I am trying to do the following.
foo="foo:foo1"
cc= `$foo | cut -f2 -d:`
I understand why this would not work but I am at a loss as to do this.
Thanks in advance.
Try this:
foo="foo:foo1"
cc=`echo $foo | cut -f2 -d:`
There are 2 changes to make:
You need to echo the value of shell
variable foo and then cut it.
You must not have white spaces around =
when assigning a value to a shell
variable.
in bourne, you can use set. No external command needed.
$ foo="foo:foo1"
$ IFS=":"
$ set -- $foo
$ echo $2
foo1
Related
How can I remove all text after a character, in this case a colon (":"), in bash? Can I remove the colon, too? I have no idea how to.
In Bash (and ksh, zsh, dash, etc.), you can use parameter expansion with % which will remove characters from the end of the string or # which will remove characters from the beginning of the string. If you use a single one of those characters, the smallest matching string will be removed. If you double the character, the longest will be removed.
$ a='hello:world'
$ b=${a%:*}
$ echo "$b"
hello
$ a='hello:world:of:tomorrow'
$ echo "${a%:*}"
hello:world:of
$ echo "${a%%:*}"
hello
$ echo "${a#*:}"
world:of:tomorrow
$ echo "${a##*:}"
tomorrow
An example might have been useful, but if I understood you correctly, this would work:
echo "Hello: world" | cut -f1 -d":"
This will convert Hello: world into Hello.
$ echo 'hello:world:again' |sed 's/:.*//'
hello
I know some solutions:
# Our mock data:
A=user:mail:password
With awk and pipe:
$ echo $A | awk -v FS=':' '{print $1}'
user
Via bash variables:
$ echo ${A%%:*}
user
With pipe and sed:
$ echo $A | sed 's#:.*##g'
user
With pipe and grep:
$ echo $A | egrep -o '^[^:]+'
user
With pipe and cut:
$ echo $A | cut -f1 -d\:
user
egrep -o '^[^:]*:'
trim off everything after the last instance of ":"
grep -o '^.*:' fileListingPathsAndFiles.txt
and if you wanted to drop that last ":"
grep -o '^.*:' file.txt | sed 's/:$//'
#kp123: you'd want to replace : with / (where the sed colon should be \/)
Let's say you have a path with a file in this format:
/dirA/dirB/dirC/filename.file
Now you only want the path which includes four "/". Type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-4 -d"/"
and your output will be
/dirA/dirB/dirC
The advantage of using cut is that you can also cut out the uppest directory as well as the file (in this example), so if you type
$ echo "/dirA/dirB/dirC/filename.file" | cut -f1-3 -d"/"
your output would be
/dirA/dirB
Though you can do the same from the other side of the string, it would not make that much sense in this case as typing
$ echo "/dirA/dirB/dirC/filename.file" | cut -f2-4 -d"/"
results in
dirA/dirB/dirC
In some other cases the last case might also be helpful. Mind that there is no "/" at the beginning of the last output.
Variable b has a string. Awk retrieves a substring which I want to assign to variable c. This is what I did:
#!/bin/bash
b=$(llsubmit multiple.cmd)
echo $b | c=$(awk '{
ret=match($0,".in.")
rwt=match($0,"\" has")
rqt=rwt-(ret+4)
subs=substr($0,(ret+4),rqt)
}')
... but I get a blank output for echo $c:
You can't pipe into an assignment.
c=$(echo "$b" | awk '{
ret=match($0,".in.")
rwt=match($0,"\" has")
rqt=rwt-(ret+4)
subs=substr($0,(ret+4),rqt)
}')
(Notice also the quoting around $b.)
But your Awk script looks rather complex. And it doesn't produce any output. Should it print something at the end? Without access to sample output from llsubmit this is mildly speculative, but I'm guessing something like this could work:
c=$(echo "b" | sed -n 's/.*\(\.in\.[^"]*\)" has .*/\1/p')
(Notice also the backslashes to make the dots match literally.)
You should properly then use double quotes in echo "$c" too (unless you are completely sure that the output cannot contain any shell metacharacters).
... And, of course, very often you don't want or need to store results in a variable in shell scripts if you can refactor your code into a pipeline. Perhaps you are really looking for something like
llsubmit multiple.cmd |
sed -n 's/.*\(\.in\.[^"]\)" has .*/p' |
while read -r job; do
: things with "$job"
done
It's hard to tell from your question since you didn't provide sample input and expected output but is this what you're trying to do:
$ b='foo .in.bar has'
$ c="${b% has*}"
$ c="${c#*.in.}"
$ echo "$c"
bar
Hi I'm using GNU awk version 3.1.5 and I've specified 2 variables in a KSH script
PKNAME= ls -lt /var/db/pkg | tr -s " " | cut -d" " -f9
PKDATE= ls -lt /var/db/pkg/$PKNAME/ | tr -s " " | cut -d" " -f6-8
I'm trying to prove that I'm getting the correct output, by running a test using
echo bar
awk -F, -v pkname="$PKNAME" -v pkdate="$PKDATE" 'BEGIN{print pkname, pkdate, "foo"; exit}'
echo baz
The output from this results in 2 blank spaces and foo, like so
bar
foo
baz
I have tried, double quoting the variables, single quotes and back ticks. Also tried double quotes with back ticks.
Any ideas why the variables are not being executed? I'm fairly new to awk and appreciate any help! Thanks
I suppose it is possible that it is not possible to run a sub shell comand within an awk statement. Is this true?
This has nothing to do with awk. The problem is in the way you're assigning your variables. Your lines should be like this:
PKNAME=$(ls -lt /var/db/pkg | tr -s " " | cut -d" " -f9)
There can be no spaces around either side of an assignment in the shell.
At the moment, you're running the command ls -lt ... with a variable PKNAME temporarily assigned to an empty string. In subsequent commands the variable remains unset.
Your awk command should remain unchanged, i.e. the shell variables should be passed like -v pkname="$PKNAME". As an aside, it's generally considered bad practice to use uppercase variable names, as these should be reserved for internal use by the shell.
I have a bash variable: agent1.ip with 192.168.100.137 as its value. When I refer to it in echo like this:
echo $agent1.ip
the result is:
.ip
How can I access the value?
UPDATE: my variables are:
Bash itself doesn't understand variable names with dots in them, but that doesn't mean you can't have such a variable in your environment. Here's an example of how to set it and get it all in one:
env 'agent1.ip=192.168.100.137' bash -c 'env | grep ^agent1\\.ip= | cut -d= -f2-'
Since bash.ip is not a valid identifier in bash, the environment string bash.ip=192.168.100.37 is not used to create a shell variable on shell startup.
I would use awk, a standard tool, to extract the value from the environment.
bash_ip=$(awk 'BEGIN {print ENVIRON["bash.ip"]}')
The cleanest solution is:
echo path.data | awk '{print ENVIRON[$1]}'
Try this:
export myval=`env | grep agent1.port | awk -F'=' '{print $2}'`;echo $myval
Is your code nested, and using functions or scripts that use ksh?
Dotted variable names are an advanced feature in ksh93. A simple case is
$ a=1
$ a.b=123
$ echo ${a.b}
123
$ echo $a
1
If you first attempt to assign to a.b, you'll get
-ksh: a.b=123: no parent
IHTH
I have this:
ls */file
dir1/file dir2/file dir3/file
But I need just the first directory name, like this: dir1
I did this:
IFS="/" read foo bar <<< "$(ls */file 2>/dev/null)"
echo $foo
dir1
And it works, but now I have a problem with subshell expansion over ssh. Is there a more elegant way (without subshells or sed) to do this?
If not, I'll then post a question regarding a completely different issue - expanding subshells over ssh.
for F in */file; do
D=${F%%/*}
break
done
Another:
F=(*/file); D=${F%%/*}
Try
ls */file | cut -d"/" -f1
Use / as a separator.
You can use the tricky Double quotes!
Like so:
LIST=`ls */file`
echo "$LIST" | cut -d/ -f1
or
echo "$LIST" | awk -F/ {'print $1'}
You can use builtin read bulletin with -d option:
read -d '/' a < <(echo */file)
echo "$a"
dir1
If you just need the name of the folder you can use :
$ls -1 | awk 'NR==n'
Where n=1 is the first directory, you can change the value of n to get the nth Directory.