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Old 07-05-2011, 08:37 AM   #1
alirz
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how to get free memory stat for the past few days


im trying to collect some data.
Is there a way to get the free memory stats over the past few days from some file...
i looked into the sar log files but they dont have date stamps in them!
 
Old 07-05-2011, 08:56 AM   #2
druuna
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Hi,

Sar, if set up correctly, places 31 files in /var/log/sa. Each file represents a previous date/day.

Code:
sar -f /var/log/sa/sa04
The above example will show yesterdays info.

Hope this helps.
 
Old 07-05-2011, 08:57 AM   #3
alirz
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Or if somehow its possible to collect some memory data for the past few days from the sar files that linux stores at /var/log/sa/

for a given day's sar file(its a very long file) e.g sar21, somewhere in the file the memory is listed as:

00:00:01 kbmemfree kbmemused %memused kbbuffers kbcached kbswpfree kbswpused %swpused kbswpcad
00:10:01 1385912 48065600 97.20 475672 23967824 16478504 301348 1.80 288644
00:20:01 2993564 46457948 93.95 464720 22337572 16478504 301348 1.80 288708

and towards the end of the day it also lists the average:

23:50:01 1003292 48448220 97.97 666296 25555924 16478500 301352 1.80 290164
Average: 1096417 48355095 97.78 555984 25517802 16478500 301352 1.80 290096

I'm trying to find a way to be able to get just the Average for the kbcached column along with the date that sar file was created as there is no date information inside the file, just time.
all my awk, grep attempts are not giving me what i need. Could someone please help?

Im trying to get my output to look something like:

Jul 1 23454545
Jul 2 78645654
Jul 3 76735443
jul 4 63537644
 
Old 07-05-2011, 08:59 AM   #4
alirz
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Quote:
Originally Posted by druuna View Post
Hi,

Sar, if set up correctly, places 31 files in /var/log/sa. Each file represents a previous date/day.

Code:
sar -f /var/log/sa/sa04
The above example will show yesterdays info.

Hope this helps.
thanks druuna but that gives me cpu info and thats not what im trying to get . Thanks though.
 
Old 07-05-2011, 09:20 AM   #5
druuna
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Hi,

The example I gave was just that; An example. It wasn't clear at the time of replying that you knew /var/log/sa holds all the daily files

About your problem:

The date _is_ in the file (first line). A quick one-liner I came up with:
Code:
S_TIME_FORMAT=ISO sar -r -f /var/log/sa/sa22 | awk 'NR == 1 { printf $4 } ; /Average:/ { print " : " $6 }'

# output
2011-06-22 : 1364496
Not exactly what you want/need, but I hope it helps.
 
Old 07-05-2011, 09:36 AM   #6
alirz
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Quote:
Originally Posted by druuna View Post
Hi,

The example I gave was just that; An example. It wasn't clear at the time of replying that you knew /var/log/sa holds all the daily files

About your problem:

The date _is_ in the file (first line). A quick one-liner I came up with:
Code:
S_TIME_FORMAT=ISO sar -r -f /var/log/sa/sa22 | awk 'NR == 1 { printf $4 } ; /Average:/ { print " : " $6 }'

# output
2011-06-22 : 1364496
Not exactly what you want/need, but I hope it helps.
well the result it gives me for that day is good. Shouldnt it be possible to run this command on all sa files in the directory?
Also any difference between the sa and the sar files?
 
Old 07-05-2011, 09:47 AM   #7
druuna
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Hi,
Quote:
Originally Posted by alirz View Post
well the result it gives me for that day is good. Shouldnt it be possible to run this command on all sa files in the directory?
Yes, that is possible:
Code:
for FILE in /var/log/sa/sa[0-9]*; do S_TIME_FORMAT=ISO sar -r -f $FILE | awk 'NR == 1 { printf $4 } ; /Average:/ { print " : " $6 }'; done
Quote:
Also any difference between the sa and the sar files?
Sa files are data files, sar files are reports.
 
  


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