本文介绍了无法将变量作为地址传递到谷歌地图MVC 4的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
am trying to load google map by passing city name or address.
1 ) Map not displaying
2 ) How to load google map passing address or city name
3 ) How to load different location on mouse-over() that pass dynamic address
<script src="~/Scripts/jquery-2.1.4.min.js"></script>
<script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"></script>
<script type="text/javascript">
function codeAddress(address) {
geocoder = new google.maps.Geocoder();
geocoder.geocode({
'address': address
}, function (results, status) {
if (status == google.maps.GeocoderStatus.OK) {
var myOptions = {
zoom: 8,
center: results[0].geometry.location,
mapTypeId: google.maps.MapTypeId.ROADMAP
}
map = new google.maps.Map(document.getElementById("map_canvas"), myOptions);
var marker = new google.maps.Marker({
map: map,
position: results[0].geometry.location
});
}
});
}
</script>
<script type="text/javascript">
$(".tiptext").mouseover(function () {
// debugger;
$(this).children(".description").show();
codeAddress(address);
}).mouseout(function () {
$(this).children(".description").hide();
});
</script>
CSS:
#map-canvas {
width: 50%;
height: 50%;
margin: 0;
padding:0px;
}
View:
<div id="map_canvas">
<div id="address">
@item.address
@item.city
</div>
</div>
推荐答案
CSS:
#map-canvas {
width: 50%;
height: 50%;
margin: 0;
padding:0px;
}
View:
<div id="map_canvas">
<div id="address">
@item.address
@item.city
</div>
</div>
这篇关于无法将变量作为地址传递到谷歌地图MVC 4的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!