Launch Zoom App from your App
Contents
1. Use Case2. Android3. 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:
Android
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) {startActivity(intent);}}
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) {startActivity(intent);}}
iOS
Use URL
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:
<key>LSApplicationQueriesSchemes</key><array><string>zoomus</string></array>
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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