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}
Related
The command that I'm making wants the first input to be a file and search how many times a certain pattern occurs within the file, using grep and sed.
Ex:
$ cat file1
oneonetwotwotwothreefourfive
Intended output:
$ ./command file1 one two three
one 2
two 3
three 1
The problem is the file does not have any lines and is just a long list of letters. I'm trying to use sed to replace the pattern I'm looking for with "FIND" and move the list to the next line and this continues until the end of file. Then, use $grep FIND to get the line that contains FIND. Finally, use wc -l to find a number of lines. However, I cannot find the option to move the list to the next line
Ex:
$cat file1
oneonetwosixone
Intended output:
FIND
FIND
twosixFIND
Another problem that I've been having is how to use the rest of the input, not including the file.
Failed attempt:
file=$1
for PATTERN in 2 3 4 5 ... N
do
variable=$(sed 's/$PATTERN/find/g' $file | grep FIND $file | wc -l)
echo $PATTERN $variable
exit
Another failed attempt:
file=$1
PATTERN=$($2,$3 ... $N)
for PATTERN in $*
do variable=$(sed 's/$PATTERN/FIND/g' $file | grep FIND $file | wc-1)
echo $PATTERN $variable
exit
Any suggestions and help will be greatly appreciated. Thank you in advance.
Non-portable solution with GNU grep:
file=$1
shift
for pattern in "$#"; do
echo "$pattern" $(grep -o -e "$pattern" <"$file" | wc -l)
done
If you want to use sed and your "patterns" are actually fixed strings (which don't contain characters that have special meaning to sed), you could do something like:
file=$1
shift
for pattern in "$#"; do
echo "$pattern" $(
sed "s/$pattern/\n&\n/g" "$file" |\
grep -e "$pattern" | wc -l
)
done
Your code has several issues:
you should quote use of variables where word splitting may happen
don't use ALLCAPS variable names - they are reserved for use by the shell
if you put a string in single-quotes, variable expansion does not happen
if you give grep a file, it won't read standard input
your for loop has no terminating done
This might work for you (GNU bash,sed and uniq):
f(){ local file=$1;
shift;
local args="$#";
sed -E 's/'${args// /|}'/\n&\n/g
s/(\n\S+)\n\S+/\1/g
s/\n+/\n/g
s/.(.*)/echo "\1"|uniq -c/e
s/ *(\S+) (\S+)/\2 \1/mg' $file; }
Separate arguments into file and remaining arguments.
Apply arguments as alternation within a sed substitution command which splits words into lines separated by a newline either side.
Remove unwanted words and unwanted newlines.
Evaluate the manufactured file within a sed substitution using the uniq command with the -c option.
Rearrange the output and print the result.
The problem is the file does not have any lines
Great! So the problem reduces to putting newlines.
func() {
file=$1
shift
rgx=$(printf "%s\\|" "$#" | sed 's#\\|$##');
# put the newline between words
sed 's/\('"$rgx"'\)/&\n/g' "$file" |
# it's just standard here
sort | uniq -c |
# filter only input - i.e. exclude fourfive
grep -xf <(printf " *[0-9]\+ %s\n" "$#")
};
func <(echo oneonetwotwotwothreefourfive) one two three
outputs:
2 one
1 three
3 two
Imagine there are these 3 subdirectories inside my directory:
dfcg7 yhjs6 gbggse3
Inside each of this subdirectories there is a txt file, which I would like to use in another program, so I would like to print all the relative paths to this files.
I am trying:
for file in /mnt/lustre/mydir*
do
printf "$file/*.txt \t"
done
and I also tried:
for file in /mnt/lustre/mydir*
do
printf "$file"/*.txt "\t"
done
but in both cases, my output is this:
/mnt/lustre/mydir/dfcg7/*txt/mnt/lustre/mydir/yhjs6/*txt/mnt/lustre/mydir/gbggse3/*txt
My output is no tab separated
It is printing the full path, instead of the relative
It is not printing the file name inside each subdirectory
So, my desired output would be this:
dfcg7/fileA.txt yhjs6/fileB.txt gbggse3/fileC.txt
How can I solve this?
You could store the path prefix in a variable:
prefix=/mnt/lustre/mydir
Assign the files to an array:
files=("$prefix"/*/*.txt)
And then print the array, tab separated, while removing the prefix from each element:
$ (IFS=$'\t'; printf '%s\n' "${files[*]/#"$prefix"\/}")
dfcg7/fileA.txt gbggse3/fileC.txt yhjs6/fileB.txt
This uses a subshell to contain the scope of the modified IFS.
This is a quick solution that I've been able to come up with
RESULT=""
for file in /path/to/you/files/*; do
OUTPUT=$(ls $file/*txt)
RESULT="$RESULT:$OUTPUT"
done
# replace every : with a tab symbol and remove the first tab
echo $RESULT | tr ':' '\t' | gsed -r 's/^\s+//g'
Notice that I've used gsed here. It is a GNU sed available for MacOS. If you are using Linux you can simply use sed.
You can try.
for file in /mnt/lustre/mydir/*/*.txt; do
printf '%s\t%s\t%s' "${file#*/*/*/*/}"
done
Output
dfcg7/fileA.txt gbggse3/fileC.txt yhjs6/fileB.txt
Try:
cd /mnt/lustre/mydir
for file in */*.txt
do
printf "%s\t" "$file"
done
printf "\n"
Or easier:
(cd /mnt/lustre/mydir; ls -d */*.txt | tr '\n' '\t'); echo
The last echo is to append a \n at the end, but if you don't need it, you can leave it out.
How do i iterate through a file and print the first word only. The line is colon separated. example
root:01:02:toor
the file contains several lines. And this is what i've done so far but it does'nt work.
FILE=$1
k=1
while read line; do
echo $1 | awk -F ':'
((k++))
done < $FILE
I'm not good with bash-scripting at all. So this is probably very trivial for one of you..
edit: variable k is to count the lines.
Use cut:
cut -d: -f1 filename
-d specifies the delimiter
-f specifies the field(s) to keep
If you need to count the lines, just
count=$( wc -l < filename )
-l tells wc to count lines
awk -F: '{print $1}' FILENAME
That will print the first word when separated by colon. Is this what you are looking for?
To use a loop, you can do something like this:
$ cat test.txt
root:hello:1
user:bye:2
test.sh
#!/bin/bash
while IFS=':' read -r line || [[ -n $line ]]; do
echo $line | awk -F: '{print $1}'
done < test.txt
Example of reading line by line in bash: Read a file line by line assigning the value to a variable
Result:
$ ./test.sh
root
user
A solution using perl
%> perl -F: -ane 'print "$F[0]\n";' [file(s)]
change the "\n" to " " if you don't want a new line printed.
You can get the first word without any external commands in bash like so:
printf '%s' "${line%%:*}"
which will access the variable named line and delete everything that matches the glob :* and do so greedily, so as close to the front (that's the %% instead of a single %).
Though with this solution you do need to do the loop yourself. If this is the only thing you want to do with the variable the cut solution is better so you don't have to do the file iteration yourself.
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}'
I have two strings and I want to split with space and use them two by two:
namespaces="Calc Fs"
files="calc.hpp fs.hpp"
for example, I want to use like this: command -q namespace[i] -l files[j]
I'm a noob in Bourne Shell.
Put them into an array like so:
#!/bin/bash
namespaces="Calc Fs"
files="calc.hpp fs.hpp"
i=1
j=0
name_arr=( $namespaces )
file_arr=( $files )
command -q "${name_arr[i]}" -l "${file_arr[j]}"
echo "hello world" | awk '{split($0, array, " ")} END{print array[2]}'
is how you would split a simple string.
if what you want to do is loop through combinations of the two split strings, then you want something like this:
for namespace in $namespaces
do
for file in $files
do
command -q $namespace -l $file
done
done
EDIT:
or to expand on the awk solution that was posted, you could also just do:
echo $foo | awk '{print $'$i'}'
EDIT 2:
Disclaimer: I don not profess to be any kind of expert in awk at all, so there may be small errors in this explanation.
Basically what the snippet above does is pipe the contents of $foo into the standard input of awk. Awk reads from it's standard in line by line, separating each line into fields based on a field separator, which is any number of spaces by default. Awk executes the program that it is given as an argument. In this case, the shell expands '{ print $'$1' }' into { print $1 } which simply tells awk to print field number 1 of each line of its input.
If you want to learn more I think that this blog post does a pretty good job of describing the basics (as well as the basics of sed and grep) if you skip past the more theoretical stuff at the start (unless you're into that kind of thing).
I wanted to find a way to do it without arrays, here it is:
paste -d " " <(tr " " "\n" <<< $namespaces) <(tr " " "\n" <<< $files) |
while read namespace file; do
command -q $namespace -l $file
done
Two special usage here: process substitution (<(...)) and here strings (<<<). Here strings are a shortcut for echo $namespaces | tr " " "\n". Process substitution is a shortcut for fifo creation, it allows paste to be run using the output of commands instead of files.
If you are using zsh this could be very easy:
files="calc.hpp fs.hpp"
# all elements
print -l ${(s/ /)files}
# just the first one
echo ${${(s/ /)files}[1]} # just the first one