AJAX發佈到php請求問題

Question

嗨我想從ajax js文件發佈一個變量到一個php文件。這是我迄今的嘗試。

 var request = createRequest(); 
     var deletenode = node.id; 
    window.alert("nodeid=" + deletenode); 
     var vars = "deletenode="+deletenode; 
     request.open("POST", "deletenode.php", true); 
     request.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); 
     request.onreadystatechange = function() { 
    handleRequest(request); 
}; 
     request.send("deletenode=" + encodeURIComponent(deletenode)); 

這裏是我的PHP文件

<?php 
print_r($_POST); 
$node = $_POST['deletenode']; 
print "node to be deleted: $node"; 
?> 

沒有在我的PHP文件出現後，什麼都可以的問題。我的ajax請求是完好無損的，也是正確的。謝謝你，這是我的處理要求。

function handleRequest(request) { 
    // we only care for now about when we get to readyState 4 
    // which means the request completed and we have the response back 
    if(request.readyState == 4){ 
     //alert("response: " + request.responseText); // check to see what 
     // we got back just for testing 
     // now get response's TEXT and put into document (specify where) 
     // below we have an html element with the id as timeLoc 
     json= eval ("(" + request.responseText + ")");; 

     //alert ("json="+json); //tests what was recieved 
     //clicking the close button 
     closeButton.onclick = function() { 
      node.setData('alpha', 0, 'end'); 
      node.eachAdjacency(function(adj) { 
       adj.setData('alpha', 0, 'end'); 
       var request = createRequest(); 
       var deletenode = node.id; 
       window.alert("nodeid=" + deletenode); 
       var vars = "deletenode="+deletenode; 
       request.open("POST", "deletenode.php", true); 
       request.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); 
       request.onreadystatechange = function() { 
        handleRequest(request); 
       }; 
       request.send("deletenode=" + encodeURIComponent(deletenode)); 
      });  
     }// end readystate=4 
    }//end handle request 

Answer 1

記住，你必須將數據發送鍵/值對 - request.send("deletenode=" + encodeURIComponent(deletenode));

Answer 2

我不知道如果這是一個複製/粘貼錯誤，但如果這是你的實際代碼，您似乎丟失一些花括號關閉。並有一個雙重的;;在你的評估線的末尾。這是用額外的大括號整理的代碼，這是否工作？

function handleRequest(request) { 
    if(request.readyState == 4){ 
     json= eval ("(" + request.responseText + ")"); 
     closeButton.onclick = function() { 
      node.setData('alpha', 0, 'end'); 
      node.eachAdjacency(function(adj) { 
       adj.setData('alpha', 0, 'end'); 
       var request = createRequest(); 
       var deletenode = node.id; 
       window.alert("nodeid=" + deletenode); 
       var vars = "deletenode="+deletenode; 
       request.open("POST", "deletenode.php", true); 
       request.setRequestHeader("Content-type", "application/x-www-form-urlencoded"); 
       request.onreadystatechange = function() { 
        handleRequest(request); 
       }; 
       request.send("deletenode=" + encodeURIComponent(deletenode)); 
      }); 
     } 
    } 
}