How to write the regex to capture alphanumeric value between square brackets?-CodePudding

Please help in writing the regular expression for below line in bash. There are muliple lines like this in my file, and I want to only capture the below values from the file. How can I do so?

value to be extracted -

Y324256
vmigha3
idea

Note - The value to be extracted is always alphanumeric. Square brackets on the left and right side will always have numeric values which I don't want to extract.

File looks like this-

2021-05-13 15:35:31,804 [16] [Y324256] [341745] DEBUG Server - End Webservice method GetProcessStatus
2021-05-13 15:35:32,587 [11] [vmigha3] [341749] DEBUG Domain - Reading user permissions from the database
2022-03-03 09:08:10,699 [31] [idea] [80387] INFO Server - Begin Webservice call
2022-04-06 01:18:33,822 [MonitorThread] INFO  - Reading user permissions from the database
2022-04-06 01:18:33,845 [None] DEBUG -Begin Webservice call

code

grep -oP '\[\K\d*[A-Za-z][\dA-Za-z]*(?=])' file > output_file

But this gives me below result -

Y324256
vmigha3
idea
MonitorThread
None

I am wondering if this can be done by first finding the lines which begins with 2022-04-06 01:18:33,845 this date format , then capture the 4th record.

CodePudding user response：

Assumptions/Understandings:

if there are 3x sets of [...] in the line then print the contents of the 2nd [...]
otherwise skip the line

One awk idea using dual field delimiters of ] and [ (and no need for a regex):

$ awk -F'[][]' 'NF>6 {print $4}' file
Y324256
vmigha3
idea

CodePudding user response：

Does it have to be grep? I mean:

cut -d[ -f3 file | cut -d] -f1

If it has to be, then just match the brackets.

grep -oP '^[^[]*\[[^[]*\[\K[^]]*'