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windowsbatch-filecommandfindstr

Windows Findstr

I'm trying to find files in a folder with specific pattern like:

abcd201 abcd001 abcd004

The folder contains files named

abcd(3 numbers)

I'm trying to use the pattern:

abcd[0,2][0][1,4] but currently not working.

DIR /b C:\Folder\abcd"[0,2][0][1,4]".txt

Thanks!

Solution

  • dir command does not support regular expressions. You need to filter the output with findstr

    dir /b "c:\folder\abcd*.txt" | findstr /r /c:"^abcd[02]0[14]\.txt$"

    That is, use dir command to obtain a first approximation of what you are searching and then filter the list (pipe the dir command to findstr) to obtain only the list of required files.

    The regular expression (/r) in findstr means: filter the lines, starting at the start of the line (initial ^), followed by abcd, followed by any character in the set [02], followed by a 0, followed by any character in the set [14], followed by a dot (a single dot means any character, so, it needs to be escaped \.), followed by the string txt and the end of the line ($).

    Maybe you will need to add a /i switch to findstr to indicate it must ignore case when matching.