I want to extract the output of a command run through shell script in a variable but I am not able to do it. I am using grep command for the same. Please help me in getting the desired output in a variable.
x=$(pwd)
pw=$(grep '\(.*\)/bin' $x)
echo "extracted is:"
echo $pw
The output of the pwd command is /opt/abc/bin/ and I want only /root/abc part of it. Thanks in advance.
Use dirname to get the path and not the last segment of the path.
You can use:
x=$(pwd)
pw=`dirname $x`
echo $pw
Or simply:
pw=`dirname $(pwd)`
echo $pw
All of what you're doing can be done in a single echo:
echo "${PWD%/*}"
$PWD variable represents current directory and %/* removes last / and part after last /.
For your case it will output: /root/abc
The second (and any subsequent) argument to grep is the name of a file to search, not a string to perform matching against.
Furthermore, grep prints the matching line or (with -o) the matching string, not whatever the parentheses captured. For that, you want a different tool.
Minimally fixing your code would be
x=$(pwd)
pw=$(printf '%s\n' "$x" | sed 's%\(.*\)/bin.*%\1%')
(If you only care about Bash, not other shells, you could do sed ... <<<"$x" without the explicit pipe; the syntax is also somewhat more satisfying.)
But of course, the shell has basic string manipulation functions built in.
pw=${x%/bin*}
Related
The rest of my bash script works, just having trouble using grep. On each file I am using the following command:
ls -l $filepath | grep "^.r..r..r.*${2}$"
How can I properly use the second argument in the regular expression? What I am trying to do is print the file if it can be read by anyone and the owner is who is passed by the second argument.
Using:
ls -l $filepath | grep "^.r..r..r"
Will print the information successfully based on the read permissions. What I am trying to do is print based on... [read permission][any characters in between][ending with the owner's name]
The immediate problem with your attempt is the final $ which anchors the search to the end of the line, which is the end of the file name, not the owner field. A better solution would replace grep with Awk instead, which has built-in support for examining only specific fields. But actually don't use ls for this, or really in scripts at all.
Unfortuntately, the stat command's options are not entirely portable, but for Linux, try
case $(stat -c %a:%u "$filepath") in
[4-7][4-7][4-7]:"$2") ls -l "$filepath";;
esac
or maybe more portably
find "$filepath" -user "$2" -perm /444 -ls
Sadly, the -perm /444 predicate is not entirely portable, either.
Paradoxically, the de facto most portable replacement for stat to get a file's permissions might actually be
perl -le '#s = stat($ARGV[0]); printf "%03o\n", $s[2]' "$filepath"
The stat call returns a list of fields; if you want the owner, too, the numeric UID is in $s[4] and getpwuid($s[4]) gets the user name.
I have a very simple perl script which I perform some regex on the value of the data piped to my perl command. ex:
cat /tmp/myfile.txt | perl -wnE"say for /my_pattern/gi"
Note, I do realize in my current setup that the -n option wraps my command ex:
while(<>){
say for /my_pattern/gi
}
...thus iterating over each line of input.
I'd like to change this where I can perform a regex against the final output of my cat command. All examples I find show processing input line by line.
Any help would be appreciated.
Update:
To be clear, I'm referring to any pipe, not only reading from a file as in my example (think curl, wget, echo, etc...) I'm not sure it is even possible given the fact that the originating command could be long-lived or run for an "indefinite" period of time.
To answer your question directly:
cat aaaa.txt | perl -ne 'BEGIN{local $/};print for /a/gi'
To get what your stuff work:
cat aaaa.txt | perl -ne ';print if /aaaa/gi'
This should be a basic question for a lot of people, but I am a biologist with no programming background, so please excuse my question.
What I am trying to do is rename about 100,000 gzipped data files that have existing name of a code (example: XG453834.fasta.gz). I'd like to name them to something easily readable and parseable by me (example: Xanthomonas_galactus_str_453.fasta.gz).
I've tried to use sed, rename, and mmv, to no avail. If I use any of those commands on a one-off script then they work fine, it's just when I try to incorporate variables into a shell script do I run into problems. I'm not getting any errors, just no names are changed, so I suspect it's an I/O error.
Here's what my files look like:
#! /bin/bash
# change a bunch of file names
file=names.txt
while IFS=' ' read -r r1 r2;
do
mmv ''$r1'.fasta.gz' ''$r2'.fasta.gz'
# or I tried many versions of: sed -i 's/"$r1"/"$r2"/' *.gz
# and I tried many versions of: rename -i 's/$r1/$r2/' *.gz
done < "$file"
...and here's the first lines of my txt file with single space delimiter:
cat names.txt
#find #replace
code1 name1
code2 name2
code3 name3
I know I can do this with python or perl, but since I'm stuck here working on this particular script I want to find a simple solution to fixing this bash script and figure out what I am doing wrong. Thanks so much for any help possible.
Also, I tried to cat the names file (see comment from Ashoka Lella below) and then use awk to move/rename. Some of the files have variable names (but will always start with the code), so I am looking for a find & replace option to just replace the "code" with the "name" and preserve the file name structure.
I suspect I am not escaping the variable within the single tick of the perl expression, but I have poured over a lot of manuals and I can't find the way to do this.
If you're absolutely sure than the filenames doesn't contain spaces of tabs, you can try the next
xargs -n2 < names.txt echo mv
This is for DRY run (will only print what will do) - if you satisfied with the result, remove the echo ...
If you want check the existence ot the target, use
xargs -n2 < names.txt echo mv -i
if you want NEVER allow overwriting of the target use
xargs -n2 < names.txt echo mv -n
again, remove the echo if youre satisfied.
I don't think that you need to be using mmv, a simple mv will do. Also, there's no need to specify the IFS, the default will work for you:
while read -r src dest; do mv "$src" "$dest"; done < names.txt
I have double quoted the variable names as it is generally considered good practice but in this case, a space in either of the filenames will result in read not working as you expect.
You can put an echo before the mv inside the loop to ensure that the correct command will be executed.
Note that in your file names.txt, the .fasta.gz suffix is already included, so you shouldn't be adding it inside the loop aswell. Perhaps that was your problem?
This should rename all files in column1 to column2 of names.txt. Provided they are in the same folder as names.txt
cat names.txt| awk '{print "mv "$1" "$2}'|sh
Why does GNU sed sometimes handle substitution with piped output into another sed instance differently than when multiple expressions are used with the same one?
Specifically, for msys/mingw sessions, in the /etc/profile script I have a series of manipulations that "rearrange" the order of the environment variable PATH and removes duplicate entries.
Take note that while normally sed treats each line of input seperately (and therfore can't easily substitute '\n' in the input stream, this sed statement does a substitution of ':' with '\n', so it still handles the entire input stream like one line (with '\n' characters in it). This behavior stays true for all sed expressions in the same instance of sed (basically until you redirect or pipe the output into another program).
Here's the obligatory specs:
Windows 7 Professional Service Pack 1
HP Pavilion dv7-6b78us
16 GB DDR3 RAM
MinGW-w64 (x86_64-w64-mingw32-gcc-4.7.1.2-release-win64-rubenvb) mounted on /mingw/
MSYS (20111123) mounted on / and on /usr/
$ uname -a="MINGW32_NT-6.1 CHRIV-L09 1.0.17(0.48/3/2) 2011-04-24 23:39 i686 Msys"
$ which sed="/bin/sed.exe" (it's part of MSYS)
$ sed --version="GNU sed version 4.2.1"
This is the contents of PATH before manipulation:
PATH='.:/usr/local/bin:/mingw/bin:/bin:/c/PHP:/c/Program Files (x86)/HP SimplePass 2011/x64:/c/Program Files (x86)/HP SimplePass 2011:/c/Windows/system32:/c/Windows:/c/Windows/System32/Wbem:/c/Windows/System32/WindowsPowerShell/v1.0:/c/si:/c/android-sdk:/c/android-sdk/tools:/c/android-sdk/platform-tools:/c/Program Files (x86)/WinMerge:/c/ntp/bin:/c/GnuWin32/bin:/c/Program Files/MySQL/MySQL Server5.5/bin:/c/Program Files (x86)/WinSCP:/c/Program Files (x86)/Overlook Fing 2.1/bin:/c/Program Files/7-zip:.:/c/Program Files/TortoiseGit/bin:/c/Program Files (x86)/Git/bin:/c/VS10/VC/bin/x86_amd64:/c/VS10/VC/bin/amd64:/c/VS10/VC/bin'
This is an excerpt of /etc/profile (where I have begun the PATH manipulation):
set | grep --color=never ^PATH= | sed -e "s#^PATH=##" -e "s#'##g" \
-e "s/:/\n/g" -e "s#\n\(/[^\n]*tortoisegit[^\n]*\)#\nZ95-\1#ig" \
-e "s#\n\(/[a-z]/win\)#\nZ90-\1#ig" -e "s#\n\(/[a-z]/p\)#\nZ70-\1#ig" \
-e "s#\.\n#A10-.\n#g" -e "s#\n\(/usr/local/bin\)#\nA15-\1#ig" \
-e "s#\n\(/bin\)#\nA20-\1#ig" -e "s#\n\(/mingw/bin\)#\nA25-\1#ig" \
-e "s#\n\(/[a-z]/vs10/vc/bin\)#\nA40-\1#ig"
The last sed expression in that line basically looks for lines that begins with "/c/VS10/VC/bin" and prepends them with 'A40-' like this:
...
/c/si
A40-/c/VS10/VC/bin
A40-/c/VS10/VC/bin/amd64
A40-/c/VS10/VC/bin/x86_amd64
/c/GnuWin32/bin
...
I like my sed expressions to be flexible (path structures change), but I don't want it to match the lines that end with amd64 or x86_amd64 (those are going to have a different string prepended). So I change the last expression to:
-e "s#\n\(/[a-z]/vs10/vc/bin\)\n#\nA40-\1\n#ig"
This works:
...
/c/si
A40-/c/VS10/VC/bin
/c/VS10/VC/bin/amd64
/c/VS10/VC/bin/x86_amd64
/c/GnuWin32/bin
...
Then, (to match any "line" matching the pseudocode "/x/.../bin") I change the last expression to:
-e "s#\n\(/[a-z]/.*/bin\)\n#\nA40-\1\n#ig"
Which produces:
...
/c/si
/c/VS10/VC/bin
/c/VS10/VC/bin/amd64
/c/VS10/VC/bin/x86_amd64
/c/GnuWin32/bin
...
??? - sed didn't match any character ('.') any number of times ('*') in the middle of the line ???
But, if I pipe the output into a different instance of sed (and compensate for sed handling each "line" seperately) like this:
| sed -e "s#^\(/[a-z]/.*/bin\)$#A40-\1#ig"
I get:
sed: -e expression #1, char 30: unterminated `s' command
??? How is that unterminated? It's got all three '#' characters after the s, has the modifiers 'i' and 'g' after the third '#', and the entire expression is in double quotes ('"'). Also, there are no escapes ('\') immediately preceding the delimiters, and the delimiter is not a part of either the search or the replacement. Let's try a different delimiter than '#', like '~':
I use:
| sed -e "s~^(/[a-z]/.*/bin)$~A40-\1~ig"
and, I get:
...
/c/si
A40-/c/VS10/VC/bin
/c/VS10/VC/bin/amd64
/c/VS10/VC/bin/x86_amd64
A40-/c/GnuWin32/bin
...
And, that is correct! The only thing I changed was the delimeter from '#' to '~' and it worked ???
This is not (even close to) the first time that sed has produced unexplainable results for me.
Why, oh, why, is sed NOT matching syntax in an expression in the same instance, but IS matching when piped into another instance of sed?
And, why, oh, why, do I have to use a different delimeter when I do this (in order not to get an "unterminated 's' command"?
And the real reason I'm asking: Is this a bug in sed, OR, is it correct behavior that I don't understand (and if so, can someone explain why this behavior is correct)? I want to know if I'm doing it wrong, or if I need a different/better tool (or both, they don't have to be mutually exclusive).
I'll mark a response it as the answer if someone can either prove why this behavior is correct or if they can prove why it is a bug. I'll gladly accept any advice about other tools or different methods of using sed, but those won't answer the question.
I'm going to have to get better at other text processors (like awk, tr, etc.) because sed is costing me too much time with it's unexplainable results.
P.S. This is not the complete logic of my PATH manipulation. The complete logic also finishes prepending all the lines with values from 'A00-' to 'Z99-', then pipes that output into 'sort -u -f' and back into sed to remove those same prefixes on each line and to convert the lines ('\n') back into colons (':'). Then "export PATH='" is prepended to the single line and "'" is appended to it. Then that output is redirected into a temporary file. Next, that temporary file is sourced. And, finally, that temporary file is removed.
The /etc/profile script also displays the contents of PATH before and after sorting (in case it screwed up the path).
P.P.S. I'm sure there is a much better way to do this. It started as some very simple sed manipulations, and grew into the monster you see here. Even if there is a better way, I still need to know why sed is giving me these results.
sed -e "s#^\(/[a-z]/.*/bin\)$#A40-\1#ig"
is unterminated because the shell is trying to expand "$#A". Put your expressions in single quotes to avoid this.
The expression
-e "s#\n\(/[a-z]/.*/bin\)\n#\nA40-\1\n#ig"
fails, or doesn't do what you expect, because . matches the newline in a multi-line expression. Check your whole output, the A40- is at the very beginning. Change it to
-e "s#\n\(/[a-z]/[^\n]*/bin\)\n#\nA40-\1\n#ig"
and it might be more what you expect. This may very well be the case with most of your issues with multi-line modifications.
You can also put the statements, one per line, into a standalone file and invoke sed with sed -f editscript. It might make maintenance of this a bit easier.
I need to get a username from an Unix path with this format:
/home/users/myusername/project/number/files
I just want "myusername" I've been trying for almost a hour and I'm completely clueless.
Any idea?
Thanks!
Maybe just /home/users/([a-zA-Z0-9_\-]*)/.*?
Note that the critical part [a-zA-Z0-9_\-]* has to contain all valid characters for unix usernames. I took from here, that a username should only contain digits, characters, dashes and underscores.
Also note that the extracted username is not the whole matching, but the first group (indicated by (...)).
The best answer to this depends on what you are trying to achieve. If you want to know the user who owns that file then you can use the stat command, this unfortunately has slightly different syntax dependant on the operating system however the following two commands work
Max OS/X
stat -f '%Su' /home/users/myusername/project/number/files
Redhat/Fedora/Centos
stat -c '%U' /home/users/myusername/project/number/files
If you really do want the string following /home/users then the either of the Regexes provided above will do that, you could use that in a bash script as follows (Mac OS/X)
USERNAME=$(echo '/home/users/myusername/project/number/files' | \
sed -E -e 's!^/home/users/([^/]+)/.*$!\1!g')
Check http://rubular.com/r/84zwJmV62G. The first match, not the entire match, is the username.
in a bourne shell something like :
string="/home/users/STRINGWEWANT/some/subdir/here"
echo $string | awk -F\/ '{print $3}'
would be one option, assuming its always the third element of the path. There are more lightweight that use only the shell builtins :
echo ${x#*users/}
will strip out everything up to and including 'users/'
echo ${y%%/*}
Will strip out the remainder.
So to put it all together :
export path="/home/users/STRINGWEWABT/some/other/dirs"
export y=`echo ${path#*users/}` && echo ${y%%/*}
STRINGWEWABT
Also checkout the bash manpage and search for "Parameter Expansion"
(\/home\/users\/)([^\/]+)
The 2nd capture group (index 1) will be myusername