I found some sed examples for my dilemma. current text file is like below.
076411-160DL 0000000052220420 000005212635883-0003-0
138574A 0000000120220421 000012012630637-0003-1
146054A 0000000420220422 000042012634007-0003-1
138575A 0000000240220421 000024012630638-0003-1
145683C 0000000145220422 000014512634671-0003-1
137611B 0000000130220419 000013012632286-0003-0
137618C 0000000120220420 000012012635623-0003-0
I am only changing date in the second column but I need to leave it as is and just change to the current date then out to a new file. I have the first part seperated using grep -v "\0003-0" *.txt Then I seperate the 0003-0 with grep -h "\0003-0" I then use a sed to change format and this is where it is almost there but due to numbers in front of the date it is not consistent.
sed -r "s/[2]{2}[0-8]{2}[0-9]{2}/$(date ' %y%m%d')/g"
I get it mostly how I want it but if there is a group with 8 zeros and only 1 other digit before the date I get something on the end of the new date such as 9 or 0. Any tips would be awesome. Thanks
I am only changing the date where the file ends in 0003-0 There is a space before the first column I was not aware of until now.
expected output would be
076411-160DL 0000000052220425 000005212635883-0003-0
138574A 0000000120220421 000012012630637-0003-1
146054A 0000000420220422 000042012634007-0003-1
138575A 0000000240220421 000024012630638-0003-1
145683C 0000000145220422 000014512634671-0003-1
137611B 0000000130220425 000013012632286-0003-0
137618C 0000000120220425 000012012635623-0003-0
CodePudding user response：
Adding a pattern match to OP's current sed code:
$ sed -r "/-0003-0$/s/[2]{2}[0-8]{2}[0-9]{2}/$(date ' %y%m%d')/g" file
076411-160DL 0000000052204250 000005212635883-0003-0
138574A 0000000120220421 000012012630637-0003-1
146054A 0000000420220422 000042012634007-0003-1
138575A 0000000240220421 000024012630638-0003-1
145683C 0000000145220422 000014512634671-0003-1
137611B 0000000130220425 000013012632286-0003-0
137618C 0000000120220425 000012012635623-0003-0
But this fails to update the first line.
One possible tweak to address the 'middle' field:
$ sed -r "/-0003-0$/s/( [0-9]{10})([0-9]{6}) /\1$(date ' %y%m%d') /" file
076411-160DL 0000000052220425 000005212635883-0003-0
138574A 0000000120220421 000012012630637-0003-1
146054A 0000000420220422 000042012634007-0003-1
138575A 0000000240220421 000024012630638-0003-1
145683C 0000000145220422 000014512634671-0003-1
137611B 0000000130220425 000013012632286-0003-0
137618C 0000000120220425 000012012635623-0003-0
CodePudding user response：
Suggesting single sed command:
sed -r '/0003-0$/s|([[:digit:]]{6}) |'$(date %y%m%d)' |'
CodePudding user response：
date=$(date %y%m%d)
sed -E '/0003-0$/{s/ ([^ ]{1,10})[^ ]* / \1'"$date"' /2}'
Note that your data has a single space before the first column.
Explanation:
TLDR: Make a pattern to match fields, which also captures (up to) the first ten characters of the field for a back-reference, and substitute the second field.
- First, filter for lines ending in
0003-0. - Then target the second field of each line, by using a pattern of "space, then at least one non-space character, then space".
- The
2at the very end means substitute the second (or N) instance of the match pattern. Hence our pattern targets the second field only. - The pattern captures (up to) the first ten characters of this field (using parentheses), so we can back-reference them in the replacement (using
\1) and discards the remaining (six) date characters, which we replace with$date. {1,10}needs to be a range of at least one character, because it must match both the first and second field, to enable us to specify the second field with2.
Edit:
Gives:
076411-160DL 0000000052220426 000005212635883-0003-0
138574A 0000000120220421 000012012630637-0003-1
146054A 0000000420220422 000042012634007-0003-1
138575A 0000000240220421 000024012630638-0003-1
145683C 0000000145220422 000014512634671-0003-1
137611B 0000000130220426 000013012632286-0003-0
137618C 0000000120220426 000012012635623-0003-0