如何通過javascript函數調用codeigniter函數

Question

這不是在我的codeigniter函數中工作我有id並且無法獲取該id。請幫助我。這是我的觀點，我試圖通過函數發送我的id。

<script type="text/javascript"> 
    function makeajaxcall(id) { 
     //alert(id); 
     var r = confirm("Do you want to Delete"); 
     if (r == true) { 
      window.location.href = "<?php echo site_url('controller_d/login/admin_link_delete_user?id='.id);?>"; 
     } else { 
      x = "You pressed Cancel!"; 
      alert(x); 
     } 

    } 
</script> 

Answer 1

改變這一行：

window.location.href = "<?php echo site_url('controller_d/login/admin_link_delete_user?id='.id);?>"; 

要：

window.location.href = "<?php echo site_url('controller_d/login/admin_link_delete_user');?>?id="+id; 

Answer 2

嘗試使用此

window.location.href = "<?php echo site_url('controller_d/login/admin_link_delete_user/'.id);?>"; 

假設在你的控制器，你有這樣的方法

function admin_link_delete_user($id){ 
    echo $id;// u'll get the id which you are passing through javascript  
} 

Answer 3

<script type="text/javascript"> 
function makeajaxcall(id) 
{ 
     //alert(id); 
     var r=confirm("Do you want to Delete"); 
     if (r==true) 
     { 
      $.post("<?php echo site_url('controller_d/login/admin_link_delete_user/');?>", {id:id}, 
      function(data) { 
       alert(data+"a"); 
      }, 'json'); 
     } 
     else 
     { 
      x="You pressed Cancel!"; 
      alert(x); 
     } 

} 
</script> 

在控制器

function admin_link_delete_user(){ 
    echo $this->input->post('id'); 
    //perform your code 
}