如何获取包含在返回的 firebase 动态短链接中的自定义查询字符串

How can I get custom querystring included in returned firebase dynamic shortlink

我正在使用 Firebase 动态 link post API 到 return 一个简短的link。当我 post 这个:

https://CENSORED.page.link/?link=https://www.CENSORED.co.uk/offers/friends/?utm_source=referafriend&utm_medium=ecrm&utm_campaign=cbk25&utm_term=988776

点击 returned shortlink 重定向到:

https://www.CENSORED.co.uk/offers/friends/?utm_source=referafriend

post 由客户端 js 制作。 Firebase 正在 return 工作短片 link,但缺少一些参数。

预计 url 来自点击的空头link:

https://www.CENSORED.co.uk/offers/friends/?utm_source=referafriend&utm_medium=ecrm&utm_campaign=cbk25&utm_term=988776

看起来它切断了我的大部分查询字符串 - 请问如何获得完整的查询字符串 returned 正确?

已解决:转义 url 对我有用:

参数丢失:

"https://www.test.co.uk/testing/?utm_source=jam&utm_medium=spoon&utm_campaign=jar&utm_term=lid"

参数正确返回:

"https%3A%2F%2Fwww.test.co.uk%2Ftesting%2F%3Futm_source%3Djam%26utm_medium%3Dspoon%26utm_campaign%3Djar%26utm_term%3Dlid"