I am trying to extract text from a file between a < and a >, but only on a line starting with another specific pattern.
So in a file that looks like:
XXX Something here
XXX Something more here
XXX <\Lines like this are a problem>
ZZZ something <\This is the text I need>
XXX Don't need any of this
I would like to print only the <\This is the text I need>.
If I do
sed -n '/^ZZZ/p' FILENAME
it pulls the correct lines I need to look at, but obviously prints the whole line.
sed -n '/<\/,/>/p' FILENAME prints way too much.
I have looked into grouping and tried
sed -n '/^ZZZ/{/<\/,/>/} FILENAME
but this doesn't seem to work at all.
Any suggestions? They will be much appreciated.
(Apologies for formatting, never posted on here before)
sed -n '/^ZZZ/ { s/^.*\(<.*>\).*$/\1/p }'
If it does not have to be sed and you have a fairly recent grep, you may use grep's option -o as in
grep '^ZZZ' | grep -o '<[^>]*>'
An awk version
awk -F"<|>" '/^ZZZ/ {print "<"$2">"}' file
<\This is the text I need>
Related
I have used sed and awk for little while now, but I am having a challenge with the below problem. I am asking for an experienced sed/awk guru to help.
I have a file where some lines have numbers and some lines do not, like:
afjjdjfj.uihuihi
trfg.rtyhd
0rtgfd.tjbghhh
hbvfd4.rtgbvdgf
00fhfg.fdrgf
rtygfd.ijhniuh
etc.
I would like to have exactly two files out of this one, where every line is represented in one of the two files (none are deleted).
One containing all lines with any numbers 0-9 on them so given above file result would be:
0rtgfd.tjbghhh
hbvfd4.rtgbvdgf
00fhfg.fdrgf
and another file containing the rest of the lines that do not have any numbers 0-9 on them, so given the above, file it would be:
afjjdjfj.uihuihi
trfg.rtyhd
rtygfd.ijhniuh
I've tried different strategies in both sed and awk and nothing is giving me exactly what I need.
What would be the best sed or awk one liner to solve this problem?
Thank you for your time,
Tom
Easily with Awk:
awk '/[0-9]/{print > file1; next} {print > file2}' inputfile
With single GNU sed command:
sed -ne '/[0-9]/w with_digits.txt' -e '//!w no_digits.txt' input
Results:
> cat no_digits.txt
afjjdjfj.uihuihi
trfg.rtyhd
rtygfd.ijhniuh
> cat with_digits.txt
0rtgfd.tjbghhh
hbvfd4.rtgbvdgf
00fhfg.fdrgf
w filename Write the pattern space to filename.
If you don't mind running twice over the input, you can use just grep:
grep '[0-9]' input > with_digits
grep -v '[0-9]' input > without_digits
perl -MFile::Slurp -lpe '/\d/ ? append_file("digits.txt",$_) : append_file("no_digits.txt",$_)' input.txt
I want to know how to use two 'grep' and 'sed' utilities or something else in order to replace the substring. I will explain what I want to do below.
We have the file 'test.txt' with the following string:
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='AA5', A6='keyword_A'
After searching 'keyword_A' using grep, I want to replace the value of A5 with other string, for example, "NEW".
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='NEW', A6='keyword_A'
I tried to use two commands like
grep keyword_A test.txt | sed -e 's/blabla/blabla/'
After trying all I know, I gave up at all.
Please let me know the right solution.
First, you never need grep and sed. Sed has a full regular-expression search engine, so it is a superset of grep. This command will read test.txt, change the lines that you've indicated, and print the entire result on standard output:
sed "/keyword_A/s/A5{ATTR}='[A-Z0-9]*'/A5{ATTR}='NEW'/g" < test.txt
If you want to store the results back into the file test.txt, use the -i (in-place editing) switch to sed:
sed "/keyword_A/s/A5{ATTR}='[A-Z0-9]*'/A5{ATTR}='NEW'/g" -i.bak test.txt
If you want to select only the indicated lines, modify those, and print only those lines to standard out, use a combination of the p (print) command and the -n (no output) switch.
sed "/keyword_A/s/A5{ATTR}='[A-Z0-9]*'/A5{ATTR}='NEW'/gp" -n test.txt
Using grep+sed is always the wrong approach. Here's one way to do it with GNU awk:
$ awk '/keyword_A/{ $0=gensub(/(A5({[^}]+})?=\047)[^\047]+/,"\\1NEW",1) } 1' file
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='NEW', A6='keyword_A'
Using a couple variables you could define the keyword and replacement ( if they change at all ):
q="keyword_A"
r="NEW"
Then with sed:
sed -r "s/^(.+\{.+\}=')(.+)('.+"${q}".+)$/\1"${r}"\3/" file
Result:
A1='AA1', A2='AA2', A3='AA3', A4='AA4', A5{ATTR}='NEW', A6='keyword_A'
A5="NEW"
A6="keyword_A"
# with sed
sed "s/='[^']*\(',[[:blank:]]*A6='${A6}'\)/='${A5}\1/" YourFile
# with awk
awk -F "'" -v A5="${A5}" -v A6="${A6}" '
BEGIN { OFS="\047" }
$12 == A6 { $10 = A5; $0 = $0 }
7
' YourFile
Change by the end of the string, for sed and using ' as field separator in awk instead of traditional space.
assuming there is no ' in value (or need to treat the escaping method) for awk version
We can just directly replace the fifth column when the sting keyword_A is found as shown below:
awk -F, 'BEGIN{OFS=",";}/keyword_A/{$5="A5{ATTR}='"'"NEW"'"'"}1' filename
Couple of slight alternatives:
sed -r "/keyword_A/s/(A5[^']*')[^']*/\1NEW/"
awk -F"'" '/keyword_A/{$10 = "NEW"}1' OFS="'"
Of course the negative with awk is afterwards you would have to rename the new file.
I want to search for a pattern "xxxx" in a file and delete 5 lines before this pattern and 6 lines after this match. How can i do this using Sed?
This might work for you (GNU sed):
sed ':a;N;s/\n/&/5;Ta;/xxxx/!{P;D};:b;N;s/\n/&/11;Tb;d' file
Keep a rolling window of 5 lines and on encountering the specified string add 6 more (11 in total) and delete.
N.B. This is a barebones solution and will most probably need tailoring to your specific needs. Questions such as: what if there are multiple string throughout the file? What if the string is within the first five lines or multiple strings are within five lines of each other etc etc etc.
Here's one way you could do it using awk. I assume that you also want to delete the line itself and that the file is small enough to fit into memory:
awk '{a[NR]=$0}/xxxx/{f=NR}END{for(i=1;i<=NR;++i)if(i<f-5||i>f+6)print a[i]}' file
Store every line into the array a. When the pattern /xxxx/ is matched, save the line number. After the whole file has been processed, loop through the array, only printing the lines you want to keep.
Alternatively, you can use grep to obtain the line number first:
grep -n 'xxxx' file | awk -F: 'NR==FNR{f=$1}NR<f-5||NR>f+6' - file
In both cases, the lines deleted will be surrounding the last line where the pattern is matched.
A third option would be to use grep to obtain the line number then use sed to delete the lines:
line=$(grep -nm1 'xxxx' file | cut -d: -f1)
sed "$((line-5)),$((line+6))d" file
In this case I've also added the -m switch so grep exits after finding the first match.
if you know, the line number (what is not difficult to obtain), you can use something like that:
filename="test"
start=`expr $curr_line - 5`
end=`expr $curr_line + 6`
sed "${start},${end}d" $filename (optionally sed -i)
of course, you have to remember about additional conditions like start shouldn't be less than 1 and end greater than number of lines in file.
Another - maybe more easy to follow - solution would be to use grep to find the keyword and the corresponding line:
grep -n 'KEYWORD' <file>
then use sed to get the line number only like this:
grep -n 'KEYWORD' <file> | sed 's/:.*//'
Now that you have the line number simply use sed like this:
sed -i "$(LINE_START),$(LINE_END) d" <file>
to remove lines before and/or after! With only the -i you will override the <file> (no backup).
A script example could be:
#!/bin/bash
KEYWORD=$1
LINES_BEFORE=$2
LINES_AFTER=$3
FILE=$4
LINE_NO=$(grep -n $KEYWORD $FILE | sed 's/:.*//' )
echo "Keyword found in line: $LINE_NO"
LINE_START=$(($LINE_NO-$LINES_BEFORE))
LINE_END=$(($LINE_NO+$LINES_AFTER))
echo "Deleting lines $LINE_START to $LINE_END!"
sed -i "$LINE_START,$LINE_END d" $FILE
Please note that this will work only if the keyword is found once! Adapt the script to your needs!
I am trying to grep a line from a file that starts with 'Residue XXX'. It works when I have only 'Residue' but does not work if I have 'Residue XXX' Any reasons for this behavior?
Here it is working:
grep '^Residue' log.txt
Residue XXX highDenisity
and not working:
grep '^Residue XXX' log.txt
Hold up your hand if you think there's a tab character between Residue and XXX.
Use a whitespace-agnostic regex instead.
grep '^Residue\s*XXX' log.txt
This regex covers multiple tabs or a tab and a space between the words.
Or you can just do a logical and with awk like this:
awk '/^Residue/ && /XXX/' file
This will only output lines that do start with Residue and also contain XXX
You can also make the and like this:
awk '/^Residue.*XXX/' file
The line I seek is stored in the file data.txt and is the only line of text that occurs only once.
How do I go about finding that particular line using linux?
This is a little bit old, but I think you are looking for this...
cat data.txt | sort | uniq -u
This will show the unique values that only occur once in the file. I assume you are familiar with "over the wire" if you are asking?? If so, this is what you are looking for.
To provide some context (I need more rep to comment) this is a question that features in an online "wargame" called Bandit that involves using the command line to discover passwords on an online Linux server to advance up the levels.
For those who would like to see data.txt in full I've Pastebin'd it here however it looks like this:
NN4e37KW2tkIb3dC9ZHyOPdq1FqZwq9h
jpEYciZvDIs6MLPhYoOGWQHNIoQZzE5q
3rpovhi1CyT7RUTunW30goGek5Q5Fu66
JOaWd4uAPii4Jc19AP2McmBNRzBYDAkO
JOaWd4uAPii4Jc19AP2McmBNRzBYDAkO
9WV67QT4uZZK7JHwmOH0jnhurJMwoGZU
a2GjmWtTe3tTM0ARl7TQwraPGXgfkH4f
7yJ8imXc7NNiovDuAl1ZC6xb0O0mMBx1
UsvVyFSfZZWbi6wgC7dAFyFuR6jQQUhR
FcOJhZkHlnwqcD8QbvjRyn886rCrnWZ7
E3ugYDa6Wh2y8C8xQev7vOS8O3OgG1Hw
E3ugYDa6Wh2y8C8xQev7vOS8O3OgG1Hw
ME7nnzbId4W3dajsl6Xtviyl5uhmMenv
J5lN3Qe4s7ktiwvcCj9ZHWrAJcUWEhUq
aouHvjzagN8QT2BCMB6e9rlN4ffqZ0Qq
ZRF5dlSuwuVV9TLhHKvPvRDrQ2L5ODfD
9ZjR3NTHue4YR6n4DgG5e0qMQcJjTaiM
QT8Bw9ofH4x3MeRvYAVbYvV1e1zq3Xim
i6A6TL6nqvjCAPvOdXZWjlYgyvqxmB7k
tx7tQ6kgeJnC446CHbiJY7fyRwrwuhrs
One way to do it is to use:
sort data.txt | uniq -u
The sort command is like cat in that it displays the contents of the file however it sorts the file lexicographically by lines (it reorders them alphabetically so that matching ones are together).
The | is a pipe that redirects the output from one command into another.
The uniq command reports or omits repeated lines and by passing it the -u argument we tell it to report only unique lines.
Used together like this, the command will sort data.txt lexicographically by each line, find the unique line and print it back in the terminal for you.
sort -u data.txt | while read line; do if [ $(grep -c $line data.txt) == 1 ] ;then echo $line; fi; done
was mine solution, until I saw here easy one:
sort data.txt | uniq -u
Add more information to you post.
How data.txt look like?
Like this:
11111111
11111111
pass1111
11111111
Or like this
afawfdgd
password
somethin
gelse...
And, do you know the password is in file or you search for not repeat string.
If you know password, use something like this
cat data.txt | grep 'password'
If you don`t know the password and this password is only unique line in file you must create a script.
For example in Python
file = open("data.txt","r")
f = file.read()
for line in f:
if 'pass' in line:
print pass
Of course replace pass with something else.
For example some slice from line.
And one with only one tool in use, awk:
awk '{a[$1]++}END{for(i in a){if(a[i] == 1){print i} }}' data.txt
sort data.txt | uniq -c | grep 1\ ?*
and it will print the only text that occurs only one time
do not forget to put space after the backslash
sort data.txt | uniq -c | grep 1
you will find only one that accures one time