用于性能测试的 C Sharp 中 URL 的正则表达式
Regular Expression for URL in C Sharp for Performance Testing
我正在使用 Visual Studio 2012 - Web 性能测试工具。
在这里我有一个 Referer URL 像这样。 （这大约有 30 行，但我保持简短）
https://first URL/Second/Third/Fourth/five/SearchResults?json=%7B%22SalesInventoryID%22%3Anull%2C%22BundleInventoryID%22%3Anull%2C%22InventorySalesPriceID%22%3Anull%2C%22SiteType %22%3A8%2C%22SiteName%22%3A%22Desert%20Club%22%2C%22PointsMin%22%3A1%2C%22PointsMax%22%3A999999%2C%22RoomType%22%3Anull%2C%22PriceMin%22%3A1 %2C%22PriceMax%22%3A999999%2C%22SeasonType%22%3Anull%2C%22FrequencyType%22%3Anull%2C%22CheckInType%22%3Anull%2C%22WeekMin%22%3A1%2C%22WeekMax%22%3A52%2C %22RoomMin%22%3A%221%22%2C%22RoomMax%22%3A%22999999%22%2C%22TourID%22%3A2022154%2C%22TourLocationID%22%3A54%2C%22ListResults%22%3Anull%2C%22BundledResults %22%3Anull%2C%22TourParticipantInfo%22%3A%7B%22TourID%22%3A2022154%2C%22SiteID%22%3A1%2C%22SiteName%22%3Anull%2C%22PersonID%22%3A364182%2C%22FirstName%22 %3A%22Lynn%20A.%22%2C%22LastName%22%3A%22McDougald%22%2C%22City%22%3A%22Charlotte%22%2C%22State%22%3A%22-%22%2C%22ShortState %22%3A%22-%22%2C%22Country%22%3Anull%2C%22CountryID%22%3A840%2C%22OwnerNumber%22%3A%226017104%22%2C%22OwnerStatus%22%3Anull%2C
搜索结果之后，一切都在不断变化。
有人可以告诉我吗，如果我在“/fourth”之后需要什么，我该怎么做
就像我只需要这样 - https://first URL/Second/Third/Fourth/
我试过了：
https://first URL/Second/Third/Fourth/.*
和
https://first URL/Second/Third/Fourth/(.*)
没用。
要获取第四个 URL 之后的所有内容，请使用以下
https?:\/\/(?:[^\/]+\/){4}(.*)
然后使用捕获组 1 获取第四个 / 之后的所有文本。
您可以在这里进行测试：Regex101 Demo
编辑： 来自@Doqnach 的建议
基于@degant
的正则表达式
https:\/\/[^\/]+\/[^\/]+\/[^\/]+\/[^\/]+\/(.*)
捕获第四部分以外的所有内容。以及前面部分的动态内容，使其占用 "everything after the 4th section" 而不是为每个部分使用显式内容。
我正在使用 Visual Studio 2012 - Web 性能测试工具。
在这里我有一个 Referer URL 像这样。 （这大约有 30 行，但我保持简短）
https://first URL/Second/Third/Fourth/five/SearchResults?json=%7B%22SalesInventoryID%22%3Anull%2C%22BundleInventoryID%22%3Anull%2C%22InventorySalesPriceID%22%3Anull%2C%22SiteType %22%3A8%2C%22SiteName%22%3A%22Desert%20Club%22%2C%22PointsMin%22%3A1%2C%22PointsMax%22%3A999999%2C%22RoomType%22%3Anull%2C%22PriceMin%22%3A1 %2C%22PriceMax%22%3A999999%2C%22SeasonType%22%3Anull%2C%22FrequencyType%22%3Anull%2C%22CheckInType%22%3Anull%2C%22WeekMin%22%3A1%2C%22WeekMax%22%3A52%2C %22RoomMin%22%3A%221%22%2C%22RoomMax%22%3A%22999999%22%2C%22TourID%22%3A2022154%2C%22TourLocationID%22%3A54%2C%22ListResults%22%3Anull%2C%22BundledResults %22%3Anull%2C%22TourParticipantInfo%22%3A%7B%22TourID%22%3A2022154%2C%22SiteID%22%3A1%2C%22SiteName%22%3Anull%2C%22PersonID%22%3A364182%2C%22FirstName%22 %3A%22Lynn%20A.%22%2C%22LastName%22%3A%22McDougald%22%2C%22City%22%3A%22Charlotte%22%2C%22State%22%3A%22-%22%2C%22ShortState %22%3A%22-%22%2C%22Country%22%3Anull%2C%22CountryID%22%3A840%2C%22OwnerNumber%22%3A%226017104%22%2C%22OwnerStatus%22%3Anull%2C
搜索结果之后，一切都在不断变化。
有人可以告诉我吗，如果我在“/fourth”之后需要什么，我该怎么做
就像我只需要这样 - https://first URL/Second/Third/Fourth/
我试过了：
https://first URL/Second/Third/Fourth/.*
和
https://first URL/Second/Third/Fourth/(.*)
没用。
要获取第四个 URL 之后的所有内容，请使用以下
https?:\/\/(?:[^\/]+\/){4}(.*)
然后使用捕获组 1 获取第四个 / 之后的所有文本。
您可以在这里进行测试：Regex101 Demo
编辑： 来自@Doqnach 的建议
基于@degant
的正则表达式https:\/\/[^\/]+\/[^\/]+\/[^\/]+\/[^\/]+\/(.*)
捕获第四部分以外的所有内容。以及前面部分的动态内容，使其占用 "everything after the 4th section" 而不是为每个部分使用显式内容。