Html 地理定位 HTML5 enableHighAccuracy True ， False 还是最佳选项？

Geolocation HTML5 enableHighAccuracy True , False or Best Option?

提问by Erhan H.

i have a problem about HTML5 geolocation feature. I use the code below to get location data. I use "enableHighAccuracy: false" option to work with Cell Based GPS feature. Accurancy is low but response it too fast. But some people always use Built-in GPS with their mobile phone, so this code does not work for them. Bu if i change accurency option as "enableHighAccuracy: true" it works for them. But this time, the code uses only built-in GPS. not CELL based GPS.

我有一个关于 HTML5 地理定位功能的问题。我使用下面的代码来获取位置数据。我使用“enableHighAccuracy: false”选项来处理基于蜂窝的 GPS 功能。准确率低，但反应太快。但是有些人总是在他们的手机上使用内置 GPS，所以这个代码对他们不起作用。但是，如果我将精度选项更改为“enableHighAccuracy：true”，则它对它们有效。但这一次，代码仅使用内置 GPS。不是基于 CELL 的 GPS。

The question -> How can i do that : First, try to get position from Built-in GPS with timeout (e.g. 5000ms ) if position cannot be got in this time just look for Cell Based position for timeout (e.g. 10000ms) if position cannot be get in this time, return an error message .

问题 -> 我该怎么做：首先，如果此时无法获得位置，请尝试从内置 GPS 获取位置超时（例如 5000 毫秒），如果位置不能，则查找基于单元的位置超时（例如 10000 毫秒）进入这个时候，返回一个错误信息。

Here is the code that i use now.

这是我现在使用的代码。

Thanks in advance.

提前致谢。

    function getLocationfromGoogle() {
    navigator.geolocation.getCurrentPosition(
      function(pos) {
        $("#lat_field").val(pos.coords.latitude);
        $("#long_field").val(pos.coords.longitude);
        var geocoder = new google.maps.Geocoder();
        var latLng = new google.maps.LatLng(pos.coords.latitude,pos.coords.longitude);
              geocoder.geocode({ 'latLng': latLng}, function (results, status) {
              if (status == google.maps.GeocoderStatus.OK) {
                //console.log(results[0].formatted_address);
                $("#adresim").val(results[0].formatted_address);
                }
                else {
            alert('Google convertion is not succesfully done.');  

            }
            });
      },function error(msg){

                alert('Please enable your GPS position future.');  

      },{maximumAge:600000, timeout:5000, enableHighAccuracy: false}

    ); 
    }

回答by Kevin

You should also be aware that the implementation of this varies from phone OS to phone OS - what works on Android may or may not work on iOS, BlackBerry, WindowsPhone, etc.

您还应该知道，此操作的实现因手机操作系统而异 - 在 Android 上运行的内容可能在 iOS、黑莓、WindowsPhone 等上运行，也可能不运行。

You're almost there, you just need to:

你快到了，你只需要：

1. Specify enableHighAccuracy: true(you have it set to false)
2. Handle the timeout error case in the error handler. If the error from the high accuracy query is timeout, then try it again with enableHighAccuracy: false.

1. 指定enableHighAccuracy: true（您已将其设置为false）
2. 在错误处理程序中处理超时错误情况。如果高精度查询的错误是超时，则使用enableHighAccuracy: false.

Have a look at this sample code.

看看这个示例代码。

You should also note that when testing this on a few devices, it returns location derived from WiFi even when enableHighAccuracy: true.

您还应该注意，在一些设备上进行测试时，即使enableHighAccuracy: true.

回答by peawormsworth

The code mentioned here: http://jsfiddle.net/CvSW4/did not work for me during error handling.

这里提到的代码：http: //jsfiddle.net/CvSW4/在错误处理期间对我不起作用。

The reason is that the error functions accept a parameter named 'position' but use an object in the functions called 'error'.

原因是错误函数接受名为“position”的参数，但在函数中使用名为“error”的对象。

function errorCallback_highAccuracy(position) { ... }
function errorCallback_lowAccuracy(position) { ... }

The solution to fix this was to switch the error methods to accept the input value as a parameter named 'error' and not 'position', since the error callbacks do not accept a position and throw an error object instead.

解决此问题的解决方案是切换错误方法以接受输入值作为名为“错误”而不是“位置”的参数，因为错误回调不接受位置并抛出错误对象。

function errorCallback_highAccuracy(error) { ... }
function errorCallback_lowAccuracy(error) { ... }

I mention it here, because I could not post on the resulting example page and also, this is the location where I linked through to find the code sample mentioned above.

我在这里提到它，因为我无法在生成的示例页面上发布，而且，这是我通过链接找到上面提到的代码示例的位置。