基於從php文件收到的數據如何返回真或假的ajax表單提交？

Question

形式總是被提交。我只想提交數據==「可用」，即... PHP文件返回可用。

function PostData() { 
     var xhr; 
     if (window.XMLHttpRequest) { 
      xhr = new XMLHttpRequest(); 
     } 
     else if (window.ActiveXObject) { 
      xhr = new ActiveXObject("Msxml2.XMLHTTP"); 
     } 
     else { 
      throw new Error("Ajax is not supported by this browser"); 
     } 
     var username = document.getElementById("username_input").value; 
     xhr.open('POST', 'header/onsubmit_check.php'); 
     xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); 
     xhr.send("username=" + username); 


     xhr.onreadystatechange = function() { 
       var data=xhr.responseText.trim(); 
       document.getElementById('onsubmit_feedback').innerHTML = data; 
     } 

     if (data=='Available') { 
     return true; 
     } else { 
     return false; 
     } 
    } 

Answer 1

在HTML確保您使用形式

<form id="my_form" onsubmit="return PostData()"> 

請在你的JavaScript代碼的一些變化

function PostData() { 
     var xhr,data; 
     if (window.XMLHttpRequest) { 
      xhr = new XMLHttpRequest(); 
     } 
     else if (window.ActiveXObject) { 
      xhr = new ActiveXObject("Msxml2.XMLHTTP"); 
     } 
     else { 
      throw new Error("Ajax is not supported by this browser"); 
     } 
     var username = document.getElementById("username_input").value; 
     xhr.open('POST', 'header/onsubmit_check.php'); 
     xhr.setRequestHeader("Content-Type", "application/x-www-form-urlencoded"); 
     xhr.send("username=" + username); 


     xhr.onreadystatechange = function() { 
       data=xhr.responseText.trim(); 
       document.getElementById('onsubmit_feedback').innerHTML = data; 
       if (data==='Available'){ 
        document.getElementById("my_form").submit(); 
       } 
     } 

     return false; 

    } 

函數總是返回false並提交表單時，需要AJAX回報值。