=new XMLHttpRequest(); 
     var name="rohit"; 
     if(http.open("POST", "sub.html.php?field="+name, true)) alert('yes'); 
     alert(http); 
     http.onreadystatechange = function() 
     { 
      if(http.readyState == 4 && http.status == 200) 
      { 

       alert('i m back'); 
      } 
      else 
      { 
       alert('sorry'); 
      } 
     } 

     http.send(); 
    } 

    </script> 

這是我按下提交按鈕提交表單時調用的操作。當我提交表單時,會調用此操作,並且正確的值將被刪除,但它不會將數據發送到請求的文件,即sub.html.php。它正在使對象正確。當我打印http.readyState的值時,它會給出答案1。所以我想發送name的值到請求的文件sub.html.php。我能爲此做些什麼?

回答

0

試試這個.....

<script language="javascript"> 
function onsub() 
{ 
    var name="rohit"; 
    alert(document.getElementById('source').value); 
    if(window.XMLHttpRequest) 
    { 
     http=new XMLHttpRequest(); 
    } 
    else 
    { 
    // code for IE6, IE5 
    http=new ActiveXObject("Microsoft.XMLHTTP"); 
    } 
    http.onreadystatechange=function() 
    { 
    if(http.readyState == 4 && http.status == 200) 
    { 
     alert('i m back'); 
    } 
    else 
    { 
     alert('sorry'); 
    } 
    }  
    http.open("GET", "sub.php?field="+name , true); 
    http.send(); 
    }  
    </script> 
+0

我都試過,但我沒有得到答案。當我採用http.readyState的答案時,它是0 – ucerturohit 2012-08-04 11:36:11