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Javascript 显示数据库中的数据而不重新加载_Javascript_Php_Jquery_Ajax_Laravel - Fatal编程技术网

Javascript 显示数据库中的数据而不重新加载

Javascript 显示数据库中的数据而不重新加载,javascript,php,jquery,ajax,laravel,Javascript,Php,Jquery,Ajax,Laravel,HTML <textarea>Ask a question</textarea><button class="addquestion">POST</button> Laravel控制器 public function addquestion(Request $request) { if($request->Ajax()) { $question = new Question()

HTML

<textarea>Ask a question</textarea><button class="addquestion">POST</button>
Laravel控制器

public function addquestion(Request $request)
    {
        if($request->Ajax())
        {
            $question = new Question();
            $question->question=$request->question;
            if($question->save())
            {
                   echo json_encode(TRUE);die;
            }
        }
    }

在发出警报后执行此操作

 if(res == true)
    {
       alert('question added!');
       //Add question here
    }
添加div

<textarea>Ask a question...</textarea><button class="addquestion">POST</button>
<div id="question"></div>
显示:

success: function (data) {
            var response = $.parseJSON(data);
            if(response.status == true)
            {
                var question=response.question;
                $('#question').append(question);                
            }
首先附加这个
$(this.prev().val()然后将其保存到数据库。
<textarea>Ask a question...</textarea><button class="addquestion">POST</button>
<div id="question"></div>
echo json_encode(array('status'=>TRUE,'question'=>$request->question));die;
success: function (data) {
            var response = $.parseJSON(data);
            if(response.status == true)
            {
                var question=response.question;
                $('#question').append(question);                
            }