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Using "tail" to follow a file without displaying the most recent lines

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Using “tail” to follow a file without displaying the most recent lines

How can I do the equivalent of tail -f with ls?Observe multiple log files in one outputMaking less's follow option show line movementtail -f but suck in content of the file first (aka `cat -f`)Using tail to follow daily log file in BashTail -f the most recent log fileOnly output most recent 10 (or n) lines of a lengthy command outputtail display whole file and then only changesFor a given directory, how do I concatenate the tail end of recently modified files to a new file?Using head and tail to grab different sets of lines and saving into same file

2

1

I would like use a program like tail to follow a file as it's being written to, but not display the most recent lines.

For instance, when following a new file, no text will be displayed while the file is less than 30 lines. After more than 30 lines are written to the file, lines will be written to the screen starting at line 1.

So as lines 31-40 are written to the file, lines 1-10 will be written to the screen.

If there is no easy way to do this with tail, maybe a there's a way to write to a new file a prior line from the first file each time the first file is extended by a line, and the tail that new file...

linux command-line tail

share|improve this question

asked 3 hours ago

ridthyselfridthyself

111

New contributor

ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

2

1

I would like use a program like tail to follow a file as it's being written to, but not display the most recent lines.

For instance, when following a new file, no text will be displayed while the file is less than 30 lines. After more than 30 lines are written to the file, lines will be written to the screen starting at line 1.

So as lines 31-40 are written to the file, lines 1-10 will be written to the screen.

If there is no easy way to do this with tail, maybe a there's a way to write to a new file a prior line from the first file each time the first file is extended by a line, and the tail that new file...

linux command-line tail

share|improve this question

asked 3 hours ago

ridthyselfridthyself

111

New contributor

ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

I would like use a program like tail to follow a file as it's being written to, but not display the most recent lines.

For instance, when following a new file, no text will be displayed while the file is less than 30 lines. After more than 30 lines are written to the file, lines will be written to the screen starting at line 1.

So as lines 31-40 are written to the file, lines 1-10 will be written to the screen.

If there is no easy way to do this with tail, maybe a there's a way to write to a new file a prior line from the first file each time the first file is extended by a line, and the tail that new file...

linux command-line tail

share|improve this question

asked 3 hours ago

ridthyselfridthyself

111

New contributor

ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 3 hours ago

ridthyselfridthyself

111

New contributor

ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 3 hours ago

ridthyselfridthyself

111

New contributor

ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

ridthyselfridthyself

111

New contributor

ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 3 hours ago

ridthyselfridthyself

111

3 Answers
3

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oldest

votes

2

Maybe buffer with awk:

tail -n +0 -f some/file | awk '{b[NR] = $0} NR > 30 {print b[NR-30]; delete b[NR-30]} END {for (i = NR - 29; i <= NR; i++) print b[i]}'

The awk code, expanded:

{

    b[NR] = $0 # save the current line in a buffer array

}

NR > 30 { # once we have more than 30 lines

    print b[NR-30]; # print the line from 30 lines ago

    delete b[NR-30]; # and delete it

}

END { # once the pipe closes, print the rest

    for (i = NR - 29; i <= NR; i++)

        print b[i]

}

share|improve this answer

edited 1 hour ago

answered 2 hours ago

murumuru

36.8k589163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works, but form the script I would expect it to work like tail, printing out a previous line as each new line is added to the file. Instead it prints out in spurts of ~70 lines after ~100 lines are added to the file. It does not print the most recent 30 lines, so it's pretty close...
  
  – ridthyself
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ridthyself if you have GNU awk, try adding a fflush(); after the print b[NR-30];. Maybe the output is being buffered.
  
  – muru
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ridthyself, your awk must be mawk, Try switching to gawk or pass the -W interactive option.
  
  – Stéphane Chazelas
  18 mins ago
  

  
  
  
  

add a comment | 

1

This isn't very efficient, because it will re-read the file every two seconds, but will do the job:

watch 'tail -n40 /path/to/file | head -n10'

share|improve this answer

answered 2 hours ago

l0b0l0b0

28.7k19121249

add a comment | 

0

Same as @muru's but using the modulo operator instead of storing and deleting:

tail -fn+1 some/file | awk -v n=30 '

   NR > n {print s[NR % n]}

   {s[NR % n] = $0}

   END{for (i = NR - n + 1; i <= NR; i++) print s[i % n]}'

share|improve this answer

answered 20 mins ago

Stéphane ChazelasStéphane Chazelas

312k57592948

add a comment | 

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2

Maybe buffer with awk:

tail -n +0 -f some/file | awk '{b[NR] = $0} NR > 30 {print b[NR-30]; delete b[NR-30]} END {for (i = NR - 29; i <= NR; i++) print b[i]}'

The awk code, expanded:

{

    b[NR] = $0 # save the current line in a buffer array

}

NR > 30 { # once we have more than 30 lines

    print b[NR-30]; # print the line from 30 lines ago

    delete b[NR-30]; # and delete it

}

END { # once the pipe closes, print the rest

    for (i = NR - 29; i <= NR; i++)

        print b[i]

}

share|improve this answer

edited 1 hour ago

answered 2 hours ago

murumuru

36.8k589163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works, but form the script I would expect it to work like tail, printing out a previous line as each new line is added to the file. Instead it prints out in spurts of ~70 lines after ~100 lines are added to the file. It does not print the most recent 30 lines, so it's pretty close...
  
  – ridthyself
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ridthyself if you have GNU awk, try adding a fflush(); after the print b[NR-30];. Maybe the output is being buffered.
  
  – muru
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ridthyself, your awk must be mawk, Try switching to gawk or pass the -W interactive option.
  
  – Stéphane Chazelas
  18 mins ago
  

  
  
  
  

add a comment | 

2

Maybe buffer with awk:

tail -n +0 -f some/file | awk '{b[NR] = $0} NR > 30 {print b[NR-30]; delete b[NR-30]} END {for (i = NR - 29; i <= NR; i++) print b[i]}'

The awk code, expanded:

{

    b[NR] = $0 # save the current line in a buffer array

}

NR > 30 { # once we have more than 30 lines

    print b[NR-30]; # print the line from 30 lines ago

    delete b[NR-30]; # and delete it

}

END { # once the pipe closes, print the rest

    for (i = NR - 29; i <= NR; i++)

        print b[i]

}

share|improve this answer

edited 1 hour ago

answered 2 hours ago

murumuru

36.8k589163

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This works, but form the script I would expect it to work like tail, printing out a previous line as each new line is added to the file. Instead it prints out in spurts of ~70 lines after ~100 lines are added to the file. It does not print the most recent 30 lines, so it's pretty close...
  
  – ridthyself
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ridthyself if you have GNU awk, try adding a fflush(); after the print b[NR-30];. Maybe the output is being buffered.
  
  – muru
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @ridthyself, your awk must be mawk, Try switching to gawk or pass the -W interactive option.
  
  – Stéphane Chazelas
  18 mins ago
  

  
  
  
  

add a comment | 

Maybe buffer with awk:

tail -n +0 -f some/file | awk '{b[NR] = $0} NR > 30 {print b[NR-30]; delete b[NR-30]} END {for (i = NR - 29; i <= NR; i++) print b[i]}'

The awk code, expanded:

{

    b[NR] = $0 # save the current line in a buffer array

}

NR > 30 { # once we have more than 30 lines

    print b[NR-30]; # print the line from 30 lines ago

    delete b[NR-30]; # and delete it

}

END { # once the pipe closes, print the rest

    for (i = NR - 29; i <= NR; i++)

        print b[i]

}

share|improve this answer

edited 1 hour ago

answered 2 hours ago

murumuru

36.8k589163

tail -n +0 -f some/file | awk '{b[NR] = $0} NR > 30 {print b[NR-30]; delete b[NR-30]} END {for (i = NR - 29; i <= NR; i++) print b[i]}'

{

    b[NR] = $0 # save the current line in a buffer array

}

NR > 30 { # once we have more than 30 lines

    print b[NR-30]; # print the line from 30 lines ago

    delete b[NR-30]; # and delete it

}

END { # once the pipe closes, print the rest

    for (i = NR - 29; i <= NR; i++)

        print b[i]

}

share|improve this answer

edited 1 hour ago

answered 2 hours ago

murumuru

36.8k589163

answered 2 hours ago

murumuru

36.8k589163

answered 2 hours ago

murumuru

36.8k589163

1

This isn't very efficient, because it will re-read the file every two seconds, but will do the job:

watch 'tail -n40 /path/to/file | head -n10'

share|improve this answer

answered 2 hours ago

l0b0l0b0

28.7k19121249

add a comment | 

1

This isn't very efficient, because it will re-read the file every two seconds, but will do the job:

watch 'tail -n40 /path/to/file | head -n10'

share|improve this answer

answered 2 hours ago

l0b0l0b0

28.7k19121249

add a comment | 

This isn't very efficient, because it will re-read the file every two seconds, but will do the job:

watch 'tail -n40 /path/to/file | head -n10'

share|improve this answer

answered 2 hours ago

l0b0l0b0

28.7k19121249

watch 'tail -n40 /path/to/file | head -n10'

share|improve this answer

answered 2 hours ago

l0b0l0b0

28.7k19121249

answered 2 hours ago

l0b0l0b0

28.7k19121249

answered 2 hours ago

l0b0l0b0

28.7k19121249

0

Same as @muru's but using the modulo operator instead of storing and deleting:

tail -fn+1 some/file | awk -v n=30 '

   NR > n {print s[NR % n]}

   {s[NR % n] = $0}

   END{for (i = NR - n + 1; i <= NR; i++) print s[i % n]}'

share|improve this answer

answered 20 mins ago

Stéphane ChazelasStéphane Chazelas

312k57592948

add a comment | 

0

Same as @muru's but using the modulo operator instead of storing and deleting:

tail -fn+1 some/file | awk -v n=30 '

   NR > n {print s[NR % n]}

   {s[NR % n] = $0}

   END{for (i = NR - n + 1; i <= NR; i++) print s[i % n]}'

share|improve this answer

answered 20 mins ago

Stéphane ChazelasStéphane Chazelas

312k57592948

add a comment | 

Same as @muru's but using the modulo operator instead of storing and deleting:

tail -fn+1 some/file | awk -v n=30 '

   NR > n {print s[NR % n]}

   {s[NR % n] = $0}

   END{for (i = NR - n + 1; i <= NR; i++) print s[i % n]}'

share|improve this answer

answered 20 mins ago

Stéphane ChazelasStéphane Chazelas

312k57592948

tail -fn+1 some/file | awk -v n=30 '

   NR > n {print s[NR % n]}

   {s[NR % n] = $0}

   END{for (i = NR - n + 1; i <= NR; i++) print s[i % n]}'

share|improve this answer

answered 20 mins ago

Stéphane ChazelasStéphane Chazelas

312k57592948

answered 20 mins ago

Stéphane ChazelasStéphane Chazelas

312k57592948

answered 20 mins ago

Stéphane ChazelasStéphane Chazelas

312k57592948