Suppose I have this text
The code for 233-CO is the main reason for 45-DFG and this 45-GH
Now I have this regexp \s[0-9]+-\w+ which matches 233-CO, 45-DFG and 45-GH.
How can I display just the third match 45-GH?
sed -re 's/\s[0-9]+-\w+/\3/g' file.txt
where \3 should be the third regexp match.
Is it mandatory to use sed? You could do it with grep, using arrays:
text="The code for 233-CO is the main reason for 45-DFG and this 45-GH"
matches=( $(echo "$text" | grep -o -m 3 '\s[0-9]\+-\w\+') ) # store first 3 matches in array
echo "${matches[0]} ${matches[2]}" # prompt first and third match
To find the last occurence of your pattern, you can use this:
$ sed -re 's/.*\s([0-9]+-\w+).*/\1/g' file
45-GH
if awk is accepted, there is an awk onliner, you give the No# of match you want to grab, it gives your the matched str.
awk -vn=$n '{l=$0;for(i=1;i<n;i++){match(l,/\s[0-9]+-\w+/,a);l=substr(l,RSTART+RLENGTH);}print a[0]}' file
test
kent$ echo $STR #so we have 7 matches in str
The code for 233-CO is the main reason for 45-DFG and this 45-GH,foo 004-AB, bar 005-CC baz 006-DDD and 007-AWK
kent$ n=6 #now I want the 6th match
#here you go:
kent$ awk -vn=$n '{l=$0;for(i=1;i<=n;i++){match(l,/\s[0-9]+-\w+/,a);l=substr(l,RSTART+RLENGTH);}print a[0]}' <<< $STR
006-DDD
This might work for you (GNU sed):
sed -r 's/\b[0-9]+-[A-Z]+\b/\n&\n/3;s/.*\n(.*)\n.*/\1/' file
s/\b[0-9]+-[A-Z]+\b/\n&\n/3 prepend and append \n (newlines) to the third (n) pattern in question.
s/.*\n(.*)\n.*/\1/ delete the text before and after the pattern
With grep for matching and sed for printing the occurrence:
$ egrep -o '\b[0-9]+-\w+' file | sed -n '1p'
233-CO
$ egrep -o '\b[0-9]+-\w+' file | sed -n '2p'
45-DFG
$ egrep -o '\b[0-9]+-\w+' file | sed -n '3p'
45-GH
Or with a little awk passing the occurrence to print using the variable o:
$ awk -v o=1 '{for(i=0;i++<NF;)if($i~/[0-9]+-\w+/&&j++==o-1)print $i}' file
233-CO
$ awk -v o=2 '{for(i=0;i++<NF;)if($i~/[0-9]+-\w+/&&j++==o-1)print $i}' file
45-DFG
$ awk -v o=3 '{for(i=0;i++<NF;)if($i~/[0-9]+-\w+/&&j++==o-1)print $i}' file
45-GH
Related
I have a text file, and I'm trying to get an array of strings containing between $..$ delimiters (LaTeX formulas) using bash script. My current code doesn't work, result is empty:
#!/bin/bash
array=($(grep -o '\$([^\$]*)\$' test.txt))
echo ${array[#]}
I tested this regex here, it finds the matches. I use the following test string:
b5f1e7$bfc2439c621353$d1ce0$629f$b8b5
Expected result is
bfc2439c621353 629f
But echo returns empty. Although if I use '[0-9]\+' it works:
5 1 7 2439 621353 1 0 629 8 5
What do I do wrong?
How about:
grep -o '\$[^$]*\$' test.txt | tr -d '$'
This is basically performing your original grep (but without the brackets, which were causing it to not match), then removing the first/last characters from each match.
You may use awk with input field separator as $:
s='b5f1e7$bfc2439c621353$d1ce0$629f$b8b5'
awk -F '$' '{for (i=2; i<=NF; i+=2) print $i}' <<< "$s"
Note that this awk command doesn't validate input. If you want awk to allow for only valid inputs then you may use this gnu awk command with FPAT:
awk -v FPAT='\\$[^$]*\\$' '{for (i=1; i<=NF; i++) {gsub(/\$/, "", $i); print $i}}' <<< "$s"
bfc2439c621353
629f
What about this?
grep -Eo '\$[^$]+\$' a.txt | sed 's/\$//g'
I'm using sed to replace the $.
Try escaping your braces:
tst> grep -o '\$\([^\$]*\)\$' test.txt
$bfc2439c621353$
$629f$
of course, you then have to strip out the $ signs (-o prints the entire match). You can try sed instead:
tst> sed 's/[^\$]*\$\([^\$]*\)\$[^\$]*/\1\n/g' test.txt
bfc2439c621353
629f
Why is your expected output given b5f1e7$bfc2439c621353$d1ce0$629f$b8b5 the two elements bfc2439c621353 629f rather than the three elements bfc2439c621353 d1ce0 629f?
Here's a single grep command to extract those:
$ grep -Po '\$\K[^\$]*(?=\$)' <<<'b5f1e7$bfc2439c621353$d1ce0$629f$b8b5'
bfc2439c621353
d1ce0
629f
(This requires GNU grep as compiled with libpcre for -P)
This uses \$\K (equivalent to (?<=\$)to look behind at the first $ and (?=\$) to look ahead to the next $. Since these are lookarounds, they are not absorbed by grep in the process and therefore d1ce0 is available to be found.
Here's a single POSIX sed command to extract those:
$ sed 's/^[^$]*\$//; s/\$[^$]*$//; s/\$/\n/g' \
<<<'b5f1e7$bfc2439c621353$d1ce0$629f$b8b5'
bfc2439c621353
d1ce0
629f
This does not use any GNU notation and should work on any POSIX-compatible system (such as OS X). It removes the leading and trailing portions that aren't wanted, then replaces each $ with a newline.
Using bash regex:
var="b5f1e7\$bfc2439c621353\$d1ce0\$629f\$b8b5" # string to var
while [[ $var =~ ([^$]*\$)([^$]*)\$(.*) ]] # matching
do
echo -n "${BASH_REMATCH[2]} " # 2nd element has the match
var="${BASH_REMATCH[3]}" # 3rd is the rest of the string
done
echo # trailing newline
bfc2439c621353 629f
I want to find the set of words that contain two consecutive e’s, and also contains two y’s.
So far i got to /eeyy/
Alteration with ERE:
$ echo evyyree | grep -E '.*ee.*yy|.*yy.*ee'
evyyree
$ echo eveeryy | grep -E '.*ee.*yy|.*yy.*ee'
eveeryy
If the match needs to be in the same word, you can do:
$ echo "eee yyyy" | grep -E 'ee[^[:space:]]*yy|yy[^[:space:]]*ee' # no match
$ echo "eeeyyyy" | grep -E 'ee[^[:space:]]*yy|yy[^[:space:]]*ee'
eeeyyyy
Then only that word:
$ echo 'eeeyy heelo' | grep -Eo 'ee[^[:space:]]*yy|yy[^[:space:]]*ee'
eeeyy
Pipe it:
$ echo eennmmyy | grep ee | grep yy
eennmmyy
awk approach to match all words that contain both ee and yy:
s="eennmmyy heello thees-whyy someyy"
echo $s | awk '{for(i=1;i<=NF;i++) if($i~/ee/ && $i~/yy/) print $i}'
The output:
eennmmyy
thees-whyy
The only sensible and extensible way to do this is with awk:
awk '/ee/&&/yy/' file
Imagine trying to do it the grep way if you also had to find zz. Here's awk:
awk '/ee/&&/yy/&&/zz/' file
and here's grep:
grep -E 'ee.*yy.*zz|ee.*zz.*yy|yy.*ee.*zz|yy.*zz.*ee|zz.*yy.*ee|zz.*ee.*yy' file
Now add a 4th additional string to search for and see what that looks like!
I need to search in multiple files for a PATTERN, if found display the file, line and PATTERN surrounded by a few extra chars. My problem is that if the line matching the PATTERN ends with ^M (CRLF) grep prints an empty line instead.
Create a file like this, first line "a^M", second line "a", third line empty line, forth line "a" (not followed by a new line).
a^M
a
a
Without trying to match a few chars after the PATTERN all occurrences are found and displayed:
# grep -srnoEiI ".{0,2}a" *
1:a
2:a
4:a
If I try to match any chars at the end of the PATTERN, it prints an empty line instead of line one, the one ending in CRLF:
# grep -srnoEiI ".{0,2}a.{0,2}" *
2:a
4:a
How can I change this to act as expected ?
P.S. I will like to fix this grep, but I will accept other solutions for example in awk.
EDIT:
Based on the answers below I choose to strip the \r and force grep to pipe the colors to tr:
grep --color=always -srnoEiI ".{0,2}a.{0,2}" * | tr -d '\r'
Here's a simpler case that reproduces your problem:
# Output
echo $'a\r' | grep -o "a"
# No output
echo $'a\r' | grep -o "a."
This is beacuse the ^M matches like a regular character, and makes your terminal overwrite its output (this is purely cosmetic).
How you want to fix this depends on what you want to do.
# Show the output in hex format to ensure it's correct
$ echo $'a\r' | grep -o "a." | od -t x1 -c
0000000 61 0d 0a
a \r \n
# Show the output in visually less ambiguous format
$ echo $'a\r' | grep -o "a." | cat -v
a^M
# Strip the carriage return
$ echo $'a\r' | grep -o "a." | tr -d '\r'
a
awk -v pattern="a" '$0 ~ pattern && !/\r$/ {print NR ": " $0}' file
or
sed -n '/a/{/\r$/!{=;p}}' ~/tmp/srcfile | paste -d: - -
Both of these do: find the pattern, see if the line does not end in a carriage return, print the line number and the line. For the sed, the line number is on its own line, so we have to join two consecutive lines with a colon.
You could use pcregrep:
pcregrep -n '.{0,2}a.{0,2}' inputfile
For your sample input:
$ printf $'a\r\na\n\na\n' | pcregrep -n '.{0,2}a.{0,2}'
1:a
2:a
4:a
A couple more ways:
Use the dos2unix utility to convert the dos-style line endings to unix-style:
dos2unix myfile.txt
Or preprocess the file using tr to remove the CR characters, then pipe to grep:
$ tr -d '\r' < myfile.txt | grep -srnoEiI ".{0,2}a.{0,2}"
1:a
2:a
4:a
$
Note dos2unix may need to be installed on whatever OS you are using. More than likely tr will be available on any POSIX-compliant OS.
You can use awk with a custom field separator:
awk -F '[[:blank:]\r]' '/.{0,2}a.{0,2}/{print FILENAME, NR, $1}' OFS=':' file
TESTING:
Your grep command:
grep -srnoEiI ".{0,2}a.{0,2}" file|cat -vte
file:1:a^M$
file:2:a$
file:4:a$
Suggested awk commmand:
awk -F '[[:blank:]\r]' '/.{0,2}a.{0,2}/{print FILENAME, NR, $1}' OFS=':' file|cat -vte
file:1:a$
file:2:a$
file:4:a$
Let A,B,C,D are the words
Input File :
..
A/B/C/D
W/B/C/Z
L/B/C/O
..
Output file:
..
A/B/C/A
W/B/C/W
L/B/C/L
..
Replace the word D with word A one the same line, only if /B/C/ delimiter present in the line and like wise for the other lines
Any sed/awk/perl oneliner to accomplish that
This is a awk solution:
awk -F/ -v OFS=/ '$2=="B" && $3=="C" {$4=$1}1' input.txt
You can do:
sed -re 's/^([^/]*)(\/B\/C\/)([^/]*)$/\1\2\1/' file
Demo:
$ cat file
A/B/C/D
W/B/C/Z
L/B/C/O
$ sed -re 's/^([^/]*)(\/B\/C\/)([^/]*)$/\1\2\1/' file
A/B/C/A
W/B/C/W
L/B/C/L
pearl.306> echo "A/B/C/D"|awk '{split($0,a,"/");print a[1]"/"a[2]"/"a[3]"/"a[1]}'
A/B/C/A
pearl.307>
another way is:
pearl.309> echo "A/B/C/D" | awk -F"/" '{OFS="/"}{$NF=$1;print}'
A/B/C/A
pearl.310>
pearl.318> cat file1
A/B/C/D
W/B/C/Z
L/B/C/O
pearl.319> awk -F"/" '{OFS="/"}{$NF=$1;print}' file1
A/B/C/A
W/B/C/W
L/B/C/L
pearl.320>
This might work for you:
sed 's|^\(\(.\)/B/C/\).|\1\2|' file
if A/B/C/D are real words e.g. wordA/wordB/wordC/wordD, then:
sed 's/|^\(\([^/]*\)/wordB/wordC/\).*|\1\2|' file
This should do the trick. perl -p -e 's/D/A/g'
In sed sed -e 's/D/A/'
perl -pe 's#(/B/C/)(.*)#$1$`#' file
this should work +
I have a file with a string on each line... ie.
test.434
test.4343
test.4343t34
test^tests.344
test^34534/test
I want to find any line containing a "^" and replace entire line with a blank.
I was trying to use sed:
sed -e '/\^/s/*//g' test.file
This does not seem to work, any suggestions?
sed -e 's/^.*\^.*$//' test.file
For example:
$ cat test.file
test.434
test.4343
test.4343t34
test^tests.344
test^34534/test
$ sed -e 's/^.*\^.*$//' test.file
test.434
test.4343
test.4343t34
$
To delete the offending lines entirely, use
$ sed -e '/\^/d' test.file
test.434
test.4343
test.4343t34
other ways
awk
awk '!/\^/' file
bash
while read -r line
do
case "$line" in
*"^"* ) continue;;
*) echo "$line"
esac
done <"file"
and probably the fastest
grep -v "\^" file