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source your script:

. ./run.sh



It runs in a subshell.

Dear cola,id -unwill give you the text name of the first user listed in /etc/passwd with your uid number.
id -gn will give you the text name of the first group in /etc/group that matches your...

sed 's|[^ ][^ ]*|&_label|2' infile

I think alister's is the most elegant, though it leaves str open to wildcard expansion by the shell if you are using any.
Using sed:
sed 's|[^ ][^ ]*|$1/&|g'
similar to ctsgnb's suggestion

It is because the * gets interpreted and expanded by the shell to the content of the current directory. Just go into a directory and issue:

var=*
echo $var
If you put quotes around $var, you...

sed "s/ *[^ ]*/&_label/" file doesn't have a global flag, therefore it matches only the first instance in each line.
sed "s/ *[^ ]*/&_label/g" file
Will just do what you originally thought

echo "file1.txt file2.txt ... filen.txt" | sed 's/\([^ ][^ ]*\)/\$1\1/g'

perl -ne 'print unless $a{$_}++' file

perl -i -pe 's/^\s+$//g' file

echo "qwertmyuiop" |awk 'BEGIN{FS=OFS=""}{print $(NF/2+1)}'

echo "Youcaneasilydothisbyhighlightingyourcode." |awk 'BEGIN{FS=OFS=""}{for (i=1;i<=NF;i++) {if (i%3==0) $i=$i " "}}1'

It removes all non-slash characters and then calculates the line length

Match operator (m//) returns all the matched strings as array. So @n contains all the "/"s matched in current line. $#n is index of array's last element, and because Perl starts indexing from 0, we...

Using sed you still need to count, e.g.:
$ echo ghw//yw/hw///??u | sed s#[^/]##g | wc -L
6

:a is a label ( label "a" )

ta means jump to label a is the previous command was successfull (if the s command managed to do a replacement)

So this creates a loop where after each iteration the...

Or you can just use &:
sed 's/.\{3\}/& /g'

Shorter one ;)
awk -F: '$0=$1' file
or
cut -d: -f1 file

echo "Youcaneasilydothisbyhighlightingyourcode." | sed 's/\(.\{3\}\)/\1 /g'

[2addr]s/regular expression/replacement/flags
Substitute the replacement string for the first instance of the regular expression in the pattern space.


If you leave out the .* then...

If you remove the ".*" part from the regex, that means you want to match a line that has only 16 characters, this simply does not match anything, so the original string is output.
...

awk -F: '{print $1}' infile

There are punctuations in your string, so:
awk -F"[ ,.]" '{for(i=1;i<=NF;++i) if($i!="")a[$i]} END {for(s in a) print s}' infile
If you insist on using FS, try this, it has the same output as the...

Aptly:


$ sed "s/\(.\{16\}\).*/\1/" file1
You can easily


I take 0 to 15 to mean 16 characters.

sed "s/:.*//"

Assuming that you have the
file in an array already, then;
${ARRAY[@]:45:5}
will extract the arrays starting at 45
with a length of 5.