I want to extract the slug between (%) and (?) from a String url?
This is the url as string.
https://xyz.page.link/product/prdt%3D29c1118b344a53949824990eec6bd267?amv=24&apn=com.example.example&ibi=com.example.example&imv=1.0.1&isi=1613285148&link=https%3A%2F%2Fxyz.page.link%2F
I want to extract the string between % and ? from this part
prdt%3D29c1118b344a53949824990eec6bd267?
Is that possible?
Thanks