在android浏览器中如果安装了app，则打开app，如果没有则跳转到应用市场下载。

这个方式也兼容ios。

先看html

<!doctype html><html><head>    <meta charset="utf-8">    <title>App Redirection</title></head><body><script>    var IS_IPAD = navigator.userAgent.match(/iPad/i) != null,            IS_IPHONE = !IS_IPAD && ((navigator.userAgent.match(/iPhone/i) != null) || (navigator.userAgent.match(/iPod/i) != null)),            IS_IOS = IS_IPAD || IS_IPHONE,            IS_ANDROID = !IS_IOS && navigator.userAgent.match(/android/i) != null,            IS_MOBILE = IS_IOS || IS_ANDROID;    function open() {        // If it's not an universal app, use IS_IPAD or IS_IPHONE        if (IS_IOS) {            window.location = "myapp://view?id=123";            setTimeout(function() {                // If the user is still here, open the App Store                if (!document.webkitHidden) {                    // Replace the Apple ID following '/id'                    window.location = 'http://itunes.apple.com/app/id1234567';                }            }, 25);        } else if (IS_ANDROID) {            // Instead of using the actual URL scheme, use 'intent://' for better UX            window.location = 'intent://view?xxxx=123#Intent;package=com.example.myapp;scheme=vip;launchFlags=268435456;end;';        }    }    open();</script></body></html>

android app中androidminifest.xml的activity的设置，具体看MyActivity

<?xml version="1.0" encoding="utf-8"?><manifest xmlns:android="http://schemas.android.com/apk/res/android"          package="com.example.myapp"          android:versionCode="1"          android:versionName="1.0">    <uses-sdk android:minSdkVersion="4" android:targetSdkVersion="19"/>    <application android:label="@string/app_name" android:icon="@drawable/ic_launcher" android:hardwareAccelerated="true">        <activity android:name=".MyActivity">            <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="vip" android:path="/view"  />            </intent-filter>        </activity>    </application>    <uses-permission android:name="android.permission.INTERNET"/></manifest>

在activity获取传入的参数xxxx的值，需要解析Uri方式来获取

package com.example.myapp;import org.apache.http.protocol.UriPatternMatcher;import java.net.URI;import java.net.URISyntaxException;public class MyActivity extends Activity {    @Override    public void onCreate(Bundle savedInstanceState) {        super.onCreate(savedInstanceState);        setContentView(R.layout.main);        String s = getIntent().toURI();        Uri parse = Uri.parse(s);        String xxxx = parse.getQueryParameter("xxxx");        Log.d("MyActivity", xxxx);    }    @Override    protected void onNewIntent(Intent intent) {        super.onNewIntent(intent);        Log.d("MyActivity", "xx" + getIntent().toURI());    }}

Reference：https://gist.github.com/FokkeZB/6635236 我使用的是这篇文章中js的实现方式，这篇文章也有后端php的实现方式。

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