我对json和ajax完全陌生。我试图从mysql查询中得到一个结果，并用result更新div。我使用不同的片段发现网上，可以得到结果的json编码，但似乎不能做什么呢？
从mysql获取并以json输出的代码或文件名api.php是：

include 'db.php';  
$sth = mysqli_query($con,"SELECT * FROM $tableName");
$rows = array();
while($r = mysqli_fetch_assoc($sth)) {
    $rows[] = $r;
}

 echo json_encode($rows);

这个输出是这样的

[{"user_id":"320326","type":"BUSINESS","business":"Business name","f_name":"DAVE","l_name":"TEST","email":"test@test.com","token":"","l_number":"0","m_number":"","password":"work","joined":"2018-07-06","sms_opt":"","instant_opt":"","instant_id":"","email_opt":"","offers_opt":"]

加载时要更改的页面上的html为：

<div id="output">this element will be accessed by jquery and this text replaced</div>

加载它并获取信息的代码是：

$(function () 
  {
    $.ajax({                                      
      url: 'api.php',                  //the script to call to get data          
      data: "",                        //you can insert url argumnets here to pass to api.php
                                       //for example "id=5&parent=6"
      dataType: 'json',                //data format      
      success: function(data)          //on recieve of reply
      {
        var user_id = data[0];              //get id
        var f_name = data[2];           //get name

        $('#output').html("<b>id: </b>"+user_id+"<b> name: </b>"+f_name); //Set output element html
        //recommend reading up on jquery selectors they are awesome 

      } 
    });
  });

理想情况下，我想测试这一点时，一个按钮被点击，然后使用相应的数据，我不知道我哪里出错，并尝试了一些事情，但很多在线例子太复杂，这是相当基本的外观。提前谢谢。