Google街景应用的网址方案(不是Google地图) [英] URL scheme for the Google Street View App (Not Google Maps)

问题描述

我正在将Google Cardboard与Street View应用程序配合使用.我希望能够创建一个链接，将用户直接发送到街景应用中以查看特定位置.

I'm using Google Cardboard with the Street View App. I want to be able to create a link that sends users directly into the Street View App to view a specific location.

在Street View应用程序中，我可以从某个位置创建链接，如果它是应用程序中的重要位置，则它生成的链接可以完美地工作，并且看起来像这样: http://www.google.com/maps/streetview/#us-highlights/法尼尔·霍尔·波士顿

Within the Street View App I can create a link from a location, if it's a featured location within the app, the link it generates works perfectly and looks something like this: http://www.google.com/maps/streetview/#us-highlights/faneuil-hall-boston

如果它不是精选位置，则它生成的URL是不同的，并且会在google地图(而不是街景应用)中打开，看起来像这样: https://www.google.com/maps/@/data=！ 3m4！1e1！3m2！1s7lx0Oz8OSTwIweRXw7eNiA！2e0

If it's not a featured location, the URL it generates is different and opens in google maps, rather than the Street View app, looking something like this: https://www.google.com/maps/@/data=!3m4!1e1!3m2!1s7lx0Oz8OSTwIweRXw7eNiA!2e0

是否可以创建一个在街景视图"应用中打开的URL，而该URL不是重要位置.如果是这样，处理/格式是什么?

推荐答案

通过将 mapmode 参数设置为街景.

目标c

 if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:@"comgooglemaps://"]]) {

       [[UIApplication sharedApplication] openURL:[NSURL URLWithString:[NSString stringWithFormat:@"comgooglemaps://?center=46.414382,10.013988&mapmode=streetview"]]];    
  }

快捷键4

if UIApplication.shared.canOpenURL(URL(string: "comgooglemaps://")!) {


   guard let url = URL(string: "comgooglemaps://?center=46.414382,10.013988&mapmode=streetview") else {
            return //be safe
        }

    if #available(iOS 10.0, *) {
        UIApplication.shared.open(url, options: [:], completionHandler: nil)
    } else {
        UIApplication.shared.openURL(url)
    }

 }

&有关更多详细信息，请参见 Google Streetview .

& more details here on Google Streetview.

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