使用sed或awk替换每行最后一个定界符之后的数据

Question

I have a text file A.txt , all lines in which have the same number of fields separated by pipe | delimiter. I need to replace , with | for the data after the last delimiter of each line.

Example:

1,2|3,4|5|6,7,8
1,8|4|6,5,3|4,5

Desired output, (only replace , with | after last delimiter):

1,2|3,4|5|6|7|8
1,8|4|6,5,3|4|5

How to achieve this using sed or awk ?

Answer 1

With awk :

awk 'BEGIN{FS=OFS="|"} gsub(",", "|", $NF)'

• BEGIN{FS=OFS="|"} sets both the input and output field delimiter as literal |

• gsub(",", "|", $NF) substitute all ( gsub() ) , s with | in the last field ( $NF )

Example:

$ awk 'BEGIN{FS=OFS="|"} gsub(",", "|", $NF)' <<<'1,2|3,4|5|6,7,8'
1,2|3,4|5|6|7|8

$ awk 'BEGIN{FS=OFS="|"} gsub(",", "|", $NF)' <<<'1,8|4|6,5,3|4,5'
1,8|4|6,5,3|4|5

Answer 2

$ cat ip.txt 
1,2|3,4|5|6,7,8
1,8|4|6,5,3|4,5

with sed

$ sed -E ':a s/^(.*\|[^,]+),([^|]+)$/\1|\2/g; ta' ip.txt 
1,2|3,4|5|6|7|8
1,8|4|6,5,3|4|5

• :a and ta to loop sed command until there is a match found
• ^(.*\\|[^,]+) from start of line to | followed by non , characters. * and + will try to match as much as possible
• , match a comma
• ([^|]+)$ after the comma, there should not be any | character till end of line

with perl

$ perl -F'\|' -lane '$F[-1] =~ tr/,/|/; print join "|",@F' ip.txt 
1,2|3,4|5|6|7|8
1,8|4|6,5,3|4|5

• -F'\\|' split input line on | and save to @F array
• $F[-1] =~ tr/,/|/; for last element of array, replace all , with |
• print join "|",@F print the modified @F array with | as separator

And for some regex magic:

$ perl -pe 's/.*\|(*SKIP)(*F)|,/|/g' ip.txt
1,2|3,4|5|6|7|8
1,8|4|6,5,3|4|5

• .*\\|(*SKIP)(*F) skip the pattern until last |
• then replace all , with |