Using “tail” to follow a file without displaying the most recent lines
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
asked 2 hours ago
ridthyselfridthyself
111
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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
asked 2 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
asked 2 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.
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
linux command-line tail
asked 2 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 2 hours ago
ridthyselfridthyself
111
New contributor
ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago
ridthyselfridthyself
111
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ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 2 hours ago
ridthyselfridthyself
111
asked 2 hours ago
ridthyselfridthyself
111
111
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ridthyself is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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.
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Check out our Code of Conduct.
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3 Answers
3
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votes
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]
}
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 afflush();after theprint b[NR-30];. Maybe the output is being buffered.
– muru
53 mins ago
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'
answered 2 hours ago
l0b0l0b0
28.7k19121249
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]}'
answered just now
Stéphane ChazelasStéphane Chazelas
312k57592948
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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]
}
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 afflush();after theprint b[NR-30];. Maybe the output is being buffered.
– muru
53 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]
}
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 afflush();after theprint b[NR-30];. Maybe the output is being buffered.
– muru
53 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]
}
answered 2 hours ago
murumuru
36.8k589163
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]
}
answered 2 hours ago
murumuru
36.8k589163
edited 53 mins ago
answered 2 hours ago
murumuru
36.8k589163
answered 2 hours ago
murumuru
36.8k589163
answered 2 hours ago
murumuru
36.8k589163
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 afflush();after theprint b[NR-30];. Maybe the output is being buffered.
– muru
53 mins ago
add a comment |
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 afflush();after theprint b[NR-30];. Maybe the output is being buffered.
– muru
53 mins ago
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
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
53 mins ago
@ridthyself if you have GNU awk, try adding a
fflush(); after the print b[NR-30];. Maybe the output is being buffered.– muru
53 mins ago
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'
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'
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'
answered 2 hours ago
l0b0l0b0
28.7k19121249
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'
answered 2 hours ago
l0b0l0b0
28.7k19121249
answered 2 hours ago
l0b0l0b0
28.7k19121249
answered 2 hours ago
l0b0l0b0
28.7k19121249
answered 2 hours ago
l0b0l0b0
28.7k19121249
28.7k19121249
add a comment |
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]}'
answered just now
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]}'
answered just now
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]}'
answered just now
Stéphane ChazelasStéphane Chazelas
312k57592948
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]}'
answered just now
Stéphane ChazelasStéphane Chazelas
312k57592948
answered just now
Stéphane ChazelasStéphane Chazelas
312k57592948
answered just now
Stéphane ChazelasStéphane Chazelas
312k57592948
answered just now
Stéphane ChazelasStéphane Chazelas
312k57592948
312k57592948
add a comment |
add a comment |
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