英文:
I need to create a regular expression where 2 groups point to the same pattern
问题
以下是你要翻译的内容：
我有4个文件名
ABC_ALL_20230508T050011.zip
ABC_Intra1_20230508T050011.zip
ABC_Intra2_20230508T050011.zip
ABC_INT_20230508T050011.zip
我正在尝试创建一个正则表达式，用于捕获ExtractId（提取标识）、FileName（文件名）和Date（日期）。ExtractId和FileName需要在中间位置捕获相同的值，例如'ALL'、'Intra1'、'Intra2'或'INT'。
我目前有：
(?<ExtractId>[a-zA-Z]{3})_(?<FileName>[a-zA-Z0-9]*)_(?<FileDate>[0-9A-Z]{8}).*
并且结果是：
ExtractId = ABC
匹配
FileName = ALL
匹配
FileDate = 20230508
我想要的是：
ExtractId = ALL
匹配
FileName = ALL
匹配
FileDate = 20230508
我相信可以通过使用正则表达式的子表达式来实现这一点，其中可以让两个组指向相同的位置，但我以前从未使用过。
谢谢
英文:
I have 4 file names
ABC_ALL_20230508T050011.zip
ABC_Intra1_20230508T050011.zip
ABC_Intra2_20230508T050011.zip
ABC_INT_20230508T050011.zip
I am trying to create a regex that captures ExtractId, FileName and Date. The ExtractId and FileName need to capture the same values in the middle position eg 'ALL', 'Intra1', 'Intra2' or 'INT'
I currently have:
(?<ExtractId>[a-zA-Z]{3})_(?<FileName>[a-zA-Z0-9]*)_(?<FileDate>[0-9A-Z]{8}).*
and results to:
ExtractId = ABC
Matched
FileName = ALL
Matched
FileDate = 20230508
I am after:
ExtractId = ALL
Matched
FileName = ALL
Matched
FileDate = 20230508
I believe there is a way of achieving this using regex sub expression where you can have 2 groups point to the same position but I have never use it before.
Thanks
答案1
得分: 1
以下是您要翻译的内容：
您可以在先行断言中使用命名捕获组：
（？= [a-zA-Z] {3} （？<ExtractId> [a-zA-Z0-9] +））[a-zA-Z] {3} （？<FileName> [a-zA-Z0-9] +）（？<FileDate> [0-9A-Z] {8}）.*
解释
（？=积极的先行断言[a-zA-Z] {3} _匹配3次 a-zA-Z，然后匹配_（？<ExtractId> [a-zA-Z0-9] +）_命名组 Extractid，捕获1个以上的字符 a-zA-Z0-9，然后匹配_
）关闭先行断言[a-zA-Z] {3} _匹配3次 a-zA-Z，然后匹配_（？<FileName> [a-zA-Z0-9] +）命名组 FileName，捕获1个以上的字符 a-zA-Z0-9_字面匹配（？<FileDate> [0-9A-Z] {8}）命名组 FileDate，捕获8个字符 0-9A-Z.*匹配字符串的其余部分（如果您需要的话，否则可以省略此部分）
查看 正则表达式演示。
如果您想将字符串锚定到开头，可以添加 ^，如下所示：
^（？=[a-zA-Z] {3} （？<ExtractId> [a-zA-Z0-9] +））[a-zA-Z] {3} （？<FileName> [a-zA-Z0-9] +）（？<FileDate> [0-9A-Z] {8}）.*
英文:
You can use a named capture group in a lookahead assertion:
(?=[a-zA-Z]{3}_(?<ExtractId>[a-zA-Z0-9]+)_)[a-zA-Z]{3}_(?<FileName>[a-zA-Z0-9]+)_(?<FileDate>[0-9A-Z]{8}).*
Explanation
(?=Positive lookahead assertion[a-zA-Z]{3}_Match 3 times a-zA-Z and then match_(?<ExtractId>[a-zA-Z0-9]+)_Named group Extractid, capture 1+ chars a-zA-Z0-9 and then match_
)Close the lookahead[a-zA-Z]{3}_Match 3 times a-zA-Z and then match_(?<FileName>[a-zA-Z0-9]+)Named group FileName, capture 1+ chars a-zA-Z0-9_Match literally(?<FileDate>[0-9A-Z]{8})Named group FileDate, capture 8 chars 0-9A-Z.*Match the rest of the string (if you need that, else you can omit this part)
See a regex demo.
If you want to anchor the strings to the start, you can prepend ^ like:
^(?=[a-zA-Z]{3}_(?<ExtractId>[a-zA-Z0-9]+)_)[a-zA-Z]{3}_(?<FileName>[a-zA-Z0-9]+)_(?<FileDate>[0-9A-Z]{8}).*
答案2
得分: 0
将重叠捕获组中的前一个部分放入一个前瞻断言中，以便为最后一个重叠捕获组留出匹配的缓冲区：
_(?=(?<ExtractId>[^_]+))(?<FileName>[^_]+)_(?<FileDate>[0-9A-Z]{8})
演示：https://regex101.com/r/YhFyIn/2
英文:
You can put the former of the overlapping capture groups in a lookahead assertion to leave the buffer for the last overlapping capture group to match:
_(?=(?<ExtractId>[^_]+))(?<FileName>[^_]+)_(?<FileDate>[0-9A-Z]{8})
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