Open AnshuBansal6 opened 4 months ago
While sharing on safari browser, we have default popup as blocked. Now there is no method returning that popup is blocked and that is the reason its not sharing.
Can you help
I think probably we need some onOpen method, like
onOpen
<TwitterIcon onOpen={(popup) => { // check popup variable which is result of window.open }} />
While sharing on safari browser, we have default popup as blocked. Now there is no method returning that popup is blocked and that is the reason its not sharing.
Can you help