I have bash script that use username and password as argument, so I what to delete that command from history,
I've tried to add:
history -d ${history 1 | awk '{print $1}'}
But I get:
${history 1 | awk '{print $1}'}: bad substitution
Then I tried:
history -d ${history 1 | awk '{print $1}'}
But nothig is happened.
I also try to wite the last command's number to another file and then cat
history -d $(cat tst.txt)
and also nothing happened...
Is there any suggestion for doing that?
Thanks a lot!
I can see two ways of achieving this.
your_script; history -d $(history 1)
unset HISTFILE; your_script; HISTFILE=~/.history
Related
The following command below does not succeed.
for i in {1..5} ; do cat /etc/fstab | egrep "(ext3|ext4|xfs)" | awk '{print $2}' | cut -d"/" -f1-$i ; done
It seems that $i is ignored completely. It always returns instead result of
cut -d"/" -f1-
Any idea why it fails?
Thanks in advance!
The command itself is a part of a script that should help me to auto re-arrange fstab lines to match the right mount order (like /test/subfolder must come after /test was mounted and not before).
I tried and it didn't work for zsh shell. BUT I tried it in bash and it does work, so if you are using zsh just run the command with bash and it should work ;)
root:x:0:0:root:/root:/bin/bash
daemon:x:1:1:daemon:/usr/sbin:/usr/sbin/nologin
sys:x:3:1:sys:/dev:/usr/sbin/nologin
games:x:5:2:games:/usr/games:/usr/sbin/nologin
mail:x:8:5:mail:/var/mail:/usr/sbin/nologin
www-data:x:33:3:www-data:/var/www:/usr/sbin/nologin
backup:x:34:2:backup:/var/backups:/usr/sbin/nologin
nobody:x:65534:1337:nobody:/nonexistent:/usr/sbin/nologin
syslog:x:101:1000::/home/syslog:/bin/false
whoopsie:x:109:99::/nonexistent:/bin/false
user:x:1000:1000:edco8700,,,,:/home/user:/bin/bash
sshd:x:116:1337::/var/run/sshd:/usr/sbin/nologin
ntp:x:117:99::/home/ntp:/bin/false
mysql:x:118:999:MySQL Server,,,:/nonexistent:/bin/false
vboxadd:x:999:1::/var/run/vboxadd:/bin/false
this is an /etc/passwd file I need to do this command on. So far I have:
cut -d: -f1,6 testPasswd.txt | grep ???
that will display all the usernames and the folder associated, but I'm stuck on how to find only the ones that start with m,w,s and print the whole line.
I've tried grep -o '^[mws]*' and different variations of it, but none have worked.
Any suggestions?
Try variations of
cut -d: -f1,6 testPasswd.txt | grep '^m\|^w\|^s'
Or to put it more concisely,
cut -d: -f1,6 testPasswd.txt | grep '^[mws]'
That's neater especially if you have a lot of patterns to match.
But of course the awk solution is much better if doing it without constraints.
Easier to do with awk:
awk 'BEGIN{FS=OFS=":"} $1 ~ /^[mws]/{print $1, $6}' testPasswd.txt
sys:/dev
mail:/var/mail
www-data:/var/www
syslog:/home/syslog
whoopsie:/nonexistent
sshd:/var/run/sshd
mysql:/nonexistent
I have a file with content as follows and want to validate the content as
1.I have entries of rec$NUM and this field should be repeated 7 times only.
for example I have rec1.any_attribute this rec1 should come only 7 times in whole file.
2.I need validating script for this.
If records for rec$NUM are less than 7 or Greater than 7 script should report that record.
FILE IS AS FOLLOWS :::
rec1:sourcefile.name=
rec1:mapfile.name=
rec1:outputfile.name=
rec1:logfile.name=
rec1:sourcefile.nodename_col=
rec1:sourcefle.snmpnode_col=
rec1:mapfile.enc=
rec2:sourcefile.name=abc
rec2:mapfile.name=
rec2:outputfile.name=
rec2:logfile.name=
rec2:sourcefile.nodename_col=
rec2:sourcefle.snmpnode_col=
rec2:mapfile.enc=
rec3:sourcefile.name=abc
rec3:mapfile.name=
rec3:outputfile.name=
rec3:logfile.name=
rec3:sourcefile.nodename_col=
rec3:sourcefle.snmpnode_col=
rec3:mapfile.enc=
Please Help
Thanks in Advance... :)
Simple awk:
awk -F: '/^rec/{a[$1]++}END{for(t in a){if(a[t]!=7){print "Some error for record: " t}}}' test.rc
grep '^rec1' file.txt | wc -l
grep '^rec2' file.txt | wc -l
grep '^rec3' file.txt | wc -l
All above should return 7.
The commands:
grep rec file2.txt | cut -d':' -f1 | uniq -c | egrep -v '^ *7'
will success if file follows your rules, fails (and returns the failing record) if it doesn't.
(replace "uniq -c" by "sort -u" if record numbers can be mixed).
I am having following syntax for one of my file.Could you please anyone explain me what is this command doing
path = /document/values.txt
where we have different username specified e.g username1 = john,username2=marry
cat ${path} | grep -e username1 | cut -d'=' -f2`
my question here is cat command is reading from the file value of username1 but why why we need to use cut command?
Cat is printing the file. The file has username1=something in one of the lines. The cut command splits this and prints out the second argument.
your command was not written well. the cat is useless.
you can do:
grep -e pattern "$path"|cut ...
you can of course do it with single process with awk if you like. anyway the line in your question smells not good.
awk example:
awk -F'=' '/pattern/{print $2}' inputFile
cut -d'=' -f2`
This cut uses -d'=' that means you use '=' as 'field delimiter' and -f2 will take only de second field.
So in this case you want only the value after the "=" .
I need a grep or another like command that will pull the usernames only from the /etc/passwd file in linux. Anything before the colon. I know this is doing with reg ex however I am not nearly experienced enough...
The following command will give all ACTUAL users, I need a way to pipe to grep or another line of code to only display the username portion.
awk -v LIMIT=500 -F: '{print $1}' '($3>=LIMIT) && ($3!=65534)' /etc/passwd
This should do:
awk -F: '$3>=LIMIT && $3!=65534 {print $1}' LIMIT=500 /etc/passwd
To do this in mostly plain bash:
limit=500
nfsnobody_id=65534
cut -d: -f1,3 /etc/passwd | while IFS=: read username uid; do
(( uid >= limit && uid != nfsnobody_id )) && echo $username
done
Get out of the habit of using VARNAMES_IN_CAPS: one day you'll write PATH=$(dirname $FILE) and then wonder why commands can no longer be found.
You simply need to rewrite the filter part of your awk command a bit. For instance, the following should work:
awk -v LIMIT=500 -F: '{if (($3>=LIMIT) && ($3!=65534)) print $1}' /etc/passwd
Otherwise, to answer your question strictly, if you want to use grep, the command would be
... | grep -o "^[^:]*"
but that would not be the way to go.