Question

• I run my application “A” normally with the launch icon in the “Applications” menu.
• I press the home button, so my application “A” is still running on my phone.
• Now I browse my files presents in my phone, and I use the share action to share this file with my application “A”.
• The file is sharing perfectly, but instead of using the instance of my application already opened, it opens a new instance of my application “A”.

If I quit this new instance, the first instance is still running and it’s a problem concerning the security goal of my application.

I try to use the FLAG_ACTIVITY_CLEAR_TOP to use the activity in the first instance, but it doesn’t work because it isn’t the same application which is launched by the OS.

Is there a way to do this ? And if yes can you give me some hints or some leads to follow ?

---

My manifest :

<?xml version="1.0" encoding="utf-8"?>

<uses-sdk
    android:minSdkVersion="8"
    android:targetSdkVersion="8" />

<uses-permission android:name="android.permission.INTERNET" />
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
<uses-permission android:name="android.permission.ACCESS_NETWORK_STATE" />

<application
    android:icon="@drawable/ic_launcher"
    android:label="@string/app_name"
    android:theme="@android:style/Theme.NoTitleBar" >
    <activity android:name=".SplashScreenActivity" >
        <intent-filter>
            <action android:name="android.intent.action.MAIN" />

            <category android:name="android.intent.category.LAUNCHER" />
        </intent-filter>
    </activity>
    <activity
        android:name=".ExplorerActivity"
        android:configChanges="orientation|keyboardHidden"
        android:theme="@android:style/Theme.Light.NoTitleBar" >
    </activity>
    <activity
        android:name=".ChooseDialogActivity"
        android:theme="@android:style/Theme.Translucent.NoTitleBar" >
        <intent-filter>
            <action android:name="android.intent.action.SEND" />

            <category android:name="android.intent.category.DEFAULT" />

            <data android:mimeType="*/*" />
        </intent-filter>
        <intent-filter>
            <action android:name="android.intent.action.SEND_MULTIPLE" />

            <category android:name="android.intent.category.DEFAULT" />

            <data android:mimeType="*/*" />
        </intent-filter>
    </activity>
</application>

Here is my manifest. Usual process : SplashScreenActivity -> ExplorerActivity

Share process : ChooseDialogActivity -> ExplorerActivity

What I want, it’s that the second ExplorerActivity has to be the same that the first ExplorerActivity if this activity already exists.

Answer 1

Please don’t use launchMode=”singleInstance”. This is not what you want. singleInstance is only for HOME-screens and similar apps.

Try instead to use Intent.FLAG_ACTIVITY_NEW_TASK when sharing. This will separate your application from the files-browsing application and may get the behaviour you want. You may also need to add FLAG_ACTIVIY_CLEAR_TOP as well, depending on how you’ve programmed your app.

EDIT

When you launch ExplorerActivity from ChooserActivity, do this (or something similar):

Intent intent = new Intent(this, ChooserActivity.class);
intent.addFlags(Intent.ACTIVITY_NEW_TASK | Intent.FLAG_ACTIVITY_CLEAR_TOP);
startActivity(intent);

If there is already a task running that contains the ChooserActivity, this should bring that task to the foreground instead of creating a new instace of ChooserActivity.