【发布时间】:2020-01-11 23:09:35
【问题描述】:
因此,我需要将 laravel 中的两个变量($event->latitude 和 $event->longitude)传递给 Javascript,因为我需要将此值用于 Google 地图。这些是 2 个值,例如... 纬度:44.4590762 和经度:26.0967384000000 .
那么,我想做什么......
var mylat = {!! json_encode($event->latitude->toArray()) !!};
var mylng = {!! json_encode($event->longitude->toArray()) !!};
or
var mylat = {!! $event->latitude !!}
var mylng = {!! $event->longitude !!}
但没有结果。我的变量仍然没有值。
这是我的事件控制器:
$event->longitude = $request['longitude'];
$event->latitude = $request['latitude'];
我的看法:
<div id="map"></div>
我的脚本:
<script>
var mylat = {!! json_encode($event->latitude->toArray()) !!};
var mylng = {!! json_encode($event->longitude->toArray()) !!};
function initMap() {
var map = new google.maps.Map(document.getElementById('map'), {
center: {lat: mylat, lng: mylng},
zoom: 18
});
var input = document.getElementById('searchInput');
map.controls[google.maps.ControlPosition.TOP_LEFT].push(input);
var autocomplete = new google.maps.places.Autocomplete(input);
autocomplete.bindTo('bounds', map);
var infowindow = new google.maps.InfoWindow();
var marker = new google.maps.Marker({
map: map,
anchorPoint: new google.maps.Point(0, -29)
});
autocomplete.addListener('place_changed', function() {
infowindow.close();
marker.setVisible(false);
var place = autocomplete.getPlace();
if (!place.geometry) {
window.alert("Autocomplete's returned place contains no geometry");
return;
}
// If the place has a geometry, then present it on a map.
if (place.geometry.viewport) {
map.fitBounds(place.geometry.viewport);
} else {
map.setCenter(place.geometry.location);
map.setZoom(17);
}
marker.setIcon(({
url: place.icon,
size: new google.maps.Size(71, 71),
origin: new google.maps.Point(0, 0),
anchor: new google.maps.Point(17, 34),
scaledSize: new google.maps.Size(35, 35)
}));
marker.setPosition(place.geometry.location);
marker.setVisible(true);
var address = '';
if (place.address_components) {
address = [
(place.address_components[0] && place.address_components[0].short_name || ''),
(place.address_components[1] && place.address_components[1].short_name || ''),
(place.address_components[2] && place.address_components[2].short_name || '')
].join(' ');
}
infowindow.setContent('<div><strong>' + place.name + '</strong><br>' + address);
infowindow.open(map, marker);
//Location details
for (var i = 0; i < place.address_components.length; i++) {
if(place.address_components[i].types[0] == 'postal_code'){
document.getElementById('postal_code').innerHTML = place.address_components[i].long_name;
}
if(place.address_components[i].types[0] == 'country'){
document.getElementById('country').innerHTML = place.address_components[i].long_name;
}
}
document.getElementById('location').innerHTML = place.formatted_address;
document.getElementById('lat').innerHTML = place.geometry.location.lat();
document.getElementById('lon').innerHTML = place.geometry.location.lng();
document.getElementById('location').innerHTML = place.geometry.location.location();
});
}
</script>
【问题讨论】:
-
"but doesn't working" 不是错误描述。请阅读How to Ask 并编辑您的问题。
-
实际生成的HTML/JS代码是什么?
-
看起来你的文件根本没有被 PHP 解析为 Laravel 模板。您最近的编辑(“我的脚本”)是浏览器收到的内容?
-
根据laravel.com/docs/5.8/blade,您应该使用
@命令或纯PHP 来执行表达式。请参阅“渲染 JSON” -
检查这个答案,它会帮助你:stackoverflow.com/questions/30074107/…
标签: javascript php laravel