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For some mysterious reason, whenever I run grep from a "standalone" shell script as opposed to a simple function, the coloring of the output is not preserved.
Why is this happening and how can I prevent it?
This is best illustrated with an example:
You should try in your script :
grep --color
But please, no need to
echo `ls` | grep ".txt"
just
ls -1 | grep --color ".txt"
or
printf '%s\n' | grep --color ".txt"
See http://porkmail.org/era/unix/award.html
EDIT
To change the defautl colors of grep, see man grep and search GREP_COLORS
Related
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Hel lo I have a df file such as
col1;col2;col3;col4
A;B;C;D
E;F;G;H
I;J;K;L
and I would like to grep I and only display the col1 and col2
and get
I;J then
because from now I only know how to do :
grep 'I' df.csv
I;J;K;L
Try this:
grep 'I' df.csv | cut -d';' -f1-2
The cut command will treat each input line as a list of fields separated by ; (-d';'), and will select only the first two fields (-f1-2) for output.
Sample session:
$ cat df.csv
col1;col2;col3;col4
A;B;C;D
E;F;G;H
I;J;K;L
$ grep 'I' df.csv | cut -d';' -f1-2
I;J
$
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Exactly what the title says: what is the bash command for printing out the longest word in a text file that appears at least 10 times.
Try this Denis:
tr -s " " "\n" < file | while read -r l; do echo "${#l} $l"; done | sort -n | awk '$1 >= 10 ' | awk '{print $2}' | tail -n1
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I have this list of inputs :
imalex
thislara
heiscarl
how to get :
alex
lara
carl
grep
Use grep to take the last four chars:
grep -o '.\{4\}$' file
The -o option makes sure only matched parts are printed.
sed
Using sed we can achieve a similar result:
sed 's/.*\(.\{4\}\)$/\1/' a
Here we capture the last four digits and replace each line with those last four digits. They are captured in a group \( \) and inserted \1.
read & tail
We can also grab the last five chars (including the newline) of each line using tail and a -c option. We do that for each line using read.
while read line; do
tail -c 5 <<< $line
done < file
2 answers using substring arithmetic
bash:
while read word; do
echo "${word:${#word}-4}"
done <<<"$list"
awk
echo "$list" | awk '{print substr($NF, length($NF)-4+1)}'
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ps -ef | grep --color "ilir1477" | awk '{print $NF}'
Please explain to me.
Is this what you are after?
$ ps -u ilir1477 -o args=
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I want to search for -c using grep
For example:
$>ls -al | grep '-c'
But grep thinks it is an option.
$>Usage: grep -hblcnsviw pattern file . . .
How can I search -c as a string?
Tell grep where your options end:
grep -- -c
You could either:
Escape the hyphen.
grep '\-c'
Use the -e (--regexp=) flag
grep -e -c
grep --regexp=-c # Not in POSIX, but supported at least in Linux and OS X.
You use the -e option:
$ ls -a1 | grep -e -c
This is of course mentioned in the documentation, thusly:
-e PATTERN, --regexp=PATTERN
Use PATTERN as the pattern. This can be used to specify multiple search patterns, or to protect a pattern beginning with a hyphen (-). (-e is specified by POSIX .)
To stay on topic, the answer to your original question is to use either -- to signal the end of options to process, or use -e to explicitly mark the option.
However, parsing the output of ls will produce an incorrect result if any of the file names contain a new line. Use find instead:
find . -depth 1 -name "-c"