Question

我想在Firebase中创建一个动态链接，该链接将在模拟器上打开Expo Client应用程序。我快到了... 目前，我正尝试在android模拟器上打开它，并避免尝试在ios模拟器上执行此操作，因为...好吧... Apple ...（尽管我在Mac上工作，但我更喜欢它）在Firebase中，我创建了一个动态链接，可以看到该链接首先在浏览器中打开，然后尝试打开设备，最后打开Android Play商店。因此，看来它需要在模拟器上打开应用程序时失败。

我的博览会app.json：

{
  "expo": {
    "name": "happy",
    "slug": "happy",
    "version": "1.0.0",
    "orientation": "portrait",
    "icon": "./assets/images/icon.png",
    "scheme": "happy",
    "userInterfaceStyle": "automatic",
    "splash": {
      "image": "./assets/images/splash.png",
      "resizeMode": "contain",
      "backgroundColor": "#ffffff"
    },
    "updates": {
      "fallbackToCacheTimeout": 0
    },
    "assetBundlePatterns": ["**/*"],
    "ios": {
      "supportsTablet": true,
      "bundleIdentifier": "com.happy.happyapp",
      "buildNumber": "1.0.0",
      "googleServicesFile": "./firebase/GoogleService-Info.plist"
    },
    "android": {
      "package": "com.happy.happyapp",
      "versionCode": 1,
      "googleServicesFile": "./firebase/google-services.json",
      "intentFilters": [
        {
          "action": "VIEW",
          "data": [
            {
              "scheme": "https",
              "host": "*.happy.com",
              "pathPrefix": "/"
            }
          ],
          "category": ["BROWSABLE", "DEFAULT"]
        }
      ]
    },
    "web": {
      "favicon": "./assets/images/favicon.png",
      "config": {
        "firebase": {
          ...
        }
      }
    }
  }
}

在Firebase中，我有：

Link name
Happy app test
Deep link
https://happy.com
Android app
com.happy.happyapp
iOS app
Long Dynamic Link
https://happy.page.link/?link=https://happy.com&apn=com.happy.happyapp
Short Dynamic Link
https://happy.page.link/go

模拟器中的Firebase深度链接打开EXPO应用程序

0 个答案: