【发布时间】:2015-01-28 21:36:09
【问题描述】:
我正在使用谷歌地图 API 来显示用户在地图上的位置。由于 initMap() 触发 onload 事件,用户在页面加载时会看到谷歌地图的一部分。但我想要的是,只显示地图 onclick sub_map 按钮。
为了得到想要的结果,这是我尝试过的,但没有成功。我在 onclick 的 sub_map 按钮之后调用了 initMap()
<form ><input type='submit' id='sub_map' name='sub_map' onclick='open_map(".$userx.");' value='See Map'></form>";
Javascript
function open_map(x){
var user_map=x;
initMap();
var ajax=new XMLHttpRequest();
ajax.open("post","search_results.php",true);
ajax.setRequestHeader("content-type","application/x-www-form-urlencoded");
ajax.onreadystatechange=function(){
if(ajax.readyState == 4 && ajax.status == 200){
document.getElementById('map').innerHTML=ajax.responseText;
}
}
ajax.send("user_map="+user_map);
}
谷歌地图部分
var icon = new google.maps.MarkerImage("http://maps.google.com/mapfiles/ms/micons/blue.png",
new google.maps.Size(40, 40), new google.maps.Point(0,0),
new google.maps.Point(16, 32));
var center = null;
var map = null;
var currentPopup;
var bounds = new google.maps.LatLngBounds();
function addMarker(lat, lng, info) {
var pt = new google.maps.LatLng(lat, lng);
bounds.extend(pt);
var marker = new google.maps.Marker({
position: pt,
icon: icon,
map: map
});
var popup = new google.maps.InfoWindow({
content: info,
maxWidth: 300
});
google.maps.event.addListener(marker, "click", function() {
if (currentPopup != null) {
currentPopup.close();
currentPopup = null;
}
popup.open(map, marker);
currentPopup = popup;
});
google.maps.event.addListener(popup, "closeclick", function() {
map.panTo(center);
currentPopup = null;
});
}
function initMap() {
map = new google.maps.Map(document.getElementById("map"), {
//map.panBy(0,30);
//map.setZoom(15);
center: new google.maps.LatLng(0,0),
zoom: 1,
mapTypeId: google.maps.MapTypeId.ROADMAP,
mapTypeControl: false,
mapTypeControlOptions: {
style: google.maps.MapTypeControlStyle.HORIZONTAL_BAR
},
navigationControl: true,
navigationControlOptions: {
style: google.maps.NavigationControlStyle.SMALL
}
});
<?php
if(isset($_POST['user_map'])){
$mapusrx=$_POST['user_map'];
$query = mysqli_query($connecti,"SELECT * FROM map WHERE user_id='$mapusrx'");
while ($row = mysqli_fetch_assoc($query)){
$name=$row['title'];
$lat=$row['lat'];
$lon=$row['lon'];
$desc=$row['address'];
echo ("addMarker($lat, $lon,'<b>$name</b><br/>$desc');\n");
}
}
?>
center = bounds.getCenter();
map.setCenter(center);
map.setZoom(16);
}
感谢您的宝贵时间。
【问题讨论】:
-
那么 search_results.php,Google 地图部分的返回结果是什么?
-
没什么,它根本没有给我显示地图
标签: javascript mysql google-maps google-maps-api-3