Launch Zoom App from your App


1. Use Case
2. Android
3. iOS

Use Case

Normally, you would embed a meeting invitation URL in your app which when clicked by a user, launches the Zoom app and lets the user join the meeting.

It is also possible to allow users to launch just the Zoom app from your app by pressing a link or a button without specifying a meeting parameter.

Here are the ways to achieve that on either Android or iOS:


For Android apps, you can either use the Package ID or the URL to enable the launch feature.

Method 1: Use Package ID

Use Android's built-in PackageManager to launch the Zoom App using this package id:us.zoom.videomeetings. This action includes a step that searches the app that matches the package ID in the system, so you might notice latency of about one second while executing.

private void launchZoomClient() {
PackageManager pm = getPackageManager();
Intent intent = pm.getLaunchIntentForPackage("us.zoom.videomeetings");
if (intent != null) {

Method 2: Use URL

Using the URL is faster as the URL scheme of our Zoom SDK has been registered in the system which enables deep links. Simply, parse the URL zoomus:// and launch it.

private void launchZoomUrl() {
Intent intent = new Intent(Intent.ACTION_VIEW, Uri.parse("zoomus://"));
if (intent.resolveActivity(getPackageManager()) != null) {



Parse the URL zoomus:// and check whether it is openable or not and then launch it. In order to be able to launch the external app on iOS ( in this case, the Zoom app), you should add LSApplicationQueriesSchemes scheme as a key in the info.plist file:


Note: If the key is missing from your app's .plist file, the following error will occur: This app is not allowed to query for scheme.

- (void)onZoomClient
NSString *Uri = @"zoomus://";
if ([[UIApplication sharedApplication] canOpenURL:[NSURL URLWithString:Uri]])
if (@available(iOS 10.0, *)) { // If the system OS is iOS 10.0 and above
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:Uri] options:@{} completionHandler:nil];
} else {
[[UIApplication sharedApplication] openURL:[NSURL URLWithString:Uri]];

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