CODEWITHSUNDEEP
linuxunixsed
jonas
jonas

Reputation: 5509

Adding text using SED

i am working with a task right now, manipulating text in Unix, but i have no idea how i will do it. I am planning to use sed. this is wat i am trying to do.

I have this text.

BS111

xxxxx
yyyyy
zzzzz

BS112

xxxxx
yyyyy
zzzzz

BS113

xxxxx
yyyyy
zzzzz

so on.. with this kind of format

and i want it to be like:

BS111 xxxxx
BS111 yyyyy
BS111 zzzzz



BS112 xxxxx
BS112 yyyyy
BS112 zzzzz



BS113 xxxxx
BS113 yyyyy
BS113 zzzzz

so on.. with this kind of format

the BS* and its data occurs almost hundred times. so i think the best way to do it is to use a script. thank you in advance for your help.

Upvotes: 3

Views: 268

Answers (4)

abasu
abasu

Reputation: 2524

try awk

awk '/^BS/{i=$0; next;} {if(length($0)) printf("%s ",i); print $0}' <filename>

Update:

'/^BS/{i=$0; next;} {if(length($0)) printf("%s ",i); print $0}' remember, in awk, {...} denotes a set of action. /pattern/{...} means if a pattern is found, do {...} actions. /^BS[0-9]/ will search for pattern in each line. If the pattern is found, i (variable in awk) will be updated with the pattern ($0 is complete line) and next will immidiately move to next line.So next set of action in {if(...} will be skipped if a pattern is found. So i will always contain the last pattern value and if a line is containing pattern, except updating i, next set of action will not take place for the line.

Next set of action is, if(length($0)) if, the line contains text, printf("%s ",i); print the last found pattern, And thenprint $0prints the original line. So for an empty line, only an empty line is printed, if some text is there, first theiorlast found pattern` is printed, then the original line is printed. Btw, sputnick's solution is also doing the same thing, only in a more compact form. So you can dissect that one also for fun. Hope i am clear, please feel free to ask anything you need to know

Upvotes: 2

jbr
jbr

Reputation: 6258

In sed you can do:

sed -n -r '/^BS[0-9]{3}/{h;d};/./G;s/(.*)\n(.*)/\2 \1/p' test.txt

-n means sed wont output anything unless told to by the p command.
-r means using extended regular expressions.

/^BS[0-9]{3}/{h;d} matches a line starting with BS and three numbers, the h command puts the pattern into the holding buffer (copies it for later insertion). The d command deletes the matched line.

/./G matches any line. The G command appends the content of the holding buffer (in this case BSxxx). s/(.*)\n(.*)/\2 \1/p matches the current and the next line, switches them and removes the newline. And finally the p command prints the resulting line

Upvotes: 1

potong
potong

Reputation: 58351

This might work for you (GNU sed):

sed -r '/^BS/{h;d};/./!b;G;s/(.*)\n(.*)/\2 \1/' file

Upvotes: 3

Gilles Qu&#233;not
Gilles Qu&#233;not

Reputation: 184965

With awk :

awk '/^BS/{v=$0;next} {print (/^$/) ? $0 : v, $0}' file.txt

BS111 xxxxx
BS111 yyyyy
BS111 zzzzz


BS112 xxxxx
BS112 yyyyy
BS112 zzzzz


BS113 xxxxx
BS113 yyyyy
BS113 zzzzz

Upvotes: 3

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