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Need to find a response time which takes from 1-3 seconds in Apache logs
extract last match from logfile till endPrint a line in stdout that matches an expression if the output contains another expressionAWK print regex patternGrep multiple patterns and print a different number of lines below each of the patterns?Grep from the last occurrence of a pattern to another patternHow can I match the date after hitting a pattern match in awk to assure the match is current?Selecting lines in a file that do not contain the value in the other fileFor a large directory, create a variable of the filenames which include lines which include the text string stored in another variableMatching columns of different csv files, not working when column value is different lengthAwk doesn't exit despite “exit 0” command
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I need to find the API response time from an Apache log file. It's like a response time which is takes between 1 to 2 secound or 2 to 3 second. $6 is response time and values comes in microseconds.
I am trying with following command but the output is always the same:
grep 17/Sep/2016:10 /access.log| awk 'print ($6 > 1000000 && 2000000 > $6)' | wc -l
awk grep
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
asked Sep 18 '16 at 5:33
roshanroshan
84
add a comment |
I need to find the API response time from an Apache log file. It's like a response time which is takes between 1 to 2 secound or 2 to 3 second. $6 is response time and values comes in microseconds.
I am trying with following command but the output is always the same:
grep 17/Sep/2016:10 /access.log| awk 'print ($6 > 1000000 && 2000000 > $6)' | wc -l
awk grep
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
asked Sep 18 '16 at 5:33
roshanroshan
84
add a comment |
I need to find the API response time from an Apache log file. It's like a response time which is takes between 1 to 2 secound or 2 to 3 second. $6 is response time and values comes in microseconds.
I am trying with following command but the output is always the same:
grep 17/Sep/2016:10 /access.log| awk 'print ($6 > 1000000 && 2000000 > $6)' | wc -l
awk grep
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
asked Sep 18 '16 at 5:33
roshanroshan
84
I need to find the API response time from an Apache log file. It's like a response time which is takes between 1 to 2 secound or 2 to 3 second. $6 is response time and values comes in microseconds.
I am trying with following command but the output is always the same:
grep 17/Sep/2016:10 /access.log| awk 'print ($6 > 1000000 && 2000000 > $6)' | wc -l
awk grep
awk grep
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
asked Sep 18 '16 at 5:33
roshanroshan
84
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
asked Sep 18 '16 at 5:33
roshanroshan
84
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
edited Sep 18 '16 at 6:13
cutrightjm
2,24121325
2,24121325
asked Sep 18 '16 at 5:33
roshanroshan
84
asked Sep 18 '16 at 5:33
roshanroshan
84
asked Sep 18 '16 at 5:33
roshanroshan
84
84
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
This question would be clearer if some lines of access.log were added as a sample. Anyway, the awk command prints a line regardless of the value of $6, so when you count the lines with wc -l you get an outcome that is determined by grep alone.
If you want to count the lines where is $6 is between two different values you can write
grep 17/Sep/2016:10 /access.log | awk '$6 > 1000000 && 2000000 > $6' | wc -l
However, this pipeline is a bit inefficient. It would almost always be preferable to combine it in a single awk command like this:
awk '/17/Sep/2016:10/ && $6 > 1000000 && 2000000 > $6 c++ ENDprint c' access.log
To include the boundaries one can do:
grep 18/Sep/2016:11 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
or equivalently
awk '/18/Sep/2016:11/ && $6>=1000000 && $6<=2000000 c++ ENDprint c' access.log
answered Sep 18 '16 at 6:23
user000001user000001
997714
hey thanks a lot , i want to say one thing i used another coommand which is gave same answer.command is here is it write or not # grep 17/Sep/2016:10 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
– roshan
Sep 18 '16 at 8:47
@roshan: Sure that should also work, if you want to include the boundary values
– user000001
Sep 18 '16 at 9:03
yes its giving same value i checked many times output is same always.
– roshan
Sep 18 '16 at 9:34
add a comment |
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1 Answer
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1 Answer
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This question would be clearer if some lines of access.log were added as a sample. Anyway, the awk command prints a line regardless of the value of $6, so when you count the lines with wc -l you get an outcome that is determined by grep alone.
If you want to count the lines where is $6 is between two different values you can write
grep 17/Sep/2016:10 /access.log | awk '$6 > 1000000 && 2000000 > $6' | wc -l
However, this pipeline is a bit inefficient. It would almost always be preferable to combine it in a single awk command like this:
awk '/17/Sep/2016:10/ && $6 > 1000000 && 2000000 > $6 c++ ENDprint c' access.log
To include the boundaries one can do:
grep 18/Sep/2016:11 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
or equivalently
awk '/18/Sep/2016:11/ && $6>=1000000 && $6<=2000000 c++ ENDprint c' access.log
answered Sep 18 '16 at 6:23
user000001user000001
997714
hey thanks a lot , i want to say one thing i used another coommand which is gave same answer.command is here is it write or not # grep 17/Sep/2016:10 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
– roshan
Sep 18 '16 at 8:47
@roshan: Sure that should also work, if you want to include the boundary values
– user000001
Sep 18 '16 at 9:03
yes its giving same value i checked many times output is same always.
– roshan
Sep 18 '16 at 9:34
add a comment |
This question would be clearer if some lines of access.log were added as a sample. Anyway, the awk command prints a line regardless of the value of $6, so when you count the lines with wc -l you get an outcome that is determined by grep alone.
If you want to count the lines where is $6 is between two different values you can write
grep 17/Sep/2016:10 /access.log | awk '$6 > 1000000 && 2000000 > $6' | wc -l
However, this pipeline is a bit inefficient. It would almost always be preferable to combine it in a single awk command like this:
awk '/17/Sep/2016:10/ && $6 > 1000000 && 2000000 > $6 c++ ENDprint c' access.log
To include the boundaries one can do:
grep 18/Sep/2016:11 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
or equivalently
awk '/18/Sep/2016:11/ && $6>=1000000 && $6<=2000000 c++ ENDprint c' access.log
answered Sep 18 '16 at 6:23
user000001user000001
997714
hey thanks a lot , i want to say one thing i used another coommand which is gave same answer.command is here is it write or not # grep 17/Sep/2016:10 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
– roshan
Sep 18 '16 at 8:47
@roshan: Sure that should also work, if you want to include the boundary values
– user000001
Sep 18 '16 at 9:03
yes its giving same value i checked many times output is same always.
– roshan
Sep 18 '16 at 9:34
add a comment |
This question would be clearer if some lines of access.log were added as a sample. Anyway, the awk command prints a line regardless of the value of $6, so when you count the lines with wc -l you get an outcome that is determined by grep alone.
If you want to count the lines where is $6 is between two different values you can write
grep 17/Sep/2016:10 /access.log | awk '$6 > 1000000 && 2000000 > $6' | wc -l
However, this pipeline is a bit inefficient. It would almost always be preferable to combine it in a single awk command like this:
awk '/17/Sep/2016:10/ && $6 > 1000000 && 2000000 > $6 c++ ENDprint c' access.log
To include the boundaries one can do:
grep 18/Sep/2016:11 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
or equivalently
awk '/18/Sep/2016:11/ && $6>=1000000 && $6<=2000000 c++ ENDprint c' access.log
answered Sep 18 '16 at 6:23
user000001user000001
997714
This question would be clearer if some lines of access.log were added as a sample. Anyway, the awk command prints a line regardless of the value of $6, so when you count the lines with wc -l you get an outcome that is determined by grep alone.
If you want to count the lines where is $6 is between two different values you can write
grep 17/Sep/2016:10 /access.log | awk '$6 > 1000000 && 2000000 > $6' | wc -l
However, this pipeline is a bit inefficient. It would almost always be preferable to combine it in a single awk command like this:
awk '/17/Sep/2016:10/ && $6 > 1000000 && 2000000 > $6 c++ ENDprint c' access.log
To include the boundaries one can do:
grep 18/Sep/2016:11 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
or equivalently
awk '/18/Sep/2016:11/ && $6>=1000000 && $6<=2000000 c++ ENDprint c' access.log
answered Sep 18 '16 at 6:23
user000001user000001
997714
edited Sep 18 '16 at 9:05
answered Sep 18 '16 at 6:23
user000001user000001
997714
answered Sep 18 '16 at 6:23
user000001user000001
997714
answered Sep 18 '16 at 6:23
user000001user000001
997714
997714
hey thanks a lot , i want to say one thing i used another coommand which is gave same answer.command is here is it write or not # grep 17/Sep/2016:10 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
– roshan
Sep 18 '16 at 8:47
@roshan: Sure that should also work, if you want to include the boundary values
– user000001
Sep 18 '16 at 9:03
yes its giving same value i checked many times output is same always.
– roshan
Sep 18 '16 at 9:34
add a comment |
hey thanks a lot , i want to say one thing i used another coommand which is gave same answer.command is here is it write or not # grep 17/Sep/2016:10 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
– roshan
Sep 18 '16 at 8:47
@roshan: Sure that should also work, if you want to include the boundary values
– user000001
Sep 18 '16 at 9:03
yes its giving same value i checked many times output is same always.
– roshan
Sep 18 '16 at 9:34
hey thanks a lot , i want to say one thing i used another coommand which is gave same answer.command is here is it write or not # grep 17/Sep/2016:10 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
– roshan
Sep 18 '16 at 8:47
hey thanks a lot , i want to say one thing i used another coommand which is gave same answer.command is here is it write or not # grep 17/Sep/2016:10 /access.log | awk ' $6>=1000000 && $6<=2000000' | wc -l
– roshan
Sep 18 '16 at 8:47
@roshan: Sure that should also work, if you want to include the boundary values
– user000001
Sep 18 '16 at 9:03
@roshan: Sure that should also work, if you want to include the boundary values
– user000001
Sep 18 '16 at 9:03
yes its giving same value i checked many times output is same always.
– roshan
Sep 18 '16 at 9:34
yes its giving same value i checked many times output is same always.
– roshan
Sep 18 '16 at 9:34
add a comment |
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