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Need Help with a sed command involving Regex

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Top Forums Shell Programming and Scripting Need Help with a sed command involving Regex
# 8  
Old 11-29-2013
In awk (as sed algo):
Code:
awk '/0-|-0/{print X"\n"$0;exit};/1234/{X=$0}' file

Regards.

---------- Post updated 29-11-13 at 09:23 AM ---------- Previous update was 28-11-13 at 09:13 PM ----------

Quote:
Originally Posted by disedorgue
Hi,
I don't all understand, but try:
Code:
sed -n '/0-/{x;p;x;p;q;};/-0/{x;p;x;p;q;};/1234/x' file

Regards.
-n : only lines explicitly selected for output are written.
here, we treat three pattern: /0-/,/-0/ and /1234/.
For the patterns /0-/ and /-0/, we do code block x;p;x;p;q where:
x : Exchange the contents of the pattern and hold space
p : Write the pattern space to standard output.
q : Quit
For the pattern /1234/ we exchange the contents of the pattern and hold space.
When sed begin a new cycle, the pattern space is empty and the new line will insert in. The hold space is inchanged.
The idea is to catch the pattern /1234/ in hold space and write to output the hold space before the pattern space when the pattern /0-/ or /-0/ are detected and to quit.

Regards.
# 9  
Old 11-29-2013
Thanks disedorgue...very well explained....appreciate the clarity of your expression
# 10  
Old 11-29-2013
If the pattern is always 0-...-0 you can shorten this to
Code:
sed -n -e '/0-.*-0/{x;p;x;p;q;}' -e '/1234/h' file

I have chosen two -e "lines" instead of a ";" separator, so it works with all sed versions.
# 11  
Old 11-30-2013
Thanks Made In Germany,
I will give a it a try to interpret it.Please comment on any mistakes on my part.
1. Look for regex 1234, if true, then pass this to hold buffer.
2. Look for pattern 0-.*-0, if true then exchange, hold buffer goes to pattern space and print pattern space. This prints line containing 1234
3.Then again exchange and print .This prints line containing 0-.*-0 .
4.Then quit. This ensures that ONLY the first instance of the occurrence gets printed.
Rgds
SShinde

---------- Post updated at 04:14 AM ---------- Previous update was at 04:07 AM ----------

Just another thought,
Supposing I were looking to print the FIRST THREE instances of the pattern.
How would the sed command look like?
Rgds
SShinde
# 12  
Old 11-30-2013
You got it, well explained!
--
sed is limited - no variables that can be used as counters.
If you have a "do it n-times" requirement then it's time for awk or perl.
BTW sed has a n-times modifier for its s command, and its buffers have multi-line capability, so there might be even an artistic solution...
Last edited by MadeInGermany; 11-30-2013 at 06:25 AM..
# 13  
Old 11-30-2013
Artistic solution as here (in red, all pattern find (five) , but print only FIRST FOUR print - see red value in sed command -)
But here, this syntax is not for all sed version:
Code:
$ cat co.yy
rwerwerwe rere
fdfefefe fsdfds
rerere ffff trtrt
aaaa 1234 asadsdsd
Xaaa 1234 asadsdsd
Aaaa 1234 asadsdsd
hfjfjfjsjfsf shfshfsjhshs
dssssdsdsdd
dsgdshgsdgdsdgssgdsgdsdhsgdshgshdshdg
akjjkjkj         0---+---0
gfhdfgdfhfgdhhfgdhfgd
sgdhsgdhsdgsdhgsdgsdhhdg
b1234hjhhh ghgsggdgdg
Yaaa 1234 asadsdsd
dddhhdjdhdh
Baaa 1234 asadsdsd
bdagdhdgahadgadg  0---0---+
dagdhdgahadgadg  00+
rwerwerwe rere
fdfefefe fsdfds
rerere ffff trtrt
caaaa 1234 asadsdsd
hfjfjfjsjfsf shfshfsjhshs
dssssdsdsdd
dsgdshgsdgdsdgssgdsgdsdhsgdshgshdshdg
cjjkjkj         0---+---0
gfhdfgdfhfgdhhfgdhfgd
sgdhsgdhsdgsdhgsdgsdhhdg
d1234hjhhh ghgsggdgdg
dddhhdjdhdh
dagdhdgahadgadg  0---0---+
dagdhdgahadgadg  00+
d1234hjhhh ghgsggdgdg
dagdhdgahadgadg  0---0---+

Code:
$ sed -n -e '/1234/H' -e '/0-.*-0/H' -e '${x' -e 's/^\n//' -e 's/$/\n/' -e ':loop' -e 's/\([^\n]*1234[^\n]*\n\)\([^\n]*1234[^\n]*\n\)/\2/' -e 't loop' -e 's/\(\([^\n]*1234[^\n]*\n[^\n]*\(0-\)*\(-0\)*[^\n]*\n\)\{4\}\).*/\1/;s/\n$//p' -e '}' co.yy
Aaaa 1234 asadsdsd
akjjkjkj         0---+---0
Baaa 1234 asadsdsd
bdagdhdgahadgadg  0---0---+
caaaa 1234 asadsdsd
cjjkjkj         0---+---0
d1234hjhhh ghgsggdgdg
dagdhdgahadgadg  0---0---+

# 14  
Old 12-02-2013
SmiliePhew!Smilie
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