点击链接跳到指定的app

1.首先在app里面MIANACTIVITY追加下面的内容

    

<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>
 
 
然后编辑脚本html
<html>
<head>
 <meta charset="utf-8">
</head>
<body>
  <script>
     function isInstalled(){
		var the_href=$(".down_app").attr("href");//获得下载链接
		window.location="apps custom url schemes";//打开某手机上的某个app应用
		setTimeout(function(){
			window.location=the_href;//如果超时就跳转到app下载页
		},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>
然后在MAINACTIVITY里面设置跳转到特定界面的方法
 
 
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);

    }
}

 
 


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转载自blog.csdn.net/dhl006009/article/details/54927885
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