javascript - 使用从 W3C 地理定位返回的位置

Question

FindMyLocation 定义如下:

function FindMyLocation(callback) {
if (navigator.geolocation) {

    var locationMarker = null;
    if (locationMarker) {
        return;
    }
    navigator.geolocation.getCurrentPosition(
        function(position){
           pos = new google.maps.LatLng(position.coords.latitude,position.coords.longitude);
           map.setCenter(new google.maps.LatLng(position.coords.latitude,position.coords.longitude));
           addMarker(
            position.coords.latitude,
            position.coords.longitude,
            "You are here"
            );
            setTimeout(function(window) {
                callback.call(pos);
            }, 1000);
        },
        function (geoPositionError) {
        switch (geoPositionError.code) {
            case 0:
                alert('An unknown error occurred, sorry');
                break;
            case 1:
                alert('Permission to use Geolocation was denied');
                break;
            case 2:
                alert('Couldn\'t find you...');
                break;
            case 3:
                alert('The Geolocation request took too long and timed out');
                break;
            default:
        }
    },
     timeoutCallback, 10, {maximumAge:60000, timeout:50000, enableHighAccuracy:true}
    );

var positionTimer = navigator.geolocation.watchPosition(
    function( position ){
        updateMarker(
            locationMarker,
            position.coords.latitude,
            position.coords.longitude,
            "Updated / Accurate Position"
        );
    }
);
setTimeout(
    function(){
        // Clear the position watcher.
        navigator.geolocation.clearWatch( positionTimer );
    },
    (1000 * 60 * 5)
);

} else {
    alert("Your phone browser doesn't support geolocation.");
}
}
function timeoutCallback(){
   alert('Hi there! we are trying to locate you but you have not answered the security    question yet.\n\nPlease choose "Share My Location" to enable us to find you.');
 }

我在我的主页中运行了以下代码:

FindMyLocation(function() {
 myPos = this;
});

当我在回调中 console.log myPos 时，我看到了该值。但是，我希望能够在此回调之外使用 myPos 来进行距离计算。现在，它不起作用。

请帮忙。

Answer 1

                    <script lang="JavaScript" type="text/javascript">
                        $.post(window.baseUrl + 'includes/tagmap.php', 
                              { name:"<?php echo $value->business_name; ?>",
                                x: "<?php echo $value->x; ?>", 
                                y: "<?php echo $value->y; ?>",
                                action:"populate"}, function(data){
                                    var locdata = $.parseJSON(data);        
                                    function ShowMyLocation() {
                                        FindMyLocation(function(myPos) {
                                            myPos = this;                                                                                                                                     addMarker(locdata.x,locdata.y,locdata.name);
                                        })
                                        FindMyLocation();
                                    }
                                    ShowMyLocation();                                  
                              });                                  
                    </script>