duan2891 2016-11-10 07:35
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单击按钮时,jQuery会自动更新视图上的值

I was create like button with current value on database and when the button has been clicked the value must be automatically updated on the view. Currently, the value was successfully updated on database, but not automatically updated on view.

Here is my jquery code:

<script type="text/javascript">
    $(document).ready(function(){
        $('#like').on('click', function(e){
            var id = '{{$news->id}}';
            $.get('{{ url('view/like')}}/'+id, function(data){
                console.log(id);
                console.log(data);
                $('#like_data').empty();
                $.each(data, function(index, element){
                    $('#like_data').append("<p>"+this.like+"</p>");
                });
            });
        });
    });
</script>

Here is my controller:

public function like($id){

    $news = News::find($id);

    $news->like += 1;
    $news->save();

    return $news;
}
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2条回答 默认 最新

  • duanshangying5102 2016-11-10 07:47
    关注

    You need to pass JSON response from your controller. Like this

    public function like($id){
    
        $news = News::find($id);
    
        $news->like += 1;
        $news->save();
    
        return response()->json([$news], 200);
    }
    

    you don't need to use $.each() for this response because you send only single object response. Just simply write this code:

    $('#like_data').html("<p>"+data.like+"</p>");
    

    if you want to use each just simply write:

     $.each(data, function (index, element) {
          $("#like_data").append("<p>" + element.like + "</p>");
     }); 
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
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