點選連結跳到指定的app
阿新 • • 發佈:2018-11-26
1.首先在app裡面MIANACTIVITY追加下面的內容
<activity android:name=".MainNovelActivity" android:resizeableActivity="false" android:screenOrientation="portrait" android:theme="@style/AppTheme.Welcome"> <intent-filter> <action android:name="android.intent.action.MAIN"/> <category android:name="android.intent.category.LAUNCHER"/> </intent-filter> <intent-filter> <action android:name="android.intent.action.VIEW"/> <category android:name="android.intent.category.DEFAULT" /> <category android:name="android.intent.category.BROWSABLE" /> <data android:scheme="hpd"/> </intent-filter> </activity>
然後編輯指令碼html
<html> <head> <meta charset="utf-8"> </head> <body> <script> function isInstalled(){var the_href=$(".down_app").attr("href");//獲得下載連結 window.location="apps custom url schemes";//開啟某手機上的某個app應用 setTimeout(function(){ window.location=the_href;//如果超時就跳轉到app下載頁 },500); } function doCallApp() { window.location="hpd://bid/info?id=20170200019" } </script> <br> <a href="hpd://bid/info?id=20170200019">app link</a> <br> <br> <button id="callApp" onclick="doCallApp()">Call App<tton> </body> <html>
然後在MAINACTIVITY裡面設定跳轉到特定介面的方法
Intent i_getvalue = getIntent(); String action = i_getvalue.getAction(); if(Intent.ACTION_VIEW.equals(action)) { Uri uri = i_getvalue.getData(); if (uri != null) { String id = uri.getQueryParameter("id"); Log.d("TAg",id); Intent intent = new Intent(); intent.setClass(this, BorrowDetailsActivity.class); intent.putExtra(EXTRA_KEY_2, id); startActivity(intent); } }