我有一个很简单的JS使用navigator。geolocation。getcurrentposition jammy。

$(document).ready(function(){
  $("#business-locate, #people-locate").click(function() {
    navigator.geolocation.getCurrentPosition(foundLocation, noLocation);
  });

  navigator.geolocation.getCurrentPosition(foundLocation, noLocation);

  function foundLocation(position) {
    var lat = position.coords.latitude;
    var lon = position.coords.longitude;
    var userLocation = lat + ', ' + lon;
    $("#business-current-location, #people-current-location").remove();
    $("#Near-Me")
      .watermark("Current Location")
      .after("<input type='hidden' name='business-current-location' id='business-current-location' value='"+userLocation+"' />");
    $("#people-Near-Me")
      .watermark("Current Location")
      .after("<input type='hidden' name='people-current-location' id='people-current-location' value='"+userLocation+"' />");
  }
  function noLocation() {
    $("#Near-Me").watermark("Could not find location");
    $("#people-Near-Me").watermark("Could not find location");
  }
})//end DocReady

基本上,这里发生的事情是我们得到当前位置,如果它得到了,两个“水印”被放置在两个字段中,表示“当前位置”,两个隐藏字段被创建,并使用latong数据作为它们的值(它们在开始时被删除,这样它们就不会每次都重复)。还有两个按钮,它们有一个点击功能,做同样的事情。 不幸的是,每隔三次左右,它就会奏效。 这里有什么问题?


当前回答

我终于找到了一个工作版本的firefox, chrome和默认导航器在android(仅4.2测试):

function getGeoLocation() {
        var options = null;
        if (navigator.geolocation) {
            if (browserChrome) //set this var looking for Chrome un user-agent header
                options={enableHighAccuracy: false, maximumAge: 15000, timeout: 30000};
            else
                options={maximumAge:Infinity, timeout:0};
            navigator.geolocation.getCurrentPosition(getGeoLocationCallback,
                    getGeoLocationErrorCallback,
                   options);
        }
    }

其他回答

我也遇到了同样的情况。我尝试了超时解决方案,但不可靠。我发现如果你只调用它两次,位置就会正确刷新

function getLocation(callback)
{   
    if(navigator.geolocation)
    {
        navigator.geolocation.getCurrentPosition(function(position)
        {
            navigator.geolocation.getCurrentPosition(callback, function(){},{maximumAge:0, timeout:10000});
        },function(){}, {maximumAge:0, timeout:10000});
    }
    return true;
}

这当然是有点慢,但我没有给我错误的位置一次。我有它击中超时几次,没有返回任何东西,但除此之外,它工作得很好。我知道这仍然有点老套,我期待有人找到真正的解决方案。

或者,如果你想确保它会一直尝试,直到你想放弃,你可以试试这样的方法。

//example
$(document).ready(function(){
    getLocation(function(position){
        //do something cool with position
        console.log(position);
    });
});


var GPSTimeout = 10; //init global var NOTE: I noticed that 10 gives me the quickest result but play around with this number to your own liking


//function to be called where you want the location with the callback(position)
function getLocation(callback)
{   
    if(navigator.geolocation)
    {
        var clickedTime = (new Date()).getTime(); //get the current time
        GPSTimeout = 10; //reset the timeout just in case you call it more then once
        ensurePosition(callback, clickedTime); //call recursive function to get position
    }
    return true;
}

//recursive position function
function ensurePosition(callback, timestamp)
{
    if(GPSTimeout < 6000)//set at what point you want to just give up
    {
        //call the geolocation function
        navigator.geolocation.getCurrentPosition(
            function(position) //on success
        {
                //if the timestamp that is returned minus the time that was set when called is greater then 0 the position is up to date
            if(position.timestamp - timestamp >= 0)
                {
                    GPSTimeout = 10; //reset timeout just in case
                    callback(position); //call the callback function you created
                }
                else //the gps that was returned is not current and needs to be refreshed
                {
                    GPSTimeout += GPSTimeout; //increase the timeout by itself n*2
                    ensurePosition(callback, timestamp); //call itself to refresh
                }
            },
            function() //error: gps failed so we will try again
            {
                GPSTimeout += GPSTimeout; //increase the timeout by itself n*2
                ensurePosition(callback, timestamp);//call itself to try again
            },
            {maximumAge:0, timeout:GPSTimeout}
        )
    }       
}

我可能在这里有一些打字和拼写错误,但我希望你能明白。如果有人有问题或者找到更好的建议,请告诉我。

@brennanyoung的回答很好,但如果你想知道在失败的情况下该怎么做,你可以使用IP地理定位API,如https://ipinfo.io(这是我的服务)。这里有一个例子:

function do_something(coords) {
    // Do something with the coords here
    // eg. show the user on a map, or customize
    // the site contents somehow
}

navigator.geolocation.getCurrentPosition(function(position) { 
    do_something(position.coords);
    },
    function(failure) {
        $.getJSON('https://ipinfo.io/geo', function(response) { 
        var loc = response.loc.split(',');
        var coords = {
            latitude: loc[0],
            longitude: loc[1]
        };
        do_something(coords);
        });  
    };
});

详情见https://ipinfo.io/developers/replacing-getcurrentposition。

这对我来说每次都管用:

navigator.geolocation.getCurrentPosition(getCoor, errorCoor, {maximumAge:60000, timeout:5000, enableHighAccuracy:true});

虽然它不是很准确。有趣的是,在同一台设备上,如果我运行这个程序,它会把我拖走大约100米(每次),但如果我去谷歌的地图,它会准确地找到我的位置。因此,尽管我认为enableHighAccuracy: true有助于它始终如一地工作,但它似乎并没有使它更准确……

我终于找到了一个工作版本的firefox, chrome和默认导航器在android(仅4.2测试):

function getGeoLocation() {
        var options = null;
        if (navigator.geolocation) {
            if (browserChrome) //set this var looking for Chrome un user-agent header
                options={enableHighAccuracy: false, maximumAge: 15000, timeout: 30000};
            else
                options={maximumAge:Infinity, timeout:0};
            navigator.geolocation.getCurrentPosition(getGeoLocationCallback,
                    getGeoLocationErrorCallback,
                   options);
        }
    }

我一直有类似的问题,并一直在研究浏览器对getCurrentPosition的调用频率有限制的可能性。似乎我可以经常得到一个位置,但如果我刷新页面马上就会超时。如果我等一会儿,我通常可以再次得到一个位置。这通常发生在FF上。在Chrome和Safari中,我还没有注意到getCurrentPosition超时。只是一个想法……

虽然我找不到任何文档来支持这一点,但这是我经过多次测试后得出的结论。也许其他人有这方面的信息?