用于计算并从给定地址返回lat和long的jQuery函数不起作用。我从表单中获取地址输入,我可以在jQuery函数中显示地址,但地理编码不返回任何内容。
<?php
echo $this->Form->input('address_street',['id' => 'street']);
echo $this->Form->input('address_suburb',['id' => 'suburb']);
echo $this->Form->input('address_postcode',['id' => 'postcode']);
echo $this->Form->input('address_state',['id' => 'state','type' => 'select', 'options' => $auStates, 'default' => 'VIC']);
?>
<hr>
<?php
echo $this->Form->input('address_lat',['id' => 'lat']);
echo $this->Form->input('address_long',['id' => 'long']);
?>
$(document).ready(function($) {
var geocoder = new google.maps.Geocoder();
$(document).on('click', '#calculate_address_lat_long', function () {
var address = '';
address += $('#street').val();
address += ' ' + $('#suburb').val();
address += ' ' + $('#postcode').val();
address += ' ' + $('#state').val();
// alert( $('#street').val());
// alert( address);
geocoder.geocode( { 'address': address}, function(results, status) {
if (status == google.maps.GeocoderStatus.OK) {
console.log(results);
var latitude = results[0].geometry.location.lat();
var longitude = results[0].geometry.location.lng();
$('#lat').val(latitude);
$('#long').val(longitude);
$('#formated_address_lat_long')
.html('<div class="alert alert-success">' + results[0].formatted_address + '</div>')
.fadeTo(100, 0.1).fadeTo(250, 1)
.css({"position": "relative","top": "15px"});
alert($('#lat').val(latitude));
} else {
alert($('#lat').val(latitude));
alert($('#long').val(longitude));
$('#formated_address_lat_long')
.fadeTo(100, 0.1).fadeTo(250, 1)
.html('<div class="alert alert-error">Address Not Found</div>')
.css({"position": "relative","top": "15px"});
}
});
});
});
我该如何解决?
答案 0 :(得分:0)
如果我包括以下行 这解决了问题
<script type="text/javascript" src="maps.google.com.au/maps /api/js?sensor=false"></…