I have a product which has a command called db2level whose output is given below
I need to extract 8.1.1.64 out of it, so far i came up with,
db2level | grep "DB2 v" | awk '{print$5}'
which gave me an output v8.1.1.64",
Please help me to fetch 8.1.1.64. Thanks
grep is enough to do that:
db2level| grep -oP '(?<="DB2 v)[\d.]+(?=", )'
Just with awk:
db2level | awk -F '"' '$2 ~ /^DB2 v/ {print substr($2,6)}'
db2level | grep "DB2 v" | awk '{print$5}' | sed 's/[^0-9\.]//g'
remove all but numbers and dot
sed is your friend for general extraction tasks:
db2level | sed -n -e 's/.*tokens are "DB2 v\([0-9.]*\)".*/\1/p'
The sed line does print no lines (the -n) but those where a replacement with the given regexp can happen. The .* at the beginning and the end of the line ensure that the whole line is matched.
Try grep with -o option:
db2level | grep -E -o "[0-9]+\.[0-9]+\.[0-9]\+[0-9]+"
Another sed solution
db2level | sed -n -e '/v[0-9]/{s/.*DB2 v//;s/".*//;p}'
This one desn't rely on the number being in a particular format, just in a particular place in the output.
db2level | grep -o "v[0-9.]*" | tr -d v
Try s.th. like db2level | grep "DB2 v" | cut -d'"' -f2 | cut -d'v' -f2
cut splits the input in parts, seperated by delimiter -d and outputs field number -f
Related
I've got the following line:
wc -l ./*.txt | sort -rn
i want to cut the file extension. So with this code i've got the output:
number filename.txt
for all my .txt-files in the .-directory. But I want the output without the file-extension, like this:
number filename
I tried a pipe with cut for different kinds of parameter, but all i got was to cut the whole filename with this command.
wc -l ./*.txt | sort -rn | cut -f 1 -d '.'
Assuming you don't have newlines in your filename you can use sed to strip out ending .txt:
wc -l ./*.txt | sort -rn | sed 's/\.txt$//'
unfortunately, cut doesn't have a syntax for extracting columns according to an index from the end. One (somewhat clunky) trick is to use rev to reverse the line, apply cut to it and then rev it back:
wc -l ./*.txt | sort -rn | rev | cut -d'.' -f2- | rev
Using sed in more generic way to cut off whatever extension the files have:
$ wc -l *.txt | sort -rn | sed 's/\.[^\.]*$//'
14 total
8 woc
3 456_base
3 123_base
0 empty_base
A better approach using proper mime type (what is the extension of tar.gz or such multi extensions ? )
#!/bin/bash
for file; do
case $(file -b $file) in
*ASCII*) echo "this is ascii" ;;
*PDF*) echo "this is pdf" ;;
*) echo "other cases" ;;
esac
done
This is a POC, not tested, feel free to adapt/improve/modify
I am trying to read/grep a particular word or content that is before a period (.).
e.g. file1 has abinaya.ashok and I want to grep whatever is before the period (.) without hardcoding anything.
if I try
grep \.\ file1
it gives abinaya.ashok.
I've tried: grep\*\.\ file1
it doesn't give anything.Can we find it using grep commands or should we do it only using awk command? Any thoughts?
Using GNU grep for PCRE regex (for non-greedy and positive look-ahead), you can do:
echo 'abinaya.ashok' | grep -oP '.*?(?=\.)'
abinaya
Using awk:
echo 'abinaya.ashok' | awk -F\. '{print $1}'
abinaya
Check the following simple examples.
Including the dot:
$ echo abinaya.ashok | grep -o '.*[.]'
abinaya.
Without the dot:
$ echo abinaya.ashok | grep -o '^[^.]\+'
abinaya
Hope I understand you correctly:
sed -n 's/\..*//p' file1 | grep whatever
sed expression will print only part before dot (lines without dot are not printed).
Now use grep to search what you need.
I need to get a list of matches with grep including filename and line number but without the match string
I know that grep -Hl will give only file names and grep -Hno will give filename with only matching string. But those not ideal for me. I need to get a list without match but with line no. For this grep -Hln doesn't work. I tried with grep -Hn 'pattern' | cut -d " " -f 1 But it doesn't cut the filename and line no properly.
awk can do that in single command:
awk '/pattern/ {print FILENAME ":" NR}' *.txt
You were pointing it well with cut, only that you need the : field separator. Also, I think you need the first and second group. Hence, use:
grep -Hn 'pattern' files* | cut -d: -f1,2
Sample
$ grep -Hn a a*
a:3:are
a:10:bar
a:11:that
a23:1:hiya
$ grep -Hn a a* | cut -d: -f1,2
a:3
a:10
a:11
a23:1
I guess you want this, just line numbers:
grep -nh PATTERN /path/to/file | cut -d: -f1
example output:
12
23
234
...
Unfortunately you'll need to use cut here. There is no way to do it with pure grep.
Try
grep -RHn Studio 'pattern' | awk -F: '{print $1 , ":", $2}'
I've a command getting the current SVN Revision and storing it in a file, is there anyway I can select the "53413" from the file to use elsewhere?
Revision: 53413
Thanks
echo "Revision: 53413" | cut -d " " -f2
cut usage: Using space as delimiter, select the second field.
This is a bit more precise, in case filename contains more than one line of data:
rev=`awk '$1=="Revision:"{print $2}' <filename>`
Then, you can use the ${rev} elsewhere in your bash script.
You could use grep:
echo "Revision: 53413" | grep -o -P "\d+"
Or if your file has more lines you could use:
cat file | grep Revision | grep -o -P "\d+"
With file data containing
dddd 2345
try following lines
$ REV=`cat data| awk '{print $2}' `
$ echo $REV
Output is
2345
Is there any way to specify a field delimiter for more spaces with the cut command? (like " "+) ?
For example: In the following string, I like to reach value '3744', what field delimiter I should say?
$ps axu | grep jboss
jboss 2574 0.0 0.0 3744 1092 ? S Aug17 0:00 /bin/sh /usr/java/jboss/bin/run.sh -c example.com -b 0.0.0.0
cut -d' ' is not what I want, for it's only for one single space.
awk is not what I am looking for either, but how to do with 'cut'?
thanks.
Actually awk is exactly the tool you should be looking into:
ps axu | grep '[j]boss' | awk '{print $5}'
or you can ditch the grep altogether since awk knows about regular expressions:
ps axu | awk '/[j]boss/ {print $5}'
But if, for some bizarre reason, you really can't use awk, there are other simpler things you can do, like collapse all whitespace to a single space first:
ps axu | grep '[j]boss' | sed 's/\s\s*/ /g' | cut -d' ' -f5
That grep trick, by the way, is a neat way to only get the jboss processes and not the grep jboss one (ditto for the awk variant as well).
The grep process will have a literal grep [j]boss in its process command so will not be caught by the grep itself, which is looking for the character class [j] followed by boss.
This is a nifty way to avoid the | grep xyz | grep -v grep paradigm that some people use.
awk version is probably the best way to go, but you can also use cut if you firstly squeeze the repeats with tr:
ps axu | grep jbos[s] | tr -s ' ' | cut -d' ' -f5
# ^^^^^^^^^^^^ ^^^^^^^^^ ^^^^^^^^^^^^^
# | | |
# | | get 5th field
# | |
# | squeeze spaces
# |
# avoid grep itself to appear in the list
I like to use the tr -s command for this
ps aux | tr -s [:blank:] | cut -d' ' -f3
This squeezes all white spaces down to 1 space. This way telling cut to use a space as a delimiter is honored as expected.
I am going to nominate tr -s [:blank:] as the best answer.
Why do we want to use cut? It has the magic command that says "we want the third field and every field after it, omitting the first two fields"
cat log | tr -s [:blank:] |cut -d' ' -f 3-
I do not believe there is an equivalent command for awk or perl split where we do not know how many fields there will be, ie out put the 3rd field through field X.
Shorter/simpler solution: use cuts (cut on steroids I wrote)
ps axu | grep '[j]boss' | cuts 4
Note that cuts field indexes are zero-based so 5th field is specified as 4
http://arielf.github.io/cuts/
And even shorter (not using cut at all) is:
pgrep jboss
One way around this is to go:
$ps axu | grep jboss | sed 's/\s\+/ /g' | cut -d' ' -f3
to replace multiple consecutive spaces with a single one.
Personally, I tend to use awk for jobs like this. For example:
ps axu| grep jboss | grep -v grep | awk '{print $5}'
As an alternative, there is always perl:
ps aux | perl -lane 'print $F[3]'
Or, if you want to get all fields starting at field #3 (as stated in one of the answers above):
ps aux | perl -lane 'print #F[3 .. scalar #F]'
If you want to pick columns from a ps output, any reason to not use -o?
e.g.
ps ax -o pid,vsz
ps ax -o pid,cmd
Minimum column width allocated, no padding, only single space field separator.
ps ax --no-headers -o pid:1,vsz:1,cmd
3443 24600 -bash
8419 0 [xfsalloc]
8420 0 [xfs_mru_cache]
8602 489316 /usr/sbin/apache2 -k start
12821 497240 /usr/sbin/apache2 -k start
12824 497132 /usr/sbin/apache2 -k start
Pid and vsz given 10 char width, 1 space field separator.
ps ax --no-headers -o pid:10,vsz:10,cmd
3443 24600 -bash
8419 0 [xfsalloc]
8420 0 [xfs_mru_cache]
8602 489316 /usr/sbin/apache2 -k start
12821 497240 /usr/sbin/apache2 -k start
12824 497132 /usr/sbin/apache2 -k start
Used in a script:-
oldpid=12824
echo "PID: ${oldpid}"
echo "Command: $(ps -ho cmd ${oldpid})"
Another way if you must use cut command
ps axu | grep [j]boss |awk '$1=$1'|cut -d' ' -f5
In Solaris, replace awk with nawk or /usr/xpg4/bin/awk
I still like the way Perl handles fields with white space.
First field is $F[0].
$ ps axu | grep dbus | perl -lane 'print $F[4]'
My approach is to store the PID to a file in /tmp, and to find the right process using the -S option for ssh. That might be a misuse but works for me.
#!/bin/bash
TARGET_REDIS=${1:-redis.someserver.com}
PROXY="proxy.somewhere.com"
LOCAL_PORT=${2:-6379}
if [ "$1" == "stop" ] ; then
kill `cat /tmp/sshTunel${LOCAL_PORT}-pid`
exit
fi
set -x
ssh -f -i ~/.ssh/aws.pem centos#$PROXY -L $LOCAL_PORT:$TARGET_REDIS:6379 -N -S /tmp/sshTunel$LOCAL_PORT ## AWS DocService dev, DNS alias
# SSH_PID=$! ## Only works with &
SSH_PID=`ps aux | grep sshTunel${LOCAL_PORT} | grep -v grep | awk '{print $2}'`
echo $SSH_PID > /tmp/sshTunel${LOCAL_PORT}-pid
Better approach might be to query for the SSH_PID right before killing it, since the file might be stale and it would kill a wrong process.