json响应在div ajax中获取数据

时间:2016-05-20 05:58:23

标签: json ajax

我在不同的div中获得json响应 json回应

 {"success":"you have not inserted","error":{"name":["The name field is required."],"detail":["The detail
 field is required."]}}

ajax代码

function postdata(){

   var name=$('#name').val();
   var detail=$('#detail').val(); 
   var token=$('#_token').val();
  $('#post').val('Submiting...');
    $.ajax({
                    type: 'GET',
                    url: 'posts',
                    data: "name="+ name + "&detail="+ detail,
                    success: function(data){
                 }  });

}

在此div

中获取json响应消息
<div class="alert alert-info col-ssm-12" id="detail"></div>

    <div class="alert alert-info col-ssm-12" id="name"></div>

1 个答案:

答案 0 :(得分:0)

通过.text()或.html()

获取值
function postdata(){

   var name=$('#name').text();
   var detail=$('#detail').text(); 
   var token=$('#_token').val();
  $('#post').val('Submiting...');
    $.ajax({
                    type: 'GET',
                    url: 'posts',
                    data: "name="+ name + "&detail="+ detail,
                    success: function(data){
                 }  });
}
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