我在 mysql 数据库中有数据，我希望能够读取 php 数组中的数据并将其用于谷歌地理定位。
因此，对于这个示例，我只想使用 SELECT * 语句，因为我不会用一些参数来打扰您...
我想要实现的是获得从 A 点到 B 点的标记线(取决于 GPS 位置的保存位置)
这是我想要在 map 上显示的链接:POLYLINE在这个例子中，数组中只有 4 个数据，我想要更多。
现在我们可以回到代码了。这是我用于连接 mysql 的 PHP 脚本。我一直在使用 mysqli，因为稍后我将在数据库中存储过程，所以不要感到困惑。
class dbMySql
{
static function Exec($query) {
// open database
$conn = mysqli_connect(
$GLOBALS['cfg_db_Server'],
$GLOBALS['cfg_db_User'],
$GLOBALS['cfg_db_Pass']
);
if($conn === false)
{
throw new Exception(mysqli_connect_error());
}
mysqli_select_db($conn,$GLOBALS['cfg_db_Name']);
$result = mysqli_query($conn,$query);
if(is_bool($result) && !$result)
{
$error = mysqli_error($conn);
mysqli_close($conn);
throw new Exception($error);
}
mysqli_close($conn);
return $result;
}
}
如何将此 php 脚本连接到 google API 页面上的示例代码，并在单击按钮时插入我的数组值而不是固定值:
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no" />
<script type="text/javascript"
src="http://maps.googleapis.com/maps/api/js?key=KEY&sensor=true">
</script>
<script type="text/javascript">
function initialize() {
var myLatLng = new google.maps.LatLng(0, -180);
var myOptions = {
zoom: 3,
center: myLatLng,
mapTypeId: google.maps.MapTypeId.TERRAIN
};
var map = new google.maps.Map(document.getElementById("map_canvas"),
myOptions);
var flightPlanCoordinates = [
new google.maps.LatLng(37.772323, -122.214897),
new google.maps.LatLng(21.291982, -157.821856),
new google.maps.LatLng(-18.142599, 178.431),
new google.maps.LatLng(-27.46758, 153.027892)
];
var flightPath = new google.maps.Polyline({
path: flightPlanCoordinates,
strokeColor: "#FF0000",
strokeOpacity: 1.0,
strokeWeight: 2
});
flightPath.setMap(map);
}
</script>
</head>
<body onload="initialize()">
<div id="map_canvas" style="width:100%; height:80%"></div>
<div><button type="button">Click Me!</button></div>
</body>
</html>
编辑:
这是我执行你的代码时得到的结果:
编辑2:
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no" />
<style type="text/css">
html { height: 100% }
body { height: 100%; margin: 0; padding: 0 }
#map_canvas { height: 100% }
</style>
<script type="text/javascript"
src="http://maps.googleapis.com/maps/api/js?key=key&sensor=true">
</script>
<script type="text/javascript">
function initialize() {
var myLatLng = new google.maps.LatLng(0, 180);
var myOptions = {
zoom: 3,
center: myLatLng,
mapTypeId: google.maps.MapTypeId.TERRAIN
};
var map = new google.maps.Map(document.getElementById("map_canvas"),
myOptions);
var flightPlanCoordinates = [ <?php echo implode(',', $coordinates) ?> ];
];
var flightPath = new google.maps.Polyline({
path: flightPlanCoordinates,
strokeColor: "#FF0000",
strokeOpacity: 1.0,
strokeWeight: 2
});
flightPath.setMap(map);
}
</script>
</script>
<?PHP
class dbMySql {
static function Exec($query) {
// open database
$conn = mysqli_connect('localhost','root','*****');
if($conn === false) {
throw new Exception(mysqli_connect_error());
}
mysqli_select_db($conn,'data_gps');
$result = mysqli_query($conn,$query);
if(is_bool($result) && !$result) {
$error = mysqli_error($conn);
mysqli_close($conn);
throw new Exception($error);
}
mysqli_close($conn);
return $result;
}
}
$coordinates = array();
$result = dbMySql::Exec('SELECT lat,lng FROM data');
while ($row = mysqli_fetch_assoc($result))
$coordinates[] = 'new google.maps.LatLng(' . $row['lat'] . ', ' . $row['lng'] . ')' ;
?>
</script>
</head>
<body onload="initialize()">
<div id="map_canvas" style="width:100%; height:80%"></div>
<div><button type="button">Click Me!</button></div>
</body>
</html>
最佳答案
我猜您需要在结果之间进行迭代，然后将它们回显到 JavaScript 中。
我假设您的数据库中存储了纬度和经度。
$coordinates = array();
$result = dbMySql::Exec('SELECT lat, lng FROM table WHERE id = 1');
while ($row = mysqli_fetch_assoc($result))
$coordinates[] = 'new google.maps.LatLng(' . $row['lat'] . ', ' . $row['lng'] . ')';
然后在你的 javascript 部分
var flightPlanCoordinates = [ <?php echo implode(',', $coordinates) ?> ];
编辑:
<!DOCTYPE html>
<?PHP
class dbMySql {
static function Exec($query) {
// open database
$conn = mysqli_connect('localhost','root','*****');
if($conn === false) {
throw new Exception(mysqli_connect_error());
}
mysqli_select_db($conn,'data_gps');
$result = mysqli_query($conn,$query);
if(is_bool($result) && !$result) {
$error = mysqli_error($conn);
mysqli_close($conn);
throw new Exception($error);
}
//mysqli_close($conn);
return $result;
}
}
$coordinates = array();
$result = dbMySql::Exec('SELECT lat,lng FROM data');
while ($row = mysqli_fetch_assoc($result))
$coordinates[] = 'new google.maps.LatLng(' . $row['lat'] . ', ' . $row['lng'] . ')';
?>
<html>
<head>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no" />
<style type="text/css">
html { height: 100% }
body { height: 100%; margin: 0; padding: 0 }
#map_canvas { height: 100% }
</style>
<script type="text/javascript"
src="http://maps.googleapis.com/maps/api/js?key=key&sensor=true">
</script>
<script type="text/javascript">
function initialize() {
var myLatLng = new google.maps.LatLng(0, 180);
var myOptions = {
zoom: 3,
center: myLatLng,
mapTypeId: google.maps.MapTypeId.TERRAIN
};
var map = new google.maps.Map(document.getElementById("map_canvas"),
myOptions);
var flightPlanCoordinates = [<?php echo implode(',', $coordinates) ?>];
var flightPath = new google.maps.Polyline({
path: flightPlanCoordinates,
strokeColor: "#FF0000",
strokeOpacity: 1.0,
strokeWeight: 2
});
flightPath.setMap(map);
}
</script>
</head>
<body onload="initialize()">
<div id="map_canvas" style="width:100%; height:80%"></div>
<div><button type="button">Click Me!</button></div>
</body>
</html>
关于php - 如何将 php 数组从 mysql 返回到 google geolocation，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/10667262/