小编use*_*387的帖子

这在我的codeigniter函数中不起作用我有id并且无法获得该id.请帮助我.这是我的观点,我试图通过该功能发送我的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>