正则表达式与shell脚本不匹配：sed的| * MOVE * \\（[^] * \\）。* | \\ 1 |

Question

I am facing pretty strange problem, below is shell script with sed command:

rec="001  MOVE 'N'    TO abcdefgef"
vname=`sed 's| *MOVE *\([^ ]*\) .*|\1|' <<<$rec`
echo "vname=$vname"

output: vname=001'N'

I am expecting it to be only as 'N' not 001'N'. Could someone please explain why 001 is coming in output? if regular exp is not matching i should be gettign complete string as ouput. And if 001 is matching how it can miss MOVE word?

Answer 1

$ rec="001  MOVE 'N'    TO abcdefgef"
$ echo "$rec" | sed 's| *MOVE *\([^ ]*\) .*|\1|'
001'N'
$ echo "$rec" | sed 's|.* *MOVE *\([^ ]*\)|\1|'
'N'    TO abcdefgef

$ echo "$rec" | sed 's|.* *MOVE *\([^ ]*\) .*|\1|'
'N'
$ echo "MOVE 'N'    TO" | sed 's|.* *MOVE *\([^ ]*\) .*|\1|'
'N'

Just like you added .* to match rest of line after capturing your required string, you need to add .* to match string before MOVE...

Also, you probably want to match the space after MOVE instead of making it optional

$ echo "MOVED TO" | sed 's|.* *MOVE *\([^ ]*\) .*|\1|'
D
$ echo "MOVED TO" | sed 's|.* *MOVE \([^ ]*\) .*|\1|'
MOVED TO

Answer 2

You can do regex matching in bash:

rec="001  MOVE 'N'    TO abcdefgef"
regex='MOVE *([^ ]+)'
[[ $rec =~ $regex ]] && declare -p BASH_REMATCH

declare -ar BASH_REMATCH='([0]="MOVE '\''N'\''" [1]="'\''N'\''")'

"BASH_REMATCH" is an array that, if the match succeeded, will be populated with the matched text (in array index 0) and the captured parts of the regex (in subsequent indices)