I suppose to strip down a substring in my shell script. I am trying as follows:
fileName="Test_VSS_TT.csv.old"
here i want to remove the string ".csv.old" and my
test=${fileName%.*}
but getting bad substitution error.
you are looking for test=${filename%%.*}
the doc for parameter expansion in bash here and in zsh here
%.* will match the first .* pattern, whereas %%.* will match the longest one
[edit]
if sed is available, you could try something like that : echo "filename.txt.bin" | sed "s/\..*//g" which yields filename
Here you go,
$ echo $f
Test_VSS_TT.csv.old
$ test=${f%%.*}
$ echo $test
Test_VSS_TT
%% will do a longest match. So it matches from the first dot upto the last and then removes the matched characters.
If your intention is to extract file name without extension, then how about this?
$ echo ${fileName}
Test_VSS_TT.csv.old
$ test=`echo ${fileName} |cut -d '.' -f1`
$ echo $test
Test_VSS_TT
echo "Test_VSS_TT.csv.old"| awk -F"." '{print $1}'
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.
I am trying to grep the folder name from full tar file. Below is the example.
example:
TEST-5.3.0.0-build1.x86_64.tar.gz
I want to grep the folder name (TEST-5.3.0.0-build1) in shell script
So i tried below command for grep
$ package_folder=$(echo TEST-5.3.0.0-build1.x86_64.tar.gz | sed -e "s/.[0-9]*[a-z]*[0-9]*.tar.gz$//" | sed -e 's/\/$//')
But I am getting below output:
$ echo $package_folder
TEST-5.3.0.0-build1.x86
Could you please anyone correct me where I am doing mistake. I need folder name as TEST-5.3.0.0-build1
Thanks in Advance!!!
In your command, you do not match _, x, etc. The [0-9]*[a-z]*[0-9]* only matches a sequence of zero or more digits, zero or more (lowercase) letters, and zero or more digits. It is better to use a [^.]* to match any chars other than . between two . chars. Also, literal dots must be escaped, or an unescaped . will match any single char.
You can use
sed 's/\.[^.]*\.tar\.gz$//'
Or, just use string manipulation if x86_64 is also a constant:
s='TEST-5.3.0.0-build1.x86_64.tar.gz'
s="${s/.x86_64.tar.gz/}"
See the online demo:
#!/bin/bash
s='TEST-5.3.0.0-build1.x86_64.tar.gz'
package_folder=$(sed 's/\.[^.]*\.tar\.gz$//' <<< "$s")
echo "${package_folder}"
# => TEST-5.3.0.0-build1
s="${s/.x86_64.tar.gz/}"
echo "$s"
# => TEST-5.3.0.0-build1
You can use uname -m in replacement part of this string:
s='TEST-5.3.0.0-build1.x86_64.tar.gz'
echo "${s%.$(uname -m)*}"
TEST-5.3.0.0-build1
Or using sed:
sed "s/\.$(uname -m).*//" <<< "$s"
TEST-5.3.0.0-build1
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.
I am working on shell script and new to it. I want to extract the string between double $$ characters, for example:
input:
$$extractabc$$
output
extractabc
I used grep and sed but not working out. Any suggestions are welcome!
You could do
awk -F"$" '{print $3}' file.txt
assuming the file contained input:$$extractabc$$ output:extractabc. awk splits your data into pieces using $ as a delimiter. First item will be input:, next will be empty, next will be extractabc.
You could use sed like so to get the same info.
sed -e 's/.*$$\(.*\)$$.*/\1/' file.txt
sed looks for information between $$s and outputs that. The goal is to type something like this .*$$(.*)$$.*. It's greedy but just stay with me.
looks for .* - i.e. any character zero or more times before $$
then the string should have $$
after $$ there'll be any character zero or more times
then the string should have another $$
and some more characters to follow
between the 2 $$ is (.*). String found between $$s is given a placeholder \1
sed finds such information and publishes it
Using grep PCRE (where available) and look-around:
$ echo '$$extractabc$$' | grep -oP "(?<=\\$\\$).*(?=\\$\\$)"
extractabc
echo '$$extractabc$$' | awk '{gsub(/\$\$/,"")}1'
extractabc
Here is an other variation:
echo "$$extractabc$$" | awk -F"$$" 'NF==3 {print $2}'
It does test of there are two set of $$ and only then prints whats between $$
Does also work for input like blabla$$some_data$$moreblabla
How about remove all the $ in the input?
$ echo '$$extractabc$$' | sed 's/\$//g'
extractabc
Same with tr
$ echo '$$extractabc$$' | tr -d '$'
extractabc
Suppose echo $PATH yields /first/dir:/second/dir:/third/dir.
Question: How does one echo the contents of $PATH one directory at a time as in:
$ newcommand $PATH
/first/dir
/second/dir
/third/dir
Preferably, I'm trying to figure out how to do this with a for loop that issues one instance of echo per instance of a directory in $PATH.
echo "$PATH" | tr ':' '\n'
Should do the trick. This will simply take the output of echo "$PATH" and replaces any colon with a newline delimiter.
Note that the quotation marks around $PATH prevents the collapsing of multiple successive spaces in the output of $PATH while still outputting the content of the variable.
As an additional option (and in case you need the entries in an array for some other purpose) you can do this with a custom IFS and read -a:
IFS=: read -r -a patharr <<<"$PATH"
printf %s\\n "${patharr[#]}"
Or since the question asks for a version with a for loop:
for dir in "${patharr[#]}"; do
echo "$dir"
done
How about this:
echo "$PATH" | sed -e 's/:/\n/g'
(See sed's s command; sed -e 'y/:/\n/' will also work, and is equivalent to the tr ":" "\n" from some other answers.)
It's preferable not to complicate things unless absolutely necessary: a for loop is not needed here. There are other ways to execute a command for each entry in the list, more in line with the Unix Philosophy:
This is the Unix philosophy: Write programs that do one thing and do it well. Write programs to work together. Write programs to handle text streams, because that is a universal interface.
such as:
echo "$PATH" | sed -e 's/:/\n/g' | xargs -n 1 echo
This is functionally equivalent to a for-loop iterating over the PATH elements, executing that last echo command for each element. The -n 1 tells xargs to supply only 1 argument to it's command; without it we would get the same output as echo "$PATH" | sed -e 'y/:/ /'.
Since this uses xargs, which has built-in support to split the input, and echoes the input if no command is given, we can write that as:
echo -n "$PATH" | xargs -d ':' -n 1
The -d ':' tells xargs to use : to separate it's input rather than a newline, and the -n tells /bin/echo to not write a newline, otherwise we end up with a blank trailing line.
here is another shorter one:
echo -e ${PATH//:/\\n}
You can use tr (translate) to replace the colons (:) with newlines (\n), and then iterate over that in a for loop.
directories=$(echo $PATH | tr ":" "\n")
for directory in $directories
do
echo $directory
done
My idea is to use echo and awk.
echo $PATH | awk 'BEGIN {FS=":"} {for (i=0; i<=NF; i++) print $i}'
EDIT
This command is better than my former idea.
echo "$PATH" | awk 'BEGIN {FS=":"; OFS="\n"} {$1=$1; print $0}'
If you can guarantee that PATH does not contain embedded spaces, you can:
for dir in ${PATH//:/ }; do
echo $dir
done
If there are embedded spaces, this will fail badly.
# preserve the existing internal field separator
OLD_IFS=${IFS}
# define the internal field separator to be a colon
IFS=":"
# do what you need to do with $PATH
for DIRECTORY in ${PATH}
do
echo ${DIRECTORY}
done
# restore the original internal field separator
IFS=${OLD_IFS}