location

When using GPX in Xcode to simulate location changes, is there a way to control the speed?

由 百般思念 提交于 2019-12-17 08:28:06
问题 I'm using the following GPX file in Xcode 4.2 to simulate a location change. It works well, but I can't control the speed of the location change. stamp seems to be not working. Does anyone have a solution for this? <?xml version="1.0"?> <gpx version="1.1" creator="Xcode"> <wpt lat="37.331705" lon="-122.030237"></wpt> <wpt lat="37.331705" lon="-122.030337"></wpt> <wpt lat="37.331705" lon="-122.030437"></wpt> <wpt lat="37.331705" lon="-122.030537"></wpt> </gpx> 回答1: I don't think (know) that

Algorithm to find all Latitude Longitude locations within a certain distance from a given Lat Lng location

由 99封情书 提交于 2019-12-17 08:02:02
问题 Given a database of places with Latitude + Longitude locations, such as 40.8120390, -73.4889650, how would I find all locations within a given distance of a specific location? It doesn't seem very efficient to select all locations from the DB and then go through them one by one, getting the distance from the starting location to see if they are within the specified distance. Is there a good way to narrow down the initially selected locations from the DB? Once I have (or don't?) a narrowed

How to detect country / location of visitor? [duplicate]

由 混江龙づ霸主 提交于 2019-12-17 05:50:33
问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: Location detecting techniques for IP addresses For our website it's important to know from exactly which country our visitor is coming from. I guess the best answer for my question would be the simple geo location feature of current browser systems, which just ask the user if the website is allowed to see his geographic location: http://dev.w3.org/geo/api/spec-source.html but I don't want to bother visitor with

Android O - Old start foreground service still working?

由 孤街醉人 提交于 2019-12-17 04:28:39
问题 So, with Android O, you need to have your service running as a foreground service if you want to receive more than just a few location updates per hour. I noticed that the old method for starting a foreground service does seem to work on O. i.e. startForeground(NOTIFICATION_ID, getNotification()); According to the behaviour changes guide here: https://developer.android.com/preview/behavior-changes.html The NotificationManager.startServiceInForeground() method starts a foreground service. The

FusedLocationApi.getLastLocation always null

由 笑着哭i 提交于 2019-12-17 03:35:10
问题 I'd like to simply retrieve device location in my Android project and in order to do so I use the play-services approach: protected synchronized void buildGoogleApiClient() { mGoogleApiClient = new GoogleApiClient.Builder( MainSearchActivity.this ) .addConnectionCallbacks(new GoogleApiClient.ConnectionCallbacks() { @Override public void onConnected( Bundle bundle ){ Location location = LocationServices.FusedLocationApi.getLastLocation(mGoogleApiClient); if( location == null ){

jquery: change the URL address without redirecting? [duplicate]

由 只愿长相守 提交于 2019-12-17 02:42:22
问题 This question already has answers here : Closed 6 years ago . Possible Duplicate: Modify Address Bar URL in AJAX App to Match Current State How can I change the URL address without redirecting the page? For instance, when I click on this link below: <a href="http://mysite.com/projects/article-1" class="get-article">link</a> I will grab the URL from the link: var path = object.attr('href'); If I do this below, the page will be redirected: window.location = path; I want to do something like

jquery: change the URL address without redirecting? [duplicate]

由 会有一股神秘感。 提交于 2019-12-17 02:41:57
问题 This question already has answers here : Closed 6 years ago . Possible Duplicate: Modify Address Bar URL in AJAX App to Match Current State How can I change the URL address without redirecting the page? For instance, when I click on this link below: <a href="http://mysite.com/projects/article-1" class="get-article">link</a> I will grab the URL from the link: var path = object.attr('href'); If I do this below, the page will be redirected: window.location = path; I want to do something like

Distance calculation from my location to destination location in android

由 二次信任 提交于 2019-12-17 02:38:10
问题 I have to made a project where i need to calculate the distance from my location to destination location and show it in a textview.Here note that this distance updated when my location change.Is it possible to make such type of project ? [NOTE: without implementing Google map i have to make it.The destination lon ,lat is known.Just have to find my location and make the calculation] 回答1: check the documentation on the google android dev page to see how to listen for position changes. http:/

getLastKnownLocation returns null

由 微笑、不失礼 提交于 2019-12-17 02:11:34
问题 I've read some questions about this, but I didn't quite find an answer that I needed. So, the case is that I have my map set up, and I want to grab my current gps location. I have checked that my variables is not NULL but yet my result from: getLastKnownLocation(provider, false); Gives me null though, so this is where I need help. I have added permissions for COARSE + FINE location. But I usually have all kinds of network data disabled for my phone, because I'm not very happy about

Nginx location规则匹配

由 大城市里の小女人 提交于 2019-12-17 02:04:57
Nginx location规则匹配 1. “= ”，字面精确匹配， 如果匹配，则跳出匹配过程。（不再进行正则匹配） 2. “^~ ”，最大前缀匹配，如果匹配，则跳出匹配过程。（不再进行正则匹配） 3. / 不带任何前缀：最大前缀匹配，举例如下： location / 代表以"/"开头的搜索匹配， 再没有正则表达式匹配的情况下才进行这个匹配（优先级最低） 4. “~ ”，大小写相关的正则匹配 5. “~* ” ， 大小写无关的正则匹配 6. “@”， Named location 不是普通的location匹配，而是用于location内部重定向的变量。 Location @apache 其中： 1、2、3 三种情况属于 location using literal string， 即使用普通字符串的location匹配； 4、5二种情况属于 location using regular expresstion，即使用正则表达式的location匹配； location 匹配的优先级(与location在配置文件中的顺序无关) = 精确匹配会第一个被处理。如果发现精确匹配，nginx停止搜索其他匹配。 普通字符匹配，正则表达式规则和长的块规则将被优先和查询匹配，也就是说如果该项匹配还需去看有没有正则表达式匹配和更长的匹配。 ^~ 则只匹配该规则，nginx停止搜索其他匹配
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