How Android jumps from the outside into the app

Blog from: http://blog.csdn.net/liuxian13183, reprint annotated source! All Rights Reserved!

This problem solved two days, because there is no complete solution on the net, after the solution to share to everyone!

There are two functions to solve this problem:

1, do not open the app directly into a page

2, realize the app to share the external, while the external access to the app's closed loop.

This topic can be divided into two aspects, on the one hand from access to the app, on the other hand, from the Web page into the app.

let's take http://my.oschina.net/liucundong/blog/354029 as an example.

mainly through scheme, host, PATH, Pathprefix and other data information to differentiate

Two programs need to add intent-filter information to the activity that jumps to, for example

<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= "Cundong" android:host= "Splash"/>

</intent-filter>

First, let's go to the app, scene: Share content from the app, and then open the app directly.

The first step, registered public platform, the relevant enterprise information, download links, logos, etc. are completed

The second step, share the use of Wxappextendobject, the link URL to assign value to extinfo

The third step, when the Onreq method of the Iwxapieventhandler interface is Req.gettype () =constantsapi.command_showmessage_from_wx, initiates your jump

The fourth step, the need for data to jump, where the data from, from the Wxentryactivity class OnCreate method Getintent Object

This is the end of the app entry.

-------------------I'm a split line------------------------

Then the web-side, due to the limitations of webview, such as Weibo, such as the use of this control, so can only share the Web page, and then into the app.

The principle is as follows:

<div>
<a id= "J-call-app" href= "javascript:;" class= "label" > Open now &gt;&gt;</a>
<input id= "J-download-app" type= "hidden" name= "Storeurl" value= "http://apk.hiapk.com/appinfo/com.tmall.wireless ">
</div>

<script>
(function () {
var ua = Navigator.userAgent.toLowerCase ();

var T;
var url=document.location;
var config = {
/*scheme: Must */
Scheme_ios: ' cundong://',
Scheme_adr: ' Cundong://splash ',
Download_url:document.getElementById (' J-download-app '). Value,
timeout:600
};

function Openclient () {
var startTime = Date.now ();

var IFR = document.createelement (' iframe ');

alert (CONFIG.SCHEME_ADR);
IFR.SRC = Ua.indexof (' os ') > 0? CONFIG.SCHEME_IOS:CONFIG.SCHEME_ADR;
Ifr.style.display = ' None ';
Document.body.appendChild (IFR);

var t = setTimeout (function () {
var endTime = Date.now ();

if (!starttime | | Endtime-starttime < config.timeout-200) {
window.location = Config.download_url;
} else {
Window.location.href=ifr.src.replace ("http://", "cundong://");;
}
}, Config.timeout);



Window.onblur = function () {
Cleartimeout (t);
}
}
Window.addeventlistener ("domcontentloaded", function () {
document.getElementById ("J-call-app"). AddEventListener (' click ', Openclient,false);

}, False);
})()
</script>

Click to open four words, the execution of a JS, within the time limit, if you can open the app is open, otherwise go to the download page

The basic idea to solve this problem is this, welcome to exchange discussion!

How Android jumps from the outside into the app