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Get number of lines after certain string

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# 1  
Old 11-27-2013
Get number of lines after certain string
Hello people,
I have a file like that:
Code:
start something
1 xxx yyy
2 zzz xxx
3 ccc vvv
4 mmm nnn
stop something
1 aaa bbb
2 ccc ddd
3 eee fff
4 qqq www
start something
1 rrr ttt
2 yyy uuu
3 iii ooo
4 ppp sss

And I need only line starting with 1 and 3 and only after start something.
Output like that:
Code:
start,1,xxx,yyy
start,3,ccc,vvv
start,1,rrr,ttt
start,3,iii,ooo

Could you please help. Thanks!
# 2  
Old 11-27-2013
for i in $(egrep(^1|3) file| sed 's/ /,/'); do echo start,$i;done
# 3  
Old 11-27-2013
Try
Code:
awk '/start/,/stop/ {if (($1-2)^2 == 1) printf "start,%s,%s,%s\n", $1, $2, $3}' file
start,1,xxx,yyy
start,3,ccc,vvv
start,1,rrr,ttt
start,3,iii,ooo

This User Gave Thanks to RudiC For This Post:
apenkov
# 4  
Old 11-27-2013
Quote:
Originally Posted by Skrynesaver
for i in $(egrep(^1|3) file| sed 's/ /,/'); do echo start,$i;done
Trying this I get:
Code:
bash: command substitution: line 1: syntax error near unexpected token `^1'
bash: command substitution: line 1: `egrep(^1|3) file.txt | sed 's/ /,/''

---------- Post updated at 05:33 PM ---------- Previous update was at 05:28 PM ----------

Quote:
Originally Posted by RudiC
Try
Code:
awk '/start/,/stop/ {if (($1-2)^2 == 1) printf "start,%s,%s,%s\n", $1, $2, $3}' file
start,1,xxx,yyy
start,3,ccc,vvv
start,1,rrr,ttt
start,3,iii,ooo

It is working! Thanks a lot!

---------- Post updated at 06:33 PM ---------- Previous update was at 05:33 PM ----------

Quote:
Originally Posted by RudiC
Try
Code:
awk '/start/,/stop/ {if (($1-2)^2 == 1) printf "start,%s,%s,%s\n", $1, $2, $3}' file
start,1,xxx,yyy
start,3,ccc,vvv
start,1,rrr,ttt
start,3,iii,ooo

One additional question, is it possible to be done in a similar way if the line starts with string? E.g.
Code:
start something
one xxx yyy
two zzz xxx
three ccc vvv
four mmm nnn
stop something
one aaa bbb
two ccc ddd
three eee fff
four qqq www
start something
one rrr ttt
two yyy uuu
three iii ooo
four ppp sss

Thanks!
# 5  
Old 11-27-2013
Try
Code:
awk '/start/,/stop/ {if ($1~/^one|^three/) printf "start,%s,%s,%s\n", $1, $2, $3}' file
start,one,xxx,yyy
start,three,ccc,vvv
start,one,rrr,ttt
start,three,iii,ooo

This User Gave Thanks to RudiC For This Post:
apenkov
# 6  
Old 11-27-2013
Using RudiC's approach

Code:
awk '
         BEGIN{
               num = "one two three four five six seven eight nine ten"
                 n = split(num,A)
               for(i=1;i<=n;i++)NUM[A[i]]=i
              }
/start/,/stop/{
               if ((NUM[tolower($1)]-2)^2 == 1)
               printf "start,%s,%s,%s\n", $1, $2, $3
              }
   ' file

Code:
start,one,xxx,yyy
start,three,ccc,vvv
start,one,rrr,ttt
start,three,iii,ooo

This User Gave Thanks to Akshay Hegde For This Post:
apenkov
# 7  
Old 11-27-2013
Quote:
Originally Posted by RudiC
Try
Code:
awk '/start/,/stop/ {if ($1~/^one|^three/) printf "start,%s,%s,%s\n", $1, $2, $3}' file
start,one,xxx,yyy
start,three,ccc,vvv
start,one,rrr,ttt
start,three,iii,ooo

Thanks a lot RudiC!

---------- Post updated at 10:41 PM ---------- Previous update was at 10:39 PM ----------

Quote:
Originally Posted by Akshay Hegde
Using RudiC's approach

Code:
awk '
         BEGIN{
               num = "one two three four five six seven eight nine ten"
                 n = split(num,A)
               for(i=1;i<=n;i++)NUM[A[i]]=i
              }
/start/,/stop/{
               if ((NUM[tolower($1)]-2)^2 == 1)
               printf "start,%s,%s,%s\n", $1, $2, $3
              }
   ' file

Code:
start,one,xxx,yyy
start,three,ccc,vvv
start,one,rrr,ttt
start,three,iii,ooo

I'm getting syntax error, will try to figure it out. Thanks!
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