如何爲表中的多個數據創建AJAX POST按鈕？

Question

我有臺這樣的

<table> 
    <tr> 
    <th>User</th> 
    <th>IP</th> 
    <th>Action</th> 
    </tr> 
    <tr> 
    <td>Jack</td> 
    <td>192.168.0.1</td> 
    <td><button id="check" name="192.168.0.1">Check</button></td> 
    </tr> 
    <tr> 
    <td>Eve</td> 
    <td>192.168.0.2</td> 
    <td><button id="check" name="192.168.0.2">Check</button></td> 
    </tr> 
<tr> 
    <td>Smith</td> 
    <td>192.168.0.3</td> 
    <td><button id="check" name="192.168.0.3">Check</button></td> 
    </tr> 
</table> 

如何實現發送POST數據IP與AJAX？ 我曾嘗試與此代碼，但不工作..

<script> 
$('#check').click(function() { 

var getIP = $('#check').name(); 
var dataIP = 'sendIP=' + getIP; 

     $.ajax({ 
      url: 'url.php', 
      type: 'POST', 
      data: dataIP; 
      success: function() { 
       alert("Success"); 
      } 
     }); 

}); 
</script> 

[更新] 這是我的項目全部代碼。

<script> 
$(document).ready(function() { 
    $('#reportrange span').on('DOMSubtreeModified', function() { 
    var dariRange = $(this).html(); 
    var SplitRange = dariRange.split("~"); 

    $('#datatable-keytable').DataTable({ 
    "destroy": true, 
    "processing": true, 
    "keys": true, 
    "order": [[ 6, "desc" ]], 
    "ajax": { 
     url: "view.php", 
     type : 'GET', 
     data : { 
     datedari : SplitRange[0].trim(), 
     datesampai : SplitRange[1].trim() 
     } 
    }, 
    "columnDefs": [ 
    { "width": "5%", "targets": 0 }, 
    ], 

    "columns": [ 
     { "data": "click_username" }, 
     { "data": "click_cid" }, 
     { "data": "click_offer" }, 
     { "data": "click_ip" }, 
     { "data": "click_isp" }, 
     { "data": "click_posttime" }, 
     { "data": "click_ip", 
     "render": function (click_ip,data,row) { 
          var clickid = data.click_cid; 
          return ('<center><button class="check" id="'+click_ip+'" name="'+clickid+'">Check</button></center>'); //This for Button check 
         } 
     },   
     ], 
    }); 
}); 
}); 
</script> 
<script type="text/javascript"> 
$(document).on('click', '.check',function() { 

     var dataID = 'sendCID=' + this.name; 
     var dataIP = this.id; 

     $.ajax({ 
      url: 'send_data.php', 
      type: 'POST', 
      data: dataID; 
      success: function() { 
       window.open('http://whatismyipaddress.com/ip/'+dataIP); 
      } 
     }); 

}); 
</script> 

我要發佈var dataID到send_data.php，然後如果成功打開新標籤頁中http://whatismyipaddress.com/ip/'+dataIP但不與此代碼的工作，

我希望有人能幫我解決這個問題，謝謝

Answer 1

你不能在一個頁面中重複ID，它們根據定義是唯一的，所以需要更改爲一個類

在事件處理程序中this是事件發生的元素so在你的情況下，this.name將是價值需要

// use class selector 
$('.check').click(function() {  

     var dataIP = 'sendIP=' + this.name; 

     $.ajax({ 
      url: 'url.php', 
      type: 'POST', 
      data: dataIP; 
      success: function() { 
       alert("Success"); 
      } 
     }); 

});