Question

我在帖子数上显示ajax点赞，但是问题是，当我单击“点赞按钮”时，点赞（不加载页面）仅显示在第一篇文章上，而当我刷新页面时，所有点赞都单独显示的地方。我想在特定帖子上显示ajax，而不仅仅是在第一个。我想我需要在成功函数中传递ID，但我不知道如何。这是代码。在我的代码中：“谣言”就是一个帖子，“赞” =批准，“不喜欢” =不批准。

查看Ajax + HTML视图的特定部分

function approve_rumor(r_id) {
  // alert("dfdfdf");
  $.ajax({
    type: "post",
    url: "<?php echo base_url('Home1/approve_rumor') ?>",
    data: {
      r_id: r_id
    },
    cache: false,
    success: function(val) {
      if (true) {
        $('#showapp').html(val);
      } else {
        // <?php //echo base_url('Login'); ?>;
        alert("You need to Login");
      }
    }

  });
}

<span class="like-listing fll approve" onclick="approve_rumor(<?php echo $rumors->rumor_id; ?>)">
  <i class="fa fa-thumbs-o-up" aria-hidden="true"></i>
</span>
<span id="showapp" class="fll"><?php echo $rumors->rumor_totalapprovals ?></span>

控制器

function approve_rumor() {

  if ($this - > session - > userdata('user_email', 'user_id')) {
    $this - > load - > model('rumor_model');
    $r_id = $this - > input - > post('r_id');
    $data['user_id'] = $this - > session - > userdata('user_id');
    $data['rumor_id'] = $r_id;
    $this - > rumor_model - > ApproveRumor($data);

    //$this->load->view('home1',['approval'=>$approval]);
  } else {
    return redirect('Login');
  }
}

function ApproveRumor($data) {
  // echo "<pre>";
  // print_r($data);
  // exit;
  $r_id = $data['rumor_id'];
  $query = $this - > db - > insert('approval', $data);
  if ($query) {
    $a = $this - > db - > query("SELECT * ,  count(rumor_id) as count_like FROM approval WHERE rumor_id = $r_id ");
    echo "<pre>";

    print_r($a - > row() - > count_like);

    $totalapprovals = $a - > row() - > count_like;
    $this - > db - > query("UPDATE rumors SET rumor_totalapprovals = $totalapprovals WHERE rumor_id = $r_id");
    // echo $totalapprovals; 
    //  $this->db->where('rumors.rumor_id',$r_id);
    //  $res = $this->db->update('rumors',$totalapprovals);

  } else {
    echo "0";
  }
}

Answer 1

我发现将数据传递到$ .ajax回调函数的一种很酷的方法是添加一个自定义选项，然后在$ .ajax回调中将其引用为this.optionProp：

$.ajax({
    type: "post",
    url: "your/url/here",
    data: {
      r_id: r_id
    },
    r_id:r_id,  //Your custom option
    cache: false,
    success: function(val) {
      console.log("r_id: " + this.r_id);  //The custom option can be referenced with this.optProp
    }

  });

让我知道这是否对您有帮助，我们将为您解决

Answer 2

您需要从服务器端函数返回一个具有id属性的对象，这样您就可以在成功后像这样调用它：

   function approve_rumor(r_id) {
   // alert("dfdfdf");
   $.ajax({
   type: "post",
   url: "<?php echo base_url('Home1/approve_rumor') ?>",
   data: {
   r_id: r_id
   },
 cache: false,
 success: function(val) {
 var id = val.id; //gets the id attribute from the returned val Object
 }
});
}

如何使用帖子ID显示像特定帖子一样的ajax

2 个答案: