$ find ~/AppData/Local/atom/ -name atom.sh -type f | grep -FzZ 'cli/atom.sh' | sed '/^\s*$/d' | cat -n
1 /c/Users/devuser/AppData/Local/atom/app-1.12.1/resources/cli/atom.sh
2 /c/Users/devuser/AppData/Local/atom/app-1.12.2/resources/cli/atom.sh
3
I tried a number of sed/awk-based options to get rid of the blank line. (#3 in the output). But I can't quite get it right...
I need to get the last line into a variable...
The below actual command I am working with fails to give an output...
find ~/AppData/Local/atom/ -name atom.sh -type f | grep -FzZ 'cli/atom.sh' | sed '/^$/d' | tail -n 1
Below sed command would remove all the empty lines.
sed '/^$/d' file
or
Below sed command would remove all the empty lines and also the lines having only spaces.
sed '/^[[:blank:]]*$/d' file
Add -i parameter to do an in-place edit.
sed -i '/^[[:blank:]]*$/d' file
The problem is the -Z flag of grep. It causes grep to terminate matched lines with a null character instead of a newline. This won't work well with sed, because sed processes input line by line, and it expects each line to be terminated by a newline. In your example grep doesn't emit any newline characters, so as far as sed is concerned, it receives a block of text without a terminating newline, so it processes it as a single line, and so the pattern /^\s*$/ doesn't match anything.
Furthermore, the -z flag would only make sense if the filenames in the output of find were terminated by null characters, that is with the -print0 flag. But you're not using that, so the -z flag in grep is pointless. And the -Z flag is pointless too, because that should be used when the next command in the pipeline expects null-terminated records, which is not your case.
Do like this:
find ~/AppData/Local/atom/ -name atom.sh -type f -print0 | grep -zF 'cli/atom.sh' | tr '\0' '\n' | tail -n 1
Not all sed's support the token \s for any whitespace character.
You can use the character class [[:blank:]] (for space or tab), or [[:space:]] (for any whitespace):
sed '/^[[:blank:]]*$/d'
sed '/^[[:space:]]*$/d'
Related
I created the following bash pipeline that will take the output of "who" and modify it to meet an assignment's requirements
This is the pipline:
who | sed -e "s/\b\(.\)/\u\1/g" | sed 's/[.]/ /g' | sed 's/ Pts\// TTY /g' | sed '1d' | sed -n 's/ .*$/ /gp'
After putting this into a sed file that looks like this:
s/\b\(.\)/\u\1/g
s/[.]/ /g
s/ Pts\// TTY /g
1d
s/ .*$/ /gp
And then running it like such:
who | sed -f sedfile
The output is correct in that everything is in the format of:
firstName lastName TTY (a number)
However each line is printed twice, where the pipeline properly printed each line once
Would anyone happen to know the issue please?
It's the gp at the last line. You're not running with sed -n (no-print), so by default you're running with sed -yesprint (or whatever). Then you hit that gp, which prints, and you get two copies.
Convert to sed -n, or change the gp to just g. Or better still, get rid of it, since the match pattern contains $, so it will only ever run in one place - the end of the line.
Say for instance I'm searching a line that is like this:
Color asdf
and I use grep to find that line, like grep asdf file.txt
How would I then display Color? Learning linux is hard.
With the command line tool sed you can replace stings by using regular expressions:
echo "Color asdf" | sed 's/\([^ ]*\).*/\1/'
This part: \([^ ]*\).* is a regular expresion. The first part of the regex: [^ ]*, matches any character except a space as many times as possible and what's between the \( and \) is being captured in the variable \1. Then you also match the remaining part of the string with .* and replace all of that with only the first word which was captured by \([^ ]*\) by using \1 in the replace part of the sed command.
Here some more info about sed:
http://linux.about.com/od/commands/a/Example-Uses-Of-Sed-Cmdsedxa.htm
You could use sed:
sed -n 's/[[:space:]][[:space:]]*asdf$//p' file.txt
Details:
The -n option tells sed not to print the pattern space automatically. Basically, it doesn't output anything unless you tell it to.
The s command of sed replaces text. Here, if a line ends with asdf, preceded by at least one whitespace character, we replace all of that with nothing and then print the line (notice the p flag at the end of the s command). The printing is only done if something was actually replaced. More information about the s command can be found e. g. in the GNU sed manual.
Edit for clarity: When using single quotes, parameter expansion does not work and thus, variables won't be replaced. To use variables, use double quotes:
search=asdf
sed -n "s/[[:space:]][[:space:]]*${search}\$//p" file.txt
If you'd really like to use grep here, you could pipe the output from grep into cut:
grep -h asdf *.txt | cut -s -d -f 1
Note that there have to be two spaces after the -d option to cut - the first tells cut to use a blank as the field delimiter (I'm assuming your fields are blank-delimited rather than tab-delimited), while the second separates the -d option from the following option (-f).
But, yeah, sed or awk are probably your friends here... :-)
you can color pattern in the line using grep
grep --colour -o 'asdf' file.txt
edit: the -o option will print only the patterns
I have the following problem.
Got a file which includes certain paths/files of a FS.
These for some reason do include the whole range of special characters, like space, single/double quotes, even sometimes the Copyright ASCII.
I need to run each line of the file and pass it to another command.
What I tried so far is:
<input_file xargs -I % command %
Which was working until I got this message from xargs
xargs: unmatched single quote; by default quotes are special to xargs unless you use the -0 option
But usinf this option did not work at all for me
xargs: argument line too long
Does anybody have a solution which does work ok with special characters.
Doesn't have to be with xargs, but I need to pass the line as it is to the command.
Many thanks in advance.
You should separate the filenames with the \0 NULL character for processing.
This can be done with
find . <args> -print0 | xargs -0
or if you must process the file with filenames, change the '\n` to '\0', e.g.
tr '\n' '\0' < filename | xargs -0 -n1 -I% echo "==%=="
the -n 1 says,
-n max-args
Use at most max-args arguments per command line.
and you should to use "%" quotes to enclosing %
The xargs -0 -n1 -I% echo "==%==" solution didn't work for me on my Mac OS X, so I found another one.
<input_with_single_quotes cat | sed "s/'/\\\'/" | xargs -I {} echo {}
This replaces the ' character with \' that works well as an input to the commands in xargs.
I need some assistance trying to build up a variable using a list of exclusions in a file.
So I have a exclude file I am using for rsync that looks like this:
*.log
*.out
*.csv
logs
shared
tracing
jdk*
8.6_Code
rpsupport
dbarchive
inarchive
comms
PR116PICL
**/lost+found*/
dlxwhsr*
regression
tmp
working
investigation
Investigation
dcsserver_weblogic_
dcswebrdtEAR_weblogic_
I need to build up a string to be used as a variable to feed into egrep -v, so that I can use the same exclusion list for rsync as I do when egrep -v from a find -ls.
So I have created this so far to remove all "*" and "/" - and then when it sees certain special characters it escapes them:
cat exclude-list.supt | while read line
do
echo $line | sed 's/\*//g' | sed 's/\///g' | 's/\([.-+_]\)/\\\1/g'
What I need the ouput too look like is this and then export that as a variable:
SEXCLUDE_supt="\.log|\.out|\.csv|logs|shared|PR116PICL|tracing|lost\+found|jdk|8\.6\_Code|rpsupport|dbarchive|inarchive|comms|dlxwhsr|regression|tmp|working|investigation|Investigation|dcsserver\_weblogic\_|dcswebrdtEAR\_weblogic\_"
Can anyone help?
A few issues with the following:
cat exclude-list.supt | while read line
do
echo $line | sed 's/\*//g' | sed 's/\///g' | 's/\([.-+_]\)/\\\1/g'
Sed reads files line by line so cat | while read line;do echo $line | sed is completely redundant also sed can do multiple substitutions by either passing them as a comma separated list or using the -e option so piping to sed three times is two too many. A problem with '[.-+_]' is the - is between . and + so it's interpreted as a range .-+ when using - inside a character class put it at the end beginning or end to lose this meaning like [._+-].
A much better way:
$ sed -e 's/[*/]//g' -e 's/\([._+-]\)/\\\1/g' file
\.log
\.out
\.csv
logs
shared
tracing
jdk
8\.6\_Code
rpsupport
dbarchive
inarchive
comms
PR116PICL
lost\+found
dlxwhsr
regression
tmp
working
investigation
Investigation
dcsserver\_weblogic\_
dcswebrdtEAR\_weblogic\_
Now we can pipe through tr '\n' '|' to replace the newlines with pipes for the alternation ready for egrep:
$ sed -e 's/[*/]//g' -e 's/\([._+-]\)/\\\1/g' file | tr "\n" "|"
\.log|\.out|\.csv|logs|shared|tracing|jdk|8\.6\_Code|rpsupport|dbarchive|...
$ EXCLUDE=$(sed -e 's/[*/]//g' -e 's/\([._+-]\)/\\\1/g' file | tr "\n" "|")
$ echo $EXCLUDE
\.log|\.out|\.csv|logs|shared|tracing|jdk|8\.6\_Code|rpsupport|dbarchive|...
Note: If your file ends with a newline character you will want to remove the final trailing |, try sed 's/\(.*\)|/\1/'.
This might work for you (GNU sed):
SEXCLUDE_supt=$(sed '1h;1!H;$!d;g;s/[*\/]//g;s/\([.-+_]\)/\\\1/g;s/\n/|/g' file)
This should work but I guess there are better solutions. First store everything in a bash array:
SEXCLUDE_supt=$( sed -e 's/\*//g' -e 's/\///g' -e 's/\([.-+_]\)/\\\1/g' exclude-list.supt)
and then process it again to substitute white space:
SEXCLUDE_supt=$(echo $SEXCLUDE_supt |sed 's/\s/|/g')
How i can delete whitespace in each line of file, using bash
For instance, file1.txt. Before:
gg g
gg g
t ttt
after:
gg g
gg g
t ttt
sed -i 's/ //g' your_file will do it, modifying the file inplace.
To delete only the whitespaces at the beginning of one single line, use sed -i 's/^ *//' your_file
In the first expression, we replace all spaces with nothing.
In the second one, we replace at the beginning using the ^ keyword
tr(delete all whitespaces):
$ tr -d ' ' <input.txt >output.txt
$ mv output.txt input.txt
sed(delete leading whitespaces)
$ sed -i 's/^ *//' input.txt
use can use perl -i for in place replacement.
perl -p -e 's/^ *//' file
To delete the white spaces before start of the line if the pattern matches. Use the following command.
For example your foo.in has pattern like this
This is a test
Lolll
blaahhh
This is a testtt
After issuing following command
sed -e '/This/s/ *//' < foo.in > foo.out
The foo.out will be
This is a test
Lolll
blaahhh
This is a testtt
"Whitespace" can include both spaces AND tabs. The solutions presented to date will only match and operate successfully on spaces; they will fail if the whitespace takes the form of a tab.
The below has been tested on the OP's specimen data set with both spaces AND tabs, matching successfully & operating on both:
sed 's/^[[:blank:]]*//g' yourFile
After testing, supply the -i switch to sed to make the changes persistent-