sed - Addressing using two strings

I am picking up sed. I am having a trouble understanding how the line addressing in sed works when a pattern is used to specify line address.

I have a sample text file named emp.lst with the following contents:

emp.lst

2233|a.k. shukla |g.m. |sales |12/12/52|6000 9876|jai sharma |director |production|12/03/50|7000 5678|sumit chakrobarty|d.g.m. |marketing |19/04/43|6000 2365|barun sengupta |director |personnel |11/05/47|7800 5423|n.k. gupta |chairman |admin |30/08/56|5400 1006|chanchal singhvi |director |sales |03/09/38|6700 6213|karuna ganguly |g.m. |accounts |05/06/62|6300 1265|s.n. dasgupta |manager |sales |12/09/63|5600 4290|jayant Choudhury |executive|production|07/09/50|6000 2476|anil aggarwal |manager |sales |01/05/59|5000 6521|lalit chowdury |director |marketing |26/09/45|8200 3212|shyam saksena |d.g.m. |accounts |12/12/55|6000 3564|sudhir Agarwal |executive|personnel |06/07/47|7500 2345|j.b. saxena |g.m. |marketing |12/03/45|8000 0110|v.k. agrawal |g.m. |marketing |31/12/40|9000

As I understand, line address can be specified either in the form of line number(s) of a pattern to match as text or regular expression.

I understand how sed -n '1p' emp.lst and sed -n '1,2p' emp.lst print line 1 and line 1 & 2 respectively without echoing all lines (-n).

sed -n '1p' emp.lst

sed -n '1,2p' emp.lst

-n

I also understand and appreciate how sed -n '/director/p' emp.lst match all the lines containing the string director, and outputs:

sed -n '/director/p' emp.lst

director

9876|jai sharma |director |production|12/03/50|7000 2365|barun sengupta |director |personnel |11/05/47|7800 1006|chanchal singhvi |director |sales |03/09/38|6700 6521|lalit chowdury |director |marketing |26/09/45|8200

Now, when I specify multiple patters as sed -n '/director/,/executive/p' emp.lst, the output shown is:

sed -n '/director/,/executive/p' emp.lst

9876|jai sharma |director |production|12/03/50|7000 5678|sumit chakrobarty|d.g.m. |marketing |19/04/43|6000 2365|barun sengupta |director |personnel |11/05/47|7800 5423|n.k. gupta |chairman |admin |30/08/56|5400 1006|chanchal singhvi |director |sales |03/09/38|6700 6213|karuna ganguly |g.m. |accounts |05/06/62|6300 1265|s.n. dasgupta |manager |sales |12/09/63|5600 4290|jayant Choudhury |executive|production|07/09/50|6000 6521|lalit chowdury |director |marketing |26/09/45|8200 3212|shyam saksena |d.g.m. |accounts |12/12/55|6000 3564|sudhir Agarwal |executive|personnel |06/07/47|7500

What does this output represent?

Is it all lines containing the pattern director and executive? Clearly no, as there are some lines not containing either one of the patterns.

director

executive

Is it all lines starting with first one matching either one of the patters till the last one matching either one of the patterns? No again, as if I go by that logic, one line (2476|anil aggarwal |manager |sales |01/05/59|5000) is missing from the output.

2476|anil aggarwal |manager |sales |01/05/59|5000

I have not been able to clearly deduce how the command sed -n '/director/,/executive/p' emp.lst is working? I have gone through the sed man page and have yet been unable to deduce.

sed -n '/director/,/executive/p' emp.lst

How do I approach understanding the working?

For context, I am running sed command built into macOS High Sierra 10.13.6 running in Bash version 4.4.

sed

Note: I am a sed newbie. Please edit any mistake or incorrect terminology that I may have used.

sed

2 Answers
2

https://www.gnu.org/software/sed/manual/sed.html#Range-Addresses:

An address range can be specified by specifying two addresses separated by a comma (,). An address range matches lines starting from where the first address matches, and continues until the second address matches (inclusively):

,

$ seq 10 | sed -n '4,6p' 4 5 6

Thus 1,2p does not mean "print lines 1 and 2" but "print all lines between line 1 and line 2". The difference becomes more clear with e.g. 3,7p, which will not just print line 3 and 7, but lines 3, 4, 5, 6, 7.

1,2p

3,7p

/director/,/executive/p prints all lines between a starting line (matching director) and an ending line (matching executive).

/director/,/executive/p

director

executive

In your case, you have two matching ranges (each starting with director and ending with executive):

director

executive

Thanks a lot. Your explanation was very helpful and I can now clearly understand the working. I experimented with various placement(s) of director and executive within the file. man pages although logically precise and concise are generally hard to understand for newbie.

– Nimesh Neema
Sep 9 '18 at 23:39

director

executive

I have a follow up query and felt it makes more sense to ask here than to write an altogether separate question. Is there a way to select just the first/last/any intermediate one of the many matching blocks?

– Nimesh Neema
Sep 10 '18 at 10:47

From man sed:

man sed

0,addr2 Start out in "matched first address" state, until addr2 is found. This is similar to 1,addr2, except that if addr2 matches the very first line of input the 0,addr2 form will be at the end of its range, whereas the 1,addr2 form will still be at the beginning of its range. This works only when addr2 is a regular expression.

Not 100% sure if this is the manual section that applies but it looks like you have 2 blocks from "director" to "executive" in your output above.
There happen to be some other "director" lines between the first "director" and first succeeding "executive".

That section does not apply. There is no 0, in OP's code.

– melpomene
Sep 9 '18 at 23:31

0,

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