Question

我得到了这段代码，它适用于

http://codepen.io/seanjacob/pen/wfcHB

但是当我在网页上试用它时却什么都没做？我是厚人还是我完全错过了什么？我只想输入数字和邮政编码并获取地址。也可以获得经度和纬度信息吗？

我的代码是

<!DOCTYPE html>
<html lang="en">
<head>
<title>Untitled 1</title>
<style>
html { background:radial-gradient(#238dbb, #114476); height:100%; text-align:center; font-family:Helvetica; font-size:1em; color:#ddd; }

h1 { font-size:4em; margin-bottom:0; text-shadow: 0 2px 0 rgba(0, 0, 0, 0.4);}

input { text-align:center; text-transform:uppercase; padding:10px; font-size:0.8em; font-family:Helvetica; border:solid 2px #95bbdc; color:#111; }

#submit { display:block; text-transform:uppercase; background:#134c7e; border:solid 2px #95bbdc; width:150px; margin:10px auto; padding:8px; font-size:0.8em }
</style>
</head>

<body>

<h1>UK Postcodes</h1>

<p id="text">Enter a property name or number and postcode.<p>

<input id="number" type="text" value="name/number" onfocus="if(this.value == 'name/number') { this.value = ''; }" />

<input id="postcode" type="text" value="postcode" onfocus="if(this.value == 'postcode') { this.value = ''; }" />

<button id="submit">Submit</button>

<script src="js/jquery.min.js"></script>

<script>
$('#submit').click(function(){  

  //Get Postcode
  var number = $('#number').val();
  var postcode = $('#postcode').val().toUpperCase();;

  //Get latitude & longitude
  $.getJSON('http://maps.googleapis.com/maps/api/geocode/json?address=' + postcode + '&sensor=false',  
            function(data) {
              var lat = data.results[0].geometry.location.lat;
              var lng = data.results[0].geometry.location.lng;

  //Get address    
  $.getJSON('http://maps.googleapis.com/maps/api/geocode/json?latlng=' + lat + ',' + lng + '&sensor=false',  
            function(data) {              
              var address = data.results[0].address_components;              
              var street = address[1].long_name;
              var town = address[2].long_name;
              var county = address[3].long_name;                        

  //Insert
  $('#text').text(number + ', ' + street + ', ' + town + ', ' + county + ', ' + postcode);

    });
  });
});
<script>

</body>

</html>

Answer 1

您收到的回复不是JSON对象。在您喜欢的浏览器开发工具的网络面板上查看它。

javascript无法处理简单请求

1 个答案: