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。

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。

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