Script

[SAM_24]

Hi,

I have a requirement. I want to grep the lines which contains SEV word and from that output I want to cut SEV to end of the line.
Ex:

SEV word1 word2
Word1 SEV Word2
Word1 word2 SEV word4 word4
word word word without SEV

Thanks.

[curt larson_1 · Accepted solution]

how about using sed

sed 's/\(.*\)\(SEV\)\(.*\)/\2\3/p' < inputfile

but if there is more then one occurance of this will print from the last occurance to the end of the line.

[curt larson_1]

or in ksh

grep SEV inputfile |
while read line
do
line=${line#*SEV}
print "SEV $line"
done

this will print from the first occurance of SEV

[curt larson_1]

or in ksh

grep SEV inputfile |
while read line
do
line=${line#*SEV}
print "SEV $line"
done

this will print from the first occurance of SEV

[SAM_24]

Hi curt,

Thanks a lot.
I tried first option. It is working fine. I had to add -n option with sed.
If possible can you tell what sed command does.
I am not getting.

Raj

[curt larson_1]

s does a substitution. s/pattern/replacement/flags
this substitutes 'replacement' for 'pattern'. the p flag prints the line if a successful substitution is done.

the regular expersion used for the pattern consists of:
'.' which matches any single character except newline.
'*' which matches any number (including zero) of the single character (including a character specified by a regular expression) that immediately precedes it.

\(\) saves the pattern enclosed between '\(' and '\)' into a special holding space (to a maximum of nine). They can be replayed in substitutions by the escape sequences \1 to \9.

\n matches the nth pattern previously saved by '\(' and '\)', where the previously saved patterns are counted from the left on the line.

so this substitution breaks the line into 3 patterns. .* for everything up to SEV. (this does means the line cannot begin with SEV). Then SEV and lastly everything after SEV (has to have at least one character). So, for the substitution to be performed the line must have a least one character before SEV and at least one character after. In which case the patterns 2 (SEV) and # (everything after SEV) will be printed.

a more exact sed command would be:
sed -e '/^SEV .*/p' -e 's/\(.*\)\(SEV .*\)/\2/p'

this will print the line if it starts with SEV, a blank and at least one character. The substitution will print from SEV to the end of the line only if SEV is followed by a blank space and at least one more character.

my previous sed command
sed 's/\(.*\)\(SEV\)\(.*\)/\2\3/p' sort of cheats a bit SEV followed by just a blank character will be printed. The cheat assumes that if there is a blank space it will be followed by a word.

[SAM_24]

Thank you. Greatly appreciated.

[H.Merijn Brand (procura]

Appreciation is expressed in points :)

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