javascript＆ajax只與confirm（）

Question

一起工作下面的javascript函數似乎只有在我有最後一個confirm（）語句時才起作用，我最初在那裏爲了調試目的。當我把它拿出來，delete_row.php似乎並沒有運行。同時，也許作爲一個提示/側面說明，當我這樣做在那裏有確認聲明，它適用於除了野生動物園所有的瀏覽器...

function deleterow(form) { 

    if (!confirm("Are you sure you want to delete?")) return false; 

    var queryString = "?ID="; 

    for (var i = 0; i < document.myForm.rows.length; i++) { 
     if (document.myForm.rows[i].checked) { 
      ID = document.myForm.rows[i].value; 
      ID = ID.slice(0, -1); 
      queryString += ID; 
      queryString += "-"; 
     } 
    } 
    queryString = queryString.slice(0, -1); 

    try{ 
     // Opera 8.0+, Firefox, Safari 
     ajaxRequest = new XMLHttpRequest(); 
    } catch (e){ 
     // Internet Explorer Browsers 
     try{ 
      ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); 
     } catch (e) { 
      try{ 
       ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); 
      } catch (e){ 
       // Something went wrong 
       alert("Your browser broke!"); 
       return false; 
      } 
     } 
    } 

    var ajaxRequest; // The variable that makes Ajax possible! 
    // Create a function that will receive data sent from the server 
    ajaxRequest.onreadystatechange = function(){ 
     if(ajaxRequest.readyState == 4){ 
      var ajaxDisplay = document.getElementById('ajaxDiv'); 
      ajaxDisplay.innerHTML = ajaxRequest.responseText; 
     } 
    } 

    ajaxRequest.open("GET", "delete_row.php" + queryString, true); 
    ajaxRequest.send(null); 
    confirm('Delete successful!'); 
} 

UPDATE解決

我是通過以下js腳本更改檢查ajaxRequest的狀態

ajaxRequest.onreadystatechange = function(){ // Create a function that will receive data sent from the server 
    if(ajaxRequest.readyState == 4 && ajaxRequest.status == 200){ 
     var ajaxDisplay = document.getElementById('ajaxDiv'); 
     ajaxDisplay.innerHTML = ajaxRequest.responseText; 
    } 
    else{ 
     alert('An error has occurred making the request'); 
     return false; 
    } 
} 

並且注意到我從服務器獲得狀態0。一些Google搜索幫助我意識到錯誤在於我如何定義調用這些函數的按鈕。

原來的代碼是：

<div style='float:left; margin-right:10px;'><input type="submit" onClick="deleterow(document.myForm)" VALUE="Delete ROWs"></div> 

解決方法是：

<div style='float:left; margin-right:10px;'><input type="button" onClick="deleterow(document.myForm)" VALUE="Delete ROWs"></div> 

（提交類型已被改爲按鈕式）

Answer 1

delete_row.php doesn't seem to run你驗證了這一點，你可以將警報添加到if(ajaxRequest.readyState == 4){我想你的JS雖然沒有形式的東西，它似乎做工精細，http://jsfiddle.net/6gjy6/你得到在谷歌鉻合金控制檯任何JS錯誤？您是否嘗試過在瀏覽器上使用基本的「GET」請求進行訪問，例如delete_row.php" + queryString,，並查看服務器如何響應而不是AJAX調用。

試試這個：

var queryString = "?ID="; 

for (var i = 0; i < document.myForm.rows.length; i++) { 
    if (document.myForm.rows[i].checked) { 
     ID = document.myForm.rows[i].value; 
     ID = ID.slice(0, -1); 
     queryString += ID; 
     queryString += "-"; 
    } 
} 
queryString = queryString.slice(0, -1); 
var ajaxRequest; 

try{ 
    // Opera 8.0+, Firefox, Safari 
    ajaxRequest = new XMLHttpRequest(); 
} catch (e){ 
    // Internet Explorer Browsers 
    try{ 
     ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP"); 
    } catch (e) { 
     try{ 
      ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP"); 
     } catch (e){ 
      // Something went wrong 
      alert("Your browser broke!"); 
      return false; 
     } 
    } 
} 

// Create a function that will receive data sent from the server 
ajaxRequest.onreadystatechange = function(){ 
    if(ajaxRequest.readyState == 4){ 
     alert("received: " + ajaxRequest.responseText); 
     var ajaxDisplay = document.getElementById('ajaxDiv'); 
     ajaxDisplay.innerHTML = ajaxRequest.responseText; 
    } 
} 

ajaxRequest.open("GET", "delete_row.php" + queryString, true); 
ajaxRequest.send(null); 

Answer 2

讓你確認（）語句存在，而在你的js頂部放

window.alert = null ; 

，並嘗試

ķ讓我查

Answer 3

我相當肯定你應該打電話open後設置onreadystatechange事件，否則處理程序被清除。