與獲取數據的Jquery /阿賈克斯

Question

我使用下面的腳本，我認爲它需要做到以下幾點：

當有人點擊與表單ID「提交」在href鏈接，它需要調用documentblock .inc.php並提醒已在documentblock.inc.php中回顯的數據

但我沒有收到警報。我做錯了什麼？

  <form action="<?php echo $domein.'/mijnpagina/bestanden/downloaden/' ?>" method="post" name="downloaditem"> 
        <label><input type="checkbox" id="checker_tiny" value="ja" name="blokkeer"/> Bestand blokkeren na downloaden</label> 
        <input name="docId" id="docId" type="hidden" value="<?php echo $fetch_bestand['id']; ?>"> 
        <input name="downloaditem" type="hidden" value="akkoord"> 
        <a href="#" id="submit" class="button-pink">Bestand downloaden</a> 
        </form> 
        <div class="clear"></div> 
       </div> 

       <script> 
       $(function() { 
        $('#submit').on('click', function() { 
         // stop form submission first 
         // GET VALUE OF APPID 
         var appid = $("#docId").val() 
          // GET JSON FROM PHP SCRIPT 
          $.ajax({ 
           type : 'POST', 
           url : '/includes/documentblock.inc.php', 
           data: {'appid':appid}, 
           success : function() { 
            alert(data); 
           }, 
           error : alert('test') 
          }); 
        }); 
       }); 
            </script>  

documentblock.inc.php：

<?php echo "test"; ?> 

Answer 1

您需要通過data您success功能：

success : function (data) { 
    alert(data); 
} 

Answer 2

難道就這樣和它的作品！

<script> 
function submitme(){ 
var docId=document.getElementById("docId").value; 
var blockId=$("input[name=blokkeer]:checked").val(); 

$.ajax({ 
     type: 'POST', 
     url: '/includes/documentblock.inc.php', 
     data: 'docId='+docId+'&blockId='+blockId, 
     success: function(msg){ 
      if(msg){ 
       document.getElementById("something2").innerHTML=msg; 
      } 
      else{ 
       return; 
      } 
     } 
    }); 
} </script>