[求助] 关于sed的命令 N,P,D

johnny_jiang

1. 
   
2. bash# sed -e 'N;P' file
   
3. one=1234
   
4. one=1234
   
5. two=ABCD
   
6. three=1614.13
   
7. three=1614.13
   
8. four=734.25
   
9. five=CDEF
   
10. five=CDEF
    
11. six=2586
    
12. seven=EFGH
    

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但是pattern space还是会被output,如果pattern space中存在内容而且没有使用-n option,对吗?

linux_now

当sed读到文件结束时，停止script处理，并打印模板：

[10:12:51 test]$ [color="SeaGreen"]sed 's/$/---header/;N;=;s/$/---tailer/' files
2
1. one---header
2. two---tailer
4
3. three---header
4. four---tailer
6
5. five---header
6. six---tailer
7. seven---header

yongjian

Post by johnny_jiang

1. 
   
2. bash# sed -e 'N;P' file
   
3. one=1234
   
4. one=1234
   
5. two=ABCD
   
6. three=1614.13
   
7. three=1614.13
   
8. four=734.25
   
9. five=CDEF
   
10. five=CDEF
    
11. six=2586
    
12. seven=EFGH
    

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但是pattern space还是会被output,如果pattern space中存在内容而且没有使用-n option,对吗?

if no "-n", pattern space and editing result will all be printed out.

johnny_jiang

Post by yongjian
if no "-n", pattern space and hold space will all be printed out.

但是yongjian兄,为什么下面的命令返回和上面的一样呢

1. 
   
2. 
   
3. bash# sed -e 'N;h;P' file
   
4. one=1234
   
5. one=1234
   
6. two=ABCD
   
7. three=1614.13
   
8. three=1614.13
   
9. four=734.25
   
10. five=CDEF
    
11. five=CDEF
    
12. six=2586
    
13. seven=EFGH
    
14. 
    

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最后的阶段,hold space中不是应该存有数据
five=CDEF
six=2586
:ask

linux_now

保存空间应该是不打印的：

[14:59:13 test]$ sed '${h;N}' files
1. one
2. two
3. three
4. four
5. five
6. six
7. seven
[14:59:52 test]$ sed '${[color="Red"]x;N}' files
1. one
2. two
3. three
4. four
5. five
6. six

[15:00:08 test]$

johnny_jiang

个人同意linux_now兄的观点

yongjian

Post by johnny_jiang
但是yongjian兄,为什么下面的命令返回和上面的一样呢

1. 
   
2. 
   
3. bash# sed -e 'N;h;P' file
   
4. one=1234
   
5. one=1234
   
6. two=ABCD
   
7. three=1614.13
   
8. three=1614.13
   
9. four=734.25
   
10. five=CDEF
    
11. five=CDEF
    
12. six=2586
    
13. seven=EFGH
    
14. 
    

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最后的阶段,hold space中不是应该存有数据
five=CDEF
six=2586
:ask

Because  it hits EOF. When sed/awk hits EOF, process will stop immediately.

johnny_jiang

Post by yongjian
Because  it hits EOF. When sed/awk hits EOF, process will stop immediately.

Yes, so my question is when it hits EOF, it just stops and does nothing or output the pattern space if any?

From the test, I think it sure outputs the pattern space and then exits.

yongjian

Post by johnny_jiang
Yes, so my question is when it hits EOF, it just stops and does nothing or output the pattern space if any?

From the test, I think it sure outputs the pattern space and then exits.

From what I understand, the last several processes seem to be
1. hit five
2. apply N
3. read in six.
4. apply h
5. put five and six into hold space
6. apply P
7. print out the first part till newline which is five.
8. reach the end of script, since there is no "-n", then run default print. pattern space "five and six" two lines will be printed. so you see "five five six". pattern space gets cleaned.
9. read a new line in which is seven
10. apply N.
11. hit EOF. I thought sed should exit immediately w/o even printing out the pattern space, but looks like it does now. So I went to an AIX unix box and tried the same thing.
On AIX with old sed

1. 
   
2. cat foo | sed -e 'N;P'
   
3. one=1234
   
4. one=1234
   
5. two=ABCD
   
6. three=1614.13
   
7. three=1614.13
   
8. four=734.25
   
9. five=CDEF
   
10. five=CDEF
    
11. six=2586
    

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On Linux with newest sed

1. 
   
2. cat foo | sed 'N;P'
   
3. one=1234
   
4. one=1234
   
5. two=ABCD
   
6. three=1614.13
   
7. three=1614.13
   
8. four=734.25
   
9. five=CDEF
   
10. five=CDEF
    
11. six=2586
    
12. seven=EFGH
    

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You can see the difference. When hitting EOF, the Linux ver looks like printing out the pattern space and then exit where old unix ver just exits immediately. Another comment that I made before is not correct, which was "without -n, pattern space and hold space will both be printed out". It should be "without -n, pattern space and the editing result will both be printed out" since output the pattern space is the default action of sed/awk. You guys are right, hold space will not be printed out.

johnny_jiang

thx, yongjian. From your explanation

8. reach the end of script, since there is no "-n", then run default print. pattern space "five and six" two lines will be printed. so you see "five five six". pattern space and hold space get cleaned.

I think you mean after each turn of script, sed outputs the content of pattern space and then clean pattern space and hold space. But the test I did seems a little different from yours. Please see below

1. 
   
2. bash# sed -e '1{ N;h }; 7x' file
   
3. one=1234
   
4. two=ABCD
   
5. three=1614.13
   
6. four=734.25
   
7. five=CDEF
   
8. six=2586
   
9. one=1234
   
10. two=ABCD
    

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You can see until last loop, hold space is filled with the content
one=1234
two=ABCD

Maybe it's another difference between some linux distribution.

I did my test under redhat and suse.