無法從Jquery發送值到PHP

Question

每次運行此代碼時，都會收到「VALUE IS EMPTY」消息。意味着變量不會從Jquery傳遞給PHP變量$value。我需要將它傳遞給PHP。幫我！

（#popup是一個div它出現在單擊表TR時）

HTML + PHP代碼：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 

<div id="popup"> 

    <?php 
     if (isset($_POST['val']) && !empty($_POST['val'])) 
     { 
     $value = $_POST['val']; 
     echo $value; 
     } 
     else{ 
      echo "VALUE IS EMPTY"; 
      //each time I get this msg.. means it won't take the variable value to PHP 
     } 
    ?> 
</div> 

的JQuery/Javascript代碼：

$(document).ready(function(){ 
    $('#tableview tbody tr').click(function(){ 

    var a=[]; 
    $(this).find('td').each(function(){ 
    a.push($(this).text()); 

    }); 

      $('#popup').show(); 
      $.ajax({ 
       url:'addfile.php', 
       data:{'val': a[1] }, 
       type:'post', 
       success:function(res){ 
       //Is there an error due to this function? or what? 
       } 

      }); 
}); 
}); 

- >沒有必要插入代碼的表..希望這可以澄清我的問題

Answer 1

OK，所以這裏有一個快速，基本原則：

你的HTML頁面，讓我們將其命名爲index.html的（PHP腳本不同！）：

// index.html 
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 

<div id="popup"></div> 
<script> 
$(document).ready(function(){ 
    $('#tableview tbody tr').click(function(){ 

     var a=[]; 
     $(this).find('td').each(function(){ 
      a.push($(this).text()); 
     }); 

     $('#popup').show(); 
     $.ajax({ 
      url:'addfile.php', 
      // HERE's A CRITICAL CHANGE: 
      // val MUST NOT be in quotes. You're creating a plain object here 
      data:{ val: a[1] }, 
      // you can change that from type to method (default is get) - though type is an alias for method, so both should work. 
      method:'post', 
      success:function(res){ 
       // here you have your data back from php, do with that what ever you need. 
       console.log(res); // will show value of a[1] in console 
       // my guess is, you want to show it in #popup 
       $('#popup').html(res); 
      }, 
      error: function(error, status, errorString) { 
       // allways check for possible errors! 
       console.log(error); 
       console.log(errorString); 
      } 
     }); 
    }); 
}); 
</script> 

你的PHP腳本

<?php 
// adfile.php 
if (isset($_POST['val']) && !empty($_POST['val'])) { 
     $value = $_POST['val']; 
     echo $value; 
} else { 
     echo "VALUE IS EMPTY"; 
} 
// don't add a closing tag.... 

最後：請閱讀有關ajax如何工作並閱讀manuals！