fiddler得到的链接，在网址上输入显示404错误，请问我怎么样才能获得真正的链接

Question

下面是header
GET https://www.amazon.cn/gp/profile/amzn1.account.AHCD2LPPD5RBR56POLK64V5KQQBA/activity_feed?review_offset=0 HTTP/1.1
Host: www.amazon.cn
Connection: keep-alive
Accept: /
X-Requested-With: XMLHttpRequest
User-Agent: Mozilla/5.0 (Windows NT 10.0; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/55.0.2883.87 Safari/537.36
Referer: https://www.amazon.cn/gp/profile/amzn1.account.AHCD2LPPD5RBR56POLK64V5KQQBA/ref=cm_cr_dp_d_gw_tr?ie=UTF8
Accept-Encoding: gzip, deflate, sdch, br
Accept-Language: zh-CN,zh;q=0.8
Cookie: x-wl-uid=1kBTeLyqptacBHIQf9ppsIpszH1MEkCY33dgF8UIEdnE+u7+06UJplqtziN+uzget7mX7l0MxiwM=; session-token="5iHqgxEg+5yFhFLFLl+69UWql9E3floBdPKRwZNGGaukjP/0X8lBsXLL9RVskRr17ZlXaLi1MIyjwIcjCmsMNKOiUWWIp/uu9qvCCZ53Y0sV5i1rJaJIRyARL1ax7RhJH9u9N151YrX56+os49SqE82EIKkl1n+++bSkwUwAyuTXYNu46aeBNoI4wjLVWTgzu0qFR+Gua7sOxMMRAF2iBdNrDpXDQDBQon5IxaHHphaIbA7y4yoOsQ=="; csm-hit=s-A1PZM68YTHV48SY8ZHGN|1521247999137; ubid-acbcn=462-8783700-2714766; session-id-time=2082729601l; session-id=461-9794329-7667546

Answer 1

访问地址是这个：https://www.amazon.cn/gp/profile/amzn1.account.AHCD2LPPD5RBR56POLK64V5KQQBA/activity_feed?review_offset=0
原来是从这个地址跳转过去的：https://www.amazon.cn/gp/profile/amzn1.account.AHCD2LPPD5RBR56POLK64V5KQQBA/ref=cm_cr_dp_d_gw_tr?ie=UTF8
直接在浏览器上输入链接访问的话由于请求头上没有Referer: 跟其他的请求头，可以试一下用postman之类的辅助插件拼接好请求头，或者用脚本拼好请求头访问...