带有日期的文件名的正则表达式模式

Question

I have files with name "data_2021_03_v1.0.zip" in server. When I tried using "data.*(ZIP|zip)" regex.It is loading all files starting with data string. I want files containing only data(exactmatch-no extra words along with it) in filename.

Ex: It is loading files with name "data_abc_efg_19-01-v2.0.zip" along with original file. Can someone help to construct a regex pattern for above file format

Answer 1

Looking at the sample I am assuming that all you need is: data_<number>_<version_info>.

Here is a regex for this:

data(_?\d)*(_v\d)?\.?\d\.(zip|ZIP)

Answer 2

You can make the pattern more specific

\bdata_\d{4}_\d{2}_v\d+(?:\.\d+)*\.(?:ZIP|zip)\b

• \bdata_ A word boundary to prevent a partial match, then match data_
• \d{4}_\d{2}_v Match 4 digits _ 2 digits _v
• \d+(?:\.\d+)* Match 1+ digits and optionally repeat a . and 1+ digits
• \.(?:ZIP|zip) Match either .zip or .ZIP
• \b A word boundary

Regex demo

You could make the pattern case insensitive, but then you would also match DATA for example

(?i)\bdata_\d{4}_\d{2}_v\d+(?:\.\d+)*\.zip\b