Add timestamp for each line on the df output command? - linux

I would like to add the date for each line of the df output.
I tried:
df -m | awk '{print `date +%Y-%m`";"$1";"$2";"$3 }'
... but it doesn't work.
How can I add the date?

Here is an alternative:
df -m | awk '{print strftime("%Y-%m"), $0}'
And here is the output from the command above:
$ df -m | awk '{print strftime("%Y-%m"), $0}'
2019-10 Filesystem 1M-blocks Used Available Use% Mounted on
2019-10 devtmpfs 9852 0 9852 0% /dev
2019-10 tmpfs 9871 132 9740 2% /dev/shm
2019-10 tmpfs 9871 2 9869 1% /run
2019-10 /dev/mapper/fedora_canvas-root 50141 14731 32834 31% /
2019-10 tmpfs 9871 1 9871 1% /tmp
2019-10 /dev/sda5 976 243 667 27% /boot
2019-10 /dev/mapper/fedora_canvas-home 1277155 217435 994777 18% /home
2019-10 tmpfs 1975 63 1912 4% /run/user/1000
$
And here is an alternative version, printing just the 3 columns you wanted on the OP:
df -m | awk '{print strftime("%Y-%m"), $1, $2, $3}' | column -t
And the corresponding output:
$ df -m | awk '{print strftime("%Y-%m"), $1, $2, $3}' | column -t
2019-10 Filesystem 1M-blocks Used
2019-10 devtmpfs 9852 0
2019-10 tmpfs 9871 132
2019-10 tmpfs 9871 2
2019-10 /dev/mapper/fedora_canvas-root 50141 14731
2019-10 tmpfs 9871 1
2019-10 /dev/sda5 976 243
2019-10 /dev/mapper/fedora_canvas-home 1277155 217435
2019-10 tmpfs 1975 63
$

You may use this way:
df -m | awk -v dt=$(date "+%Y-%m") '{print dt "::", $0}'
We use -v dt=$(date "+%Y-%m") to execute date command in shell and pass it to awk in an argument dt.
If you want only first 3 columns from df command output then use:
df -m | awk -v dt=$(date "+%Y-%m") '{print dt, $1, $2, $3}'

A Perl solution.
df -m | perl -MPOSIX=strftime -alpe '$_ = strftime("%Y-%M ", localtime) . "#F[0..2]"'
Command line options:
-M : Load thestrftime() function from the POSIX module
-a : Autosplit the input records on whitespace into #F
-l : Remove newlines from input and add them to output
-p : Put each input record into $_, execute code and then print $_
-e : Run this code for each input record
The code updates $_ by concatenating the date (strftime("%Y-%M ", localtime)) with the first three columns (#F[0 .. 2]) of the input record.

Related

keep the header of file before sorting in csh

I have the below file.rpt:
Data Manhattan Wire Wire
Delay Distance Length Diff[%]
#############################
0.558 728.282 738.89 1%
0.481 255.258 285.887 12%
0.552 652.888 677.985 4%
0.431 420.896 477.73 14%
0.506 580.288 633.018 9%
0.388 350.22 463.05 32%
0.541 622.088 672.761 8%
0.462 384.064 586.98 53%
0.565 574.016 594.434 4%
0.470 339.268 450.927 33%
0.566 664.304 672.092 1%
0.441 436.428 534.146 22%
I'm trying to sort one of the column and keep the 3 lines header without any change.
How can I do it without manipulate to other file?
I tried something similar to the below:
head -3 file.rpt ; sort -nrk4,4 file.rpt | -
I suppose this doesn't really count as a csh answer, but:
sh -c '{ sed 3q file.rpt; sed 1,3d file.rpt | sort -nrk4,4; }'
In csh, you can do:
( sed 3q file.rpt; sed 1,3d file.rpt | sort -nrk4,4; )
You can do this:
head -3 file.rpt ; tail -n +4 file.rpt | sort -nrk4,4
-n: output the last X lines
+4: output starting with line 4. Hence skip the first 3 lines
use | to send the output of tail as the input to sort.
Result:
Data Manhattan Wire Wire
Delay Distance Length Diff[%]
#############################
0.462 384.064 586.98 53%
0.470 339.268 450.927 33%
0.388 350.22 463.05 32%
0.441 436.428 534.146 22%
0.431 420.896 477.73 14%
0.481 255.258 285.887 12%
0.506 580.288 633.018 9%
0.541 622.088 672.761 8%
0.565 574.016 594.434 4%
0.552 652.888 677.985 4%
0.566 664.304 672.092 1%
0.558 728.282 738.89 1%

How do I get grep to search for any number I put in the variable?

So at the beginning of my script, I am defining "threshold," which has a number that isn't going to exist in the next part, (2) and a number that will exist in the next part (6). I'm having the result of running df -h to a file called dffile. My question, is how do I get grep in line 7 to search all of the variable "threshold" for the number that will exist in the file? It works if I have the 6 before the 2 in the variable, so it seems as if it's only searching the first number in it. Thanks!
#!/bin/bash
threshold=("2%" "6%")
df -h > dffile
grep $threshold dffile >> thresh
cat thresh | awk '{print $6}' >> finding1
LINES=()
while IFS= read -r finding1
do
find $finding1 -xdev -size +40M -exec ls -lah {} \; | head -n 10
done < "finding1"
The output of df -h on my test server is:
root#tstd0001:~/scripts# df -h
Filesystem Size Used Avail Use% Mounted on
udev 481M 0 481M 0% /dev
tmpfs 99M 616K 98M 1% /run
/dev/vda1 25G 1.3G 23G 6% /
tmpfs 493M 0 493M 0% /dev/shm
tmpfs 5.0M 0 5.0M 0% /run/lock
tmpfs 493M 0 493M 0% /sys/fs/cgroup
/dev/vda15 105M 3.4M 102M 4% /boot/efi
tmpfs 99M 0 99M 0% /run/user/0
As you can see above, "2" from my variable, does not exist, whereas "6" does. My goal is to make grep find any number that matches a number inside the variable.
Let's consider that cat output_df is the output of your df -h command:
$ cat output_df
Filesystem Size Used Avail Use% Mounted on
udev 481M 0 481M 2% /dev
tmpfs 99M 616K 98M 1% /run
/dev/vda1 25G 1.3G 23G 6% /
tmpfs 493M 0 493M 6% /dev/shm
tmpfs 5.0M 0 5.0M 2% /run/lock
tmpfs 493M 0 493M 0% /sys/fs/cgroup
/dev/vda15 105M 3.4M 102M 4% /boot/efi
tmpfs 99M 0 99M 0% /run/user/0
Then you change the first part of your script in the following way:
thresold="2%|6%"
cat output_df | awk -v VAR=$thresold '{if($5~VAR)print $6}'
of course you will have to replace cat output_df by df -h in the final script.
This will give the output:
/dev
/
/dev/shm
/run/lock
Explanations:
thresold="2%|6%" is a regex that matches 2% or 6% you can generalize it to "2%|6%|X%|Y%|...|Z%"
You pass it as a variable to awk via -v VAR=$thresold
Then you command awk to print the 6th field when the 5th field does match the regex that you have passed to it via '{if($5~VAR)print $6}'
Then you can regroup everything without using intermediate files:
thresold="2%|6%"
for f in `df -h | awk -v VAR=$thresold '{if($5~VAR)print $6}'`
do
find $f -xdev -size +40M -exec ls -lah {} \; 2>/dev/null | head -n10
done
On my box it gives the following output:
-rw-r--r-- 1 root root 47M 6月 1 06:33 /var/cache/apt/srcpkgcache.bin
-rw-r--r-- 1 root root 47M 6月 1 06:33 /var/cache/apt/pkgcache.bin
-rw-r--r-- 1 root root 62M 4月 28 02:27 /usr/lib/jvm/java-8-openjdk-amd64/jre/lib/rt.jar
-rw-r--r-- 1 root root 56M 2月 20 06:10 /usr/lib/libreoffice/program/libmergedlo.so
-rw-r--r-- 1 root root 73M 5月 22 19:31 /usr/lib/thunderbird/libxul.so
-rwxr-xr-x 1 root root 142M 5月 22 01:35 /usr/lib/chromium-browser/chromium-browser
-rw-r--r-- 1 root root 41M 5月 8 09:57 /usr/lib/x86_64-linux-gnu/libwebkit2gtk-4.0.so.37.28.2
-rw-r--r-- 1 root root 54M 11月 17 2017 /usr/lib/x86_64-linux-gnu/libLLVM-5.0.so.1
-rw-r--r-- 1 root root 99M 3月 18 2017 /usr/lib/x86_64-linux-gnu/libOxideQtCore.so.0
-rwxr-xr-x 1 root root 42M 5月 8 09:57 /usr/lib/x86_64-linux-gnu/webkit2gtk-4.0/WebKitPluginProcess2
...
Notes: 2>/dev/null this redirection is to remove all the permission errors by redirecting stderr to /dev/null (muting the stderr)
You initialize threshold as an indexed array,
threshold=("2%" "6%")
You then call grep with:
grep $threshold dffile
Since threshold is an array, to dereference all values in the array, you use the form:
${threshold[#]} ## which can be quoted to preserve whitespace in elements
When you deference the array as a normal variable, e.g. $threshold you return only the 1st element, e.g.
echo $threshold ## output '2%'
So before going further, you need to determine what you want to pass to grep, if you want to search for either 2% or 6%, then Allan has a nice explanation above. You can also construct the grep expression using printf -v, e.g.
printf -v gexp "%s\|%s" ${threshold[#]}
or to properly limit to the first 2-elements,
printf -v gexp "%s\|%s" "${threshold[0]}" "${threshold[1]}"
and then call grep with
grep "$gexp" dffile >> thresh
Let me know if you have further questions.

overall CPU usage and Memory(RAM) usage in percentage in linux/ubuntu

I want to findout overall CPU usage and RAM usage in percentage, but i dint get success
$ command for cpu usage
4.85%
$ command for memory usage
15.15%
OR
$ command for cpu and mamory usage
cpu: 4.85%
mem: 15.15%
How can I achieve this?
You can use top and/or vmstat from the procps package.
Use vmstat -s to get the amount of RAM on your machine (optional), and
then use the output of top to calculate the memory usage percentages.
%Cpu(s): 3.8 us, 2.8 sy, 0.4 ni, 92.0 id, 1.0 wa, 0.0 hi, 0.0 si, 0.0 st
KiB Mem : 24679620 total, 1705524 free, 7735748 used, 15238348 buff/cache
KiB Swap: 0 total, 0 free, 0 used. 16161296 avail Mem
You can also do this for relatively short output:
watch '/usr/bin/top -b | head -4 | tail -2'
A shell pipe that calculates the current RAM usage periodically is
watch -n 5 "/usr/bin/top -b | head -4 | tail -2 | perl -anlE 'say sprintf(\"used: %s total: %s => RAM Usage: %.1f%%\", \$F[7], \$F[3], 100*\$F[7]/\$F[3]) if /KiB Mem/'"
(CPU + Swap usages were filtered out here.)
This command prints every 5 seconds:
Every 5.0s: /usr/bin/top -b | head -4 | tail -2 | perl -anlE 'say sprintf("u... wb3: Wed Nov 21 13:51:49 2018
used: 8349560 total: 24667856 => RAM Usage: 33.8%
Please use one of the following:
$ free -t | awk 'NR == 2 {print "Current Memory Utilization is : " $3/$2*100}'
Current Memory Utilization is : 14.6715
OR
$ free -t | awk 'FNR == 2 {print "Current Memory Utilization is : " $3/$2*100}'
Current Memory Utilization is : 14.6703
CPU usage => top -bn2 | grep '%Cpu' | tail -1 | grep -P '(....|...) id,' | awk '{print 100-$8 "%"}'
Memory usage => free -m | grep 'Mem:' | awk '{ print $3/$2*100 "%"}'
For the cpu usage% you can use:
top -b -n 1| grep Cpu | awk -F "," '{print $4}' | awk -F "id" '{print $1}' | awk -F "%" '{print $1}'
One liner solution to get RAM % in-use:
free -t | awk 'FNR == 2 {printf("%.0f%"), $3/$2*100}'
Example output: 24%
for more precision, you can change the integer N inside printf(%.<N>%) from the the previous command. For example to get 2 decimal places of precision you could do:
free -t | awk 'FNR == 2 {printf("%.2f%"), $3/$2*100}'
Example output: 24.57%

How to find the highest disk space usage mount?

I'm looking for a command where only the highest disk space usage mount will be shown. So The maximum %usage mount will be shown.
Running df -h
Filesystem Size Used Avail Use% Mounted on
/dev/vx/dsk/appdg/boom-vol
1.0G 19M 943M 2% /opt/blah99
500G 349G 152G 70% /opt/blah/data
/dev/vx/dsk/isdg/boom-shared-vol
50G 1.6G 46G 4% /opt/blah99/product/shared
/dev/vx/dsk/isdg/boom-bc-vol
150G 64G 81G 45% /opt/blah99/product/a_01
/dev/vx/dsk/isdg/boom-bt-vol
150G 47G 98G 33% /opt/blah99/product/a_02
Output should be -
500G 349G 152G 70% /opt/blah/data
What you are looking for is:
df -h | grep -vw "^\/dev" | sort -k 5 -n | tail -n 2 | head -n 1
Output of df -h | grep -vw "^\/dev":
Filesystem Size Used Avail Use% Mounted on
1.0G 19M 943M 2% /opt/blah99
500G 349G 152G 70% /opt/blah/data
50G 1.6G 46G 4% /opt/blah99/product/shared
150G 64G 81G 45% /opt/blah99/product/a_01
150G 47G 98G 33% /opt/blah99/product/a_02
Sorting by column 5 in numeric order: df -h | grep -vw "^\/dev" | sort -k 5 -n:
50G 1.6G 46G 4% /opt/blah99/product/shared
1.0G 19M 943M 2% /opt/blah99
150G 47G 98G 33% /opt/blah99/product/a_02
150G 64G 81G 45% /opt/blah99/product/a_01
500G 349G 152G 70% /opt/blah/data
Filesystem Size Used Avail Use% Mounted on
Getting second row from the end: df -h | grep -vw "^\/dev" | sort -k 5 -n | tail -n 2 | head -n 1:
500G 349G 152G 70% /opt/blah/data

How to omit heading in df -k command of SunOs

Input: df -k
Output:
Filesystem kbytes used avail capacity Mounted on
/dev/dsk/c0t0d0s0 10332220 443748 9785150 5% /
/devices 0 0 0 0% /devices
ctfs 0 0 0 0% /system/contract
proc 0 0 0 0% /proc
mnttab 0 0 0 0% /etc/mnttab
swap 45475864 1688 45474176 1% /etc/svc/volatile
objfs 0 0 0 0% /system/object
sharefs 0 0 0 0% /etc/dfs/sharetab
/dev/dsk/c0t0d0s3 10332220 3513927 6714971 35% /usr
I want to omit the 1st line Filesystem kbytes used avail capacity Mounted on from the output.
I used df -k | tail -n+2 in linux to get exactly what i wanted, but in SunOs I get
zenvo% df -k | tail -n+2
usage: tail [+/-[n][lbc][f]] [file]
tail [+/-[n][l][r|f]] [file]
How can i achieve the Required output:
/dev/dsk/c0t0d0s0 10332220 443748 9785150 5% /
/devices 0 0 0 0% /devices
ctfs 0 0 0 0% /system/contract
proc 0 0 0 0% /proc
mnttab 0 0 0 0% /etc/mnttab
swap 45475864 1688 45474176 1% /etc/svc/volatile
objfs 0 0 0 0% /system/object
sharefs 0 0 0 0% /etc/dfs/sharetab
/dev/dsk/c0t0d0s3 10332220 3513927 6714971 35% /usr
Note: No. of rows might change
I know it's an old thread, but the shortest and the clearest of all:
df -k | sed 1d
I haven't used SunOS but using sed you should be able to delete the first line like this:
df -k | sed -e /Filesystem/d
edit: But you would have to be careful that the word Filesystem doesn't show up elsewhere in the output. A better solution would be:
df -k | sed -e /^Filesystem/d
If you want to omit the first line of any result, you can use tail:
<command> | tail -n +2
So in your case:
df -k | tail -n +2
https://man7.org/linux/man-pages/man1/tail.1.html
What about:
df -k | tail -$((`df -k | wc -l`-1))

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