AJAX jQUERY将POST中的数据传递给codeigniter控制器

Question

我想使用AJAX和Jquery以POST方法将数据传递给我的控制器。

看起来像这样： https : //jsfiddle.net/10bhsd2o/我有bootstrap下拉列表，显示数据库中的记录。

<li class="dropdown">


  <a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="false">Your Sites <span class="caret"></span></a>
  <ul class="dropdown-menu">
    <li ><a href="#"><?php

foreach($sites as $site)
{

echo "<li class='specialLink' id='$site->site_key'>".$site->site_key."</li>";
}?></a></li>

  </ul>
</li>

现在，我想将数据发送到控制器，即我从<li></li>标记的下拉列表中选择的项目。

这是我的脚本：

   <script type="text/javascript">



    $( ".specialLink" ).click(function() {
        var value = this.id;

        var url= "<?php echo base_url('customer/dashboard/index') ?>"; 
        alert(url) 
         //get value for throw to controller
        alert(value);  

        $("#specialLink").submit(function(){
            $.ajax({ 
                    type: "POST", //send with post 
                    url: "<?php echo base_url('customer/dashboard/index') ?>", 
                    data: {value:value}, 
                    success:function(data){ 
                    alert(data) 
                    } 
              });
        });
    });


    </script>

生成HTML

<ul class="dropdown-menu">
    <li ><a href="#"><li class='specialLink' id='54th-65hy'>54th-65hy</li><li class='specialLink' id='HT45-YT6T'>HT45-YT6T</li></a></li>

  </ul>

调节器

public function index()
{

    var_dump($_POST);
    print_r($_POST);
    $this->data['subview'] = 'customer/dashboard/index';  
    $this->load->view('customer/_layout_main',$this->data);
}

现在我的警报是正确的，我在警报中获取了正确的URL，在警报中获取了正确的数据。

但是在我的控制器中，当我执行print_r（$ _ POST）; 我得到空白数组为什么？ 我要去哪里错了？

Answer 1

您只需更改Java语言，即可删除$("#specialLink").submit(function(){和});

 <script type="text/javascript">



$( ".specialLink" ).click(function() {
    var value = this.id;

    var url= "<?php echo base_url('customer/dashboard/index') ?>"; 
    alert(url) 
     //get value for throw to controller
    alert(value);  


        $.ajax({ 
                type: "POST", //send with post 
                url: "<?php echo base_url('customer/dashboard/index') ?>", 
                data: {value:value}, 
                success:function(data){ 
                alert(data) 
                } 
          });

});


</script>

Answer 2

更改您的html标记。 因为你在锚内循环

    <li class="dropdown">
      <a href="#" class="dropdown-toggle" data-toggle="dropdown" role="button" aria-haspopup="true" aria-expanded="false">Your Sites <span class="caret"></span></a>
      <ul class="dropdown-menu">
        <?php
          foreach($sites as $site)
          {
            echo "<li class='specialLink' id='$site->site_key'><a href='#'>".$site->site_key."</a></li>";
          }
        ?>
      </ul>
    </li>

然后更改您的click功能

    $( ".specialLink" ).click(function() {
        var value = this.id;
        var url= "<?php echo base_url('customer/dashboard/index') ?>"; 
        alert(url) 
         //get value for throw to controller
        alert(value); 
        $.ajax({ 
            type: "POST", //send with post 
            url: "<?php echo base_url('customer/dashboard/index') ?>", 
            data: {value:value}, 
            success:function(data){ 
            alert(data) 
            } 
        });
    });

我不知道您是否将index函数用于其他视图但用于测试目的。 先尝试一下

    public function index()
    {
        if(!empty($_POST)){
            var_dump($_POST);
            print_r($_POST);
        }else{
            $this->data['subview'] = 'customer/dashboard/index';  
            $this->load->view('customer/_layout_main',$this->data);
        }
    }

让我知道您在第三个警报中得到了什么

Answer 3

$( ".specialLink" ).click(function() {
        var value = this.id;
        var url= "<?php echo base_url('customer/dashboard/index') ?>"; 
        alert(url) 
        alert(value); 
    $.post(url,{value:value},function(data){alert(data)});
  });