I want to delete all the rows/lines in a file that has a specific character, '?' in my case. I hope there is a single line command in Bash or AWK or Perl. Thanks
You can use sed to modify the file "in-place":
sed -i "/?/d" file
Alternatively, use grep:
grep -v "?" file > newfile.txt
Even better, just a single line using sed
sed '/?/d' input
use -i to edit file in place.
perl -i -ne'/\?/ or print' file
or
perl -i -pe's/^.*?\?.*//s' file
Here are already grep, sed and perl solutions - only for fun, pure bash one:
pattern='?'
while read line
do
[[ "$line" =~ "$pattern" ]] || echo "$line"
done
translated
for every line on the STDIN
match it for the pattern =~
and if the match is not successful || - print out the line
awk '!($0~/?/){print $0}' file_name
Related
I'm having trouble in getting sed to remove just the specific line I want. Let's say I have a file that looks like this:
testfile
testfile.txt
testfile2
Currently I'm using this to remove the line I want:
sed -i "/$1/d" file
The issue is that with this if I were to give testfile as input it would delete all three lines but I want it to only remove the first line. How do I do this?
With grep
grep -x -F -v -- "$1" file
# or
grep -xFv -- "$1" file
-F is for "fixed strings" -- turns off regex engine.
-x is to match entire line.
-v is for "everything but" the matched line(s).
-- to signal the end of options, in case $1 starts with a hyphen.
To save the file
grep -xFv -- "$1" file | sponge file # `moreutils` package
# or
tmp=$(mktemp)
grep -xFv -- "$1" file > "$tmp" && mv "$tmp" file
So match the whole line.
var=testfile
sed -i '/^'"$var"'$/d' file
# or with " quoting
sed -i "/^$var\$/d" file
You can learn regex with fun online with regex crosswords.
On a Unix system I am trying to add a new line in a file using sed or perl but it seems I am missing something.
Supposing my file has multiple lines of texts, always ending like this {TNG:}}${1:F01.
I am trying to find a to way to add a new line after the }$, in this way {1 should always start on a new line.
I tried it by escaping $ sign using this:
perl -e '$/ = "\${"; while (<>) { s/\$}\{$/}\n{/; print; }' but it does not work.
Any ideas will be appreciated.
give this a try:
sed 's/{TNG:}}\$/&\n/' file > newfile
The sed will by default use BRE, that is, the {}s are literal characters. But we must escape the $.
kent$ cat f
{TNG:}}${1:F01.
kent$ sed 's/{TNG:}}\$/&\n/' f
{TNG:}}$
{1:F01.
With perl:
$ cat input.txt
line 1 {TNG:}}${1:F01
line 2 {TNG:}}${1:F01
$ perl -pe 's/TNG:\}\}\$\K/\n/' input.txt
line 1 {TNG:}}$
{1:F01
line 2 {TNG:}}$
{1:F01
(Read up on the -p and -n options in perlrun and use them instead of trying to do what they do in a one-liner yourself)
I have the followiing input file and I need to remove all the characters from the strings that appear after the last '/'. I'll also show my expected output below.
input:
/start/one/two/stopone.js
/start/one/two/three/stoptwo.js
/start/one/stopxyz.js
expected output:
/start/one/two/
/start/one/two/three/
/start/one/
I have tried to use sed but with no luck so far.
You could simply use good old grep:
grep -o '.*/' file.txt
This simple expression takes advantage of the fact that grep is matching greedy. Meaning it will consume as much characters as possible, including /, until the last / in path.
Original Answer:
You can use dirname:
while read line ; do
echo dirname "$line"
done < file.txt
or sed:
sed 's~\(.*/\).*~\1~' file.txt
perl -lne 'print $1 if(/(.*)\//)' your_file
Try this GNU sed command,
$ sed -r 's~^(.*\/).*$~\1~g' file
/start/one/two/
/start/one/two/three/
/start/one/
Through awk,
awk -F/ '{sub(/.*/,"",$NF); print}' OFS="/" file
I have a file with many lines, in each line
there is either substring
whatever_blablablalsfjlsdjf;asdfjlds;f/watch?v=yPrg-JN50sw&,whatever_blabla
or
whatever_blablabla"/watch?v=yPrg-JN50sw&" class=whatever_blablablavwhate
I want to extract a substring, like the "yPrg-JN50s" above
the matching pattern is
the 11 characters after the string "/watch?="
how to extract the substring
I hope it is sed, awk in one line
if not, a pn line perl script is also ok
You can do
grep -oP '(?<=/watch\?v=).{11}'
if your grep knows Perl regex, or
sed 's/.*\/watch?v=\(.\{11\}\).*/\1/g'
$ cat file
/watch?v=yPrg-JN50sw&
"/watch?v=yPrg-JN50sw&" class=
$
$ awk 'match($0,/\/watch\?v=/) { print substr($0,RSTART+RLENGTH,11) }' file
yPrg-JN50sw
yPrg-JN50sw
Just with the shell's parameter expansion, extract the 11 chars after "watch?v=":
while IFS= read -r line; do
tmp=${line##*watch?v=}
echo ${tmp:0:11}
done < filename
You could use sed to remove the extraneous information:
sed 's/[^=]\+=//; s/&.*$//' file
Or with awk and sensible field separators:
awk -F '[=&]' '{print $2}' file
Contents of file:
cat <<EOF > file
/watch?v=yPrg-JN50sw&
"/watch?v=yPrg-JN50sw&" class=
EOF
Output:
yPrg-JN50sw
yPrg-JN50sw
Edit accommodating new requirements mentioned in the comments
cat <<EOF > file
<div id="" yt-grid-box "><div class="yt-lockup-thumbnail"><a href="/watch?v=0_NfNAL3Ffc" class="ux-thumb-wrap yt-uix-sessionlink yt-uix-contextlink contains-addto result-item-thumb" data-sessionlink="ved=CAMQwBs%3D&ei=CPTsy8bhqLMCFRR0fAodowXbww%3D%3D"><span class="video-thumb ux-thumb yt-thumb-default-185 "><span class="yt-thumb-clip"><span class="yt-thumb-clip-inner"><img src="//i1.ytimg.com/vi/0_NfNAL3Ffc/mqdefault.jpg" alt="Miniature" width="185" ><span class="vertical-align"></span></span></span></span><span class="video-time">5:15</span>
EOF
Use awk with sensible record separator:
awk -v RS='[=&"]' '/watch/ { getline; print }' file
Note, you should use a proper XML parser for this sort of task.
grep --perl-regexp --only-matching --regexp="(?<=/watch\\?=)([^&]{0,11})"
Assuming your lines have exactly the format you quoted, this should work.
awk '{print substr($0,10,11)}'
Edit: From the comment in another answer, I guess your lines are much longer and complicated than this, in which case something more comprehensive is needed:
gawk '{if(match($0, "/watch\\?v=(\\w+)",a)) print a[1]}'
In a file, say xyz.txt i want to replace the pattern of any number followed by a dot example:1.,2.,10.,11. etc.. with a whitespace.
How to compose a perl command on the command line to act on the file to do the above, what should be the regex to be used ?
Please Help
Thank You.
This HAS to be a Perl oneliner?
perl -i -pe 's/\d+\./ /g' <fileName>
The Perl command line options: -i is used to specify what happens to the input file. If you don't give it a file extension, the original file is lost and is replaced by the Perl munged output. For example, if I had this:
perl -i.bak -pe 's/\d+\./ /g' <fileName>
The original file would be stored with a .bak suffix and <fileName> itself would contain your output.
The -p means to enclose your Perl program in a print loop that looks SOMEWHAT like this:
while ($_ = <>) {
<Your Perl one liner>
print "$_";
}
This is a somewhat simplified explanation what's going on. You can see the actual perl loop by doing a perldoc perlrun from the command line. The main idea is that it allows you to act on each line of a file just like sed or awk.
The -e simply contains your Perl command.
You can also do file redirection too:
perl -pe 's/\d+\./ /g' < xyz.txt > xyz.txt.out
Answer (not tested):
perl -ipe "s/\d+\./ /g" xyz.txt
Both
perl -ipe "s/\d+\./ /g" xyz.txt
and
perl -pie
cannot execute on my system.
I use the following order:
perl -i -pe