weixin_33722405 2017-04-21 06:15 采纳率: 0%
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jQuery Ajax未定义

Hello I am newbie here,
I want to ask a few question about jquery
1. How can I hide undefined in the PHP
2. How to name inside name in the API

This is my coding Index.php or

mediamonitoring.comyr.com/jquery.php

Or this

 <!DOCTYPE html>
    <html>
    <head>
        <title>Tutorial</title>
        <link rel='stylesheet' href='css/css.html'/>
    </head>
    <body>
    <h1>jQuery Ajax Tutorial</h1>
    <h2>Name Age</h2>
    <ul id="tutor">

    </ul>


    </body>
    </html>
    <script src='js/jquery.js'></script>
        <script src="js/index.js"></script>
        <script src="js/main.js"></script>


    </body>
    </html>

And my Main.js

$(function(){

  var $tutor = $('#tutor');
  $.ajax({
    type: 'GET',
    url: 'http://rest.learncode.academy/api/johnbob/friends',
    success: function(tutor) {
      $.each(tutor,function(i,ex){//the new item is returned with an ID
        $tutor.append('<li>Name:'+ex.name+', Age '+ex.age+'</li>');
      });
    }
  });
});
  • 写回答

2条回答 默认 最新

  • 妄徒之命 2017-04-21 06:26
    关注

    no 1. using if, if(variable == null){ //hide }

    no 2. change your main.js with this

    $(function(){
      var $tutor = $('#tutor');
      $.ajax({
        type: 'GET',
        url: 'http://rest.learncode.academy/api/johnbob/friends',
        success: function(tutor) {
          $.each(tutor,function(i,ex){//the new item is returned with an ID
            if(ex.name != null && ex.age != null){
                console.log(ex);
                $tutor.append('<li>Name:'+ex.name+', Age '+ex.age+'</li>');
            }        
          });
        }
      });
    });
    
    评论

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