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
Yes you can use RegExp class.
Please take a look at this answer: How to use RegEx in Dart?
void main() {
var a ='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';
var regexp = RegExp(r"(?<=\%).+?(?=\?)");
print(regexp.hasMatch(a)); // true
print(regexp.firstMatch(a)?.group(0)); // 3D29c1118b344a53949824990eec6bd267
}
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