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

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.

shellksh


Solution 1:[1]

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

Solution 2:[2]

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[^]]*'

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 KamilCuk

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