Given paths like this:
/data/mirrors/third-party/centos/5/projectA/x86_64
/data/mirrors/third-party/centos/5/projectA/i386
/data/mirrors/third-party/centos/5/projectA/noarch
/data/mirrors/third-party/centos/4/projectB/x86_64
/data/mirrors/third-party/centos/4/projectB/i386
/data/mirrors/third-party/centos/4/projectB/noarch
/data/mirrors/third-party/centos/4/projectC/x86_64
/data/mirrors/third-party/centos/4/projectC/i386
/data/mirrors/third-party/centos/4/projectC/noarch
How can I grab the values from field 5 and 7 ('5' and 'x86_64') using Bash shell commands?
I have something like this so far, but I'm looking for something more elegant, and without the need to capture the 'junk*':
cd /data/mirrors/third-party/centos/5/project/x86_64
echo `pwd` | tr '/' ' ' | while read junk1 junk2 junk3 junk4 version junk5 arch; do
echo version=$version arch=$arch
done
version=5 arch=x86_64
This works for me:
pwd | awk -F'/' '{print "version=" $6 " arch=" $8}'
You can use IFS and an array to split the directory into its components:
#!/bin/bash
saveIFS=$IFS
IFS='/'
dirs=($(pwd))
IFS=$saveIFS
version=${dirs[5]}
arch=${dirs[7]}
> p=$(pwd)
> echo $p
/data/mirrors/third-party/centos/5/projectA/x86_64
> basename ${p}
x86_64
> basename ${p%/*/*}
5
You can also use something like:
echo `expr match "$p" '<regular-expression>'`
...perhaps someone might help me with that regular expression ;)
try this
echo `pwd` | cut -d'/' -f6,8 | tr '/' ' '
to display field
or to display with sring version and arch
echo `pwd` | cut -d'/' -f6,8 | sed -e 's/\(.*\)\/\(.*\)/version=\1 arch=\2/'
Related
I have a String like this
//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf
and want to get last part of
00000000957481f9-08d035805a5c94bf
Let's say you have
text="//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf"
If you know the position, i.e. in this case the 9th, you can go with
echo "$text" | cut -d'/' -f9
However, if this is dynamic and your want to split at "/", it's safer to go with:
echo "${text##*/}"
This removes everything from the beginning to the last occurrence of "/" and should be the shortest form to do it.
For more information on this see: Bash Reference manual
For more information on cut see: cut man page
The tool basename does exactly that:
$ basename //ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf
00000000957481f9-08d035805a5c94bf
I would use bash string function:
$ string="//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf"
$ echo "${string##*/}"
00000000957481f9-08d035805a5c94bf
But following are some other options:
$ awk -F'/' '$0=$NF' <<< "$string"
00000000957481f9-08d035805a5c94bf
$ sed 's#.*/##g' <<< "$string"
00000000957481f9-08d035805a5c94bf
Note: <<< is herestring notation. They do not create a subshell, however, they are NOT portable to POSIX sh (as implemented by shells such as ash or dash).
In case you want more than just the last part of the path,
you could do something like this:
echo $PWD | rev | cut -d'/' -f1-2 | rev
You can use this BASH regex:
s='//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf'
[[ "$s" =~ [^/]+$ ]] && echo "${BASH_REMATCH[0]}"
00000000957481f9-08d035805a5c94bf
This can be done easily in awk:
string="//ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf"
echo "${string}" | awk -v FS="/" '{ print $NF }'
Use "/" as field separator and print the last field.
You can try this...
echo //ABC/REC/TLC/SC-prod/1f9/20/00000000957481f9-08d035805a5c94bf |awk -F "/" '{print $NF}'
I'm trying to make a small function that removes all the chars that are not digits.
123a45a ---> will become ---> 12345
I've came up with :
temp=$word | grep -o [[:digit:]]
echo $temp
But instead of 12345 I get 1 2 3 4 5. How to I get rid of the spaces?
Pure bash:
word=123a45a
number=${word//[^0-9]}
Here's a pure bash solution
var='123a45a'
echo ${var//[^0-9]/}
12345
is this what you are looking for?
kent$ echo "123a45a"|sed 's/[^0-9]//g'
12345
grep & tr
echo "123a45a"|grep -o '[0-9]'|tr -d '\n'
12345
I would recommend using sed or perl instead:
temp="$(sed -e 's/[^0-9]//g' <<< "$word")"
temp="$(perl -pe 's/\D//g' <<< "$word")"
Edited to add: If you really need to use grep, then this is the only way I can think of:
temp="$( grep -o '[0-9]' <<< "$word" \
| while IFS= read -r ; do echo -n "$REPLY" ; done
)"
. . . but there's probably a better way. (It uses grep -o, like your solution, then runs over the lines that it outputs and re-outputs them without line-breaks.)
Edited again to add: Now that you've mentioned that you use can use tr instead, this is much easier:
temp="$(tr -cd 0-9 <<< "$word")"
What about using sed?
$ echo "123a45a" | sed -r 's/[^0-9]//g'
12345
As I read you are just allowed to use grep and tr, this can make the trick:
$ echo "123a45a" | grep -o [[:digit:]] | tr -d '\n'
12345
In your case,
temp=$(echo $word | grep -o [[:digit:]] | tr -d '\n')
tr will also work:
echo "123a45a" | tr -cd '[:digit:]'
# output: 12345
Grep returns the result on different lines:
$ echo -e "$temp"
1
2
3
4
5
So you cannot remove those spaces during the filtering, but you can afterwards, since $temp can transform itself like this:
temp=`echo $temp | tr -d ' '`
$ echo "$temp"
12345
Suppose I have the string 1:2:3:4:5 and I want to get its last field (5 in this case). How do I do that using Bash? I tried cut, but I don't know how to specify the last field with -f.
You can use string operators:
$ foo=1:2:3:4:5
$ echo ${foo##*:}
5
This trims everything from the front until a ':', greedily.
${foo <-- from variable foo
## <-- greedy front trim
* <-- matches anything
: <-- until the last ':'
}
Another way is to reverse before and after cut:
$ echo ab:cd:ef | rev | cut -d: -f1 | rev
ef
This makes it very easy to get the last but one field, or any range of fields numbered from the end.
It's difficult to get the last field using cut, but here are some solutions in awk and perl
echo 1:2:3:4:5 | awk -F: '{print $NF}'
echo 1:2:3:4:5 | perl -F: -wane 'print $F[-1]'
Assuming fairly simple usage (no escaping of the delimiter, for example), you can use grep:
$ echo "1:2:3:4:5" | grep -oE "[^:]+$"
5
Breakdown - find all the characters not the delimiter ([^:]) at the end of the line ($). -o only prints the matching part.
You could try something like this if you want to use cut:
echo "1:2:3:4:5" | cut -d ":" -f5
You can also use grep try like this :
echo " 1:2:3:4:5" | grep -o '[^:]*$'
One way:
var1="1:2:3:4:5"
var2=${var1##*:}
Another, using an array:
var1="1:2:3:4:5"
saveIFS=$IFS
IFS=":"
var2=($var1)
IFS=$saveIFS
var2=${var2[#]: -1}
Yet another with an array:
var1="1:2:3:4:5"
saveIFS=$IFS
IFS=":"
var2=($var1)
IFS=$saveIFS
count=${#var2[#]}
var2=${var2[$count-1]}
Using Bash (version >= 3.2) regular expressions:
var1="1:2:3:4:5"
[[ $var1 =~ :([^:]*)$ ]]
var2=${BASH_REMATCH[1]}
$ echo "a b c d e" | tr ' ' '\n' | tail -1
e
Simply translate the delimiter into a newline and choose the last entry with tail -1.
Using sed:
$ echo '1:2:3:4:5' | sed 's/.*://' # => 5
$ echo '' | sed 's/.*://' # => (empty)
$ echo ':' | sed 's/.*://' # => (empty)
$ echo ':b' | sed 's/.*://' # => b
$ echo '::c' | sed 's/.*://' # => c
$ echo 'a' | sed 's/.*://' # => a
$ echo 'a:' | sed 's/.*://' # => (empty)
$ echo 'a:b' | sed 's/.*://' # => b
$ echo 'a::c' | sed 's/.*://' # => c
There are many good answers here, but still I want to share this one using basename :
basename $(echo "a:b:c:d:e" | tr ':' '/')
However it will fail if there are already some '/' in your string.
If slash / is your delimiter then you just have to (and should) use basename.
It's not the best answer but it just shows how you can be creative using bash commands.
If your last field is a single character, you could do this:
a="1:2:3:4:5"
echo ${a: -1}
echo ${a:(-1)}
Check string manipulation in bash.
Using Bash.
$ var1="1:2:3:4:0"
$ IFS=":"
$ set -- $var1
$ eval echo \$${#}
0
echo "a:b:c:d:e"|xargs -d : -n1|tail -1
First use xargs split it using ":",-n1 means every line only have one part.Then,pring the last part.
Regex matching in sed is greedy (always goes to the last occurrence), which you can use to your advantage here:
$ foo=1:2:3:4:5
$ echo ${foo} | sed "s/.*://"
5
A solution using the read builtin:
IFS=':' read -a fields <<< "1:2:3:4:5"
echo "${fields[4]}"
Or, to make it more generic:
echo "${fields[-1]}" # prints the last item
for x in `echo $str | tr ";" "\n"`; do echo $x; done
improving from #mateusz-piotrowski and #user3133260 answer,
echo "a:b:c:d::e:: ::" | tr ':' ' ' | xargs | tr ' ' '\n' | tail -1
first, tr ':' ' ' -> replace ':' with whitespace
then, trim with xargs
after that, tr ' ' '\n' -> replace remained whitespace to newline
lastly, tail -1 -> get the last string
For those that comfortable with Python, https://github.com/Russell91/pythonpy is a nice choice to solve this problem.
$ echo "a:b:c:d:e" | py -x 'x.split(":")[-1]'
From the pythonpy help: -x treat each row of stdin as x.
With that tool, it is easy to write python code that gets applied to the input.
Edit (Dec 2020):
Pythonpy is no longer online.
Here is an alternative:
$ echo "a:b:c:d:e" | python -c 'import sys; sys.stdout.write(sys.stdin.read().split(":")[-1])'
it contains more boilerplate code (i.e. sys.stdout.read/write) but requires only std libraries from python.
Using OS X, I need a one line bash script to look at a client mac hostname like:
12345-BA-PreSchool-LT.local
Where the first 5 digits are an asset serial number, and the hyphens separate a business unit code from a department name followed by something like 'LT' to denote a laptop.
I guess I need to echo the hostname and use a combination of sed, awk and perhaps cut to strip characters out to leave me with:
"BA PreSchool"
Any help much appreciated. This is what I have so far:
echo $HOSTNAME | sed 's/...\(...\)//' | sed 's/.local//'
echo "12345-BA-PreSchool-LT.local" | cut -d'-' -f2,3 | sed -e 's/-/ /g'
(Not on OSX, so not sure if cut is defined)
I like to keep things simple :)
You could do it with just cut:
echo 12345-BA-PreSchool-LT.local | cut -d"-" -f2,3
BA-PreSchool
If you want to remove the hyphen you can use tr
echo 12345-BA-PreSchool-LT.local | cut -d"-" -f2,3 | tr "-" " "
BA PreSchool
How about
echo $HOSTNAME | awk 'BEGIN { FS = "-" } ; { print $2, $3 }'
Awk can solve your question easily.
echo "12345-BA-PreSchool-LT.local" | awk -F'-' '$0=$2" "$3'
BA PreSchool
bash$ string="12345-BA-PreSchool-LT.local"
bash$ IFS="-"
bash$ set -- $string
bash$ echo $2-$3
BA-PreSchool
i have variable such as "1,2,3,4"
i want to count of commas in this text in bash
any idea ?
thanks for help
This will do what you want:
echo "1,2,3" | tr -cd ',' | wc -c
Off the top of my head using pure bash:
var="1,2,3,4"
temp=${var//[^,]/}
echo ${#temp}
Isolate commas per line, count lines:
echo "$VAR"|grep -o ,|wc -l
very simply with awk
$ echo 1,2,3,4 | awk -F"," '{print NF-1}'
3
with just the shell
$ s="1,2,3,4"
$ IFS=","
$ set -- $s
$ echo $(($#-1))
3
A purely bash solution with no external programs:
$ X=1,2,3,4
$ count=$(( $(IFS=,; set -- $X; echo $#) - 1 ))
$ echo $count
3
$
Note: This destroys your positional parameters.
Another pure Bash solution:
var="bbb,1,2,3,4,a,b,qwerty,,,"
saveIFS="$IFS"
IFS=','
var=($var)x
IFS="$saveIFS"
echo $((${#var[#]} - 1))
will output "10" with the string shown.
echo '1,2,3' | grep -o ',' | wc -l