I have a text file that carries the following values
Key 1: 0e3f02b50acfe57e21ba991b39d75170d80d98e831400250d3b4813c9b305fd801
Key 2: 8e3db2b4cdfc55d91512daa9ed31b348545f6ba80fcf2c3e1dbb6ce9405f959602
I am using the following grep command to extract value of Key 1
grep -Po '(?<=Key 1=)[^"]*' abc.txt
However, it doesn't seem to work.
Please help me figure out the correct grep command
My output should be:
0e3f02b50acfe57e21ba991b39d75170d80d98e831400250d3b4813c9b305fd801
A grep+cut solution: Search for the right key, then return the third field:
$ grep '^Key 1:' abc.txt | cut -d' ' -f3
Or, equivalently in awk:
$ awk '/^Key 1:/ { print $3 }' abc.txt
Don't use grep to modify the matching string, that's pointless, messy, and non-portable when sed already does it concisely and portably:
$ sed -n 's/^Key 1: //p' file
0e3f02b50acfe57e21ba991b39d75170d80d98e831400250d3b4813c9b305fd801
If your version of grep doesn't support PCRE, you can do the same with sed, e.g.
$ sed -n '/^Key 1: [^"]/s/^Key 1: //p' file.txt
0e3f02b50acfe57e21ba991b39d75170d80d98e831400250d3b4813c9b305fd801
Explanation
-n suppress normal printing of pattern space
/^Key 1: [^"]/ find the pattern
s/^Key 1: // substitute (nothing) for pattern
p print the remainder
You have mistake in your grep (change Key 1= to Key 1:)
grep -Po '(?<=Key 1: )[^"]*' abc.txt
grep -oP '(?<=Key 1: )[^"]+' abc.txt
seems to work for me.
Related
I am trying to get a substring in a string that is in a large line of data.
The regex (INC............) matches the substring I am trying to get the value of at https://regexr.com/, but I am unable to get the value of the substring into a variable or print it out.
The part of the string around this value is
......TemplateID2":null,"Incident Number":"INC000006743193","Priority":"High","mc_ueid":null,"Assint......
I am getting the error char 26: unknown option to `s' when I try this or the entire string is printed out.
cat /tmp/file1 | sed -n 's/\(INC............\)/\1/p'
cat /tmp/file1 | sed -n 's/./*\(INC............).*/\1/'
Using sed, you need to remove what precedes and follows the string:
sed 's/.*\(INC............\).*/\1/' file
But you can also use grep, if your implementation supports the -o option:
grep -o 'INC............' file
Perl can be used, too:
perl -lne 'print $1 if /(INC............)/' file
That looks like JSON. If it's got {braces} around it which you cut out before posting (tsk tsk), you should definitely use jq if it's available. That said, this page needs some awk!
POSIX (works everywhere):
awk 'match($0, /INC[^"]+/) {print substr($0, RSTART, RLENGTH)}' /tmp/file1`
GNU (works on GNU/Linux):
gawk 'match($0, /INC[^"]+/, a) {print a[0]}' /tmp/file1
If you have more than one match per line (GNU):
gawk '{while(match($0=substr($0, RSTART+RLENGTH), /INC[0-9]+/, a)) print a[0]}' /tmp/file1
I've a file with the below header generated by certain process
Link: <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=2>; rel="next", <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=8>; rel="last"
I want to cut just the number 8 from page=8 in the above content. How to go about it? Appreciate any help.
Try this -
$ cat f
Link: <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=2>; rel="next", <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=8>; rel="last"
$ awk -F'[&=<>]' '{for(i=1;i<=NF;i++) if($i ~ /^page$/) {print $(i+1)}}' f
2
8
If it is getting appended then you will get the last value using below awk :
$ awk -F'[&=<>]' '{for(i=1;i<=NF;i++) if($i ~ /^page$/) {kk=$(i+1)}} END{print kk}' ff
8
Limitation : Currently you have page=2 and page=8 and above command
will print the last page value.
And if you always want to print the 2nd value "8" (Added extra lines to the existing url, considering that it will keep on increasing and you always need the 2nd value then use below) -
$ cat f
Link: <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=2>; rel="next", <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=8>; rel="last"
<https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=8>; rel="last"
$ awk -v k=1 -F'[&=<>]' '{for(i=1;i<=NF;i++) if(($i ~ /^page$/) && (k==2) ) {print $(i+1)} k++}' f
8
Following is an implementation using grep:
grep -Po "&page=[0-9]*" <file_name> | grep -Po "[0-9]*"
Example:
echo 'Link: <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=2>; rel="next", <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=8000>; rel="last"' | grep -Po "&page=[0-9]*" | grep -Po "[0-9]*"
This will produces the result as expected.
echo 'Link: <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=2>; rel="next", <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=12345>; rel="last"' | grep -Po "&page=[0-9]*" |grep -Po "[0-9]*"| awk '2 == NR % $ct'
In awk. reverse the text, remove first [0-9]+=egap, output and rev again:
$ rev foo | awk 'sub(/[0-9]+=egap/,"")||1' |rev
Output:
Link: <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&page=2>; rel="next", <https://rnd.corp.zoom/api/v3/repositories/99/issues?state=all&per_page=100&>; rel="last"
try:
awk '{gsub(/.*page=/,"page=");sub(/>.*/,"");print}' Input_file
Simply substitute the all line with .*page= to page= which is nothing but will go till last page string(as * is a greedy regex match), so then substitute >.*(means starting from > to till end of line) with NULL, then print the line which will be page=8 or last value of the page. Off course I am considering that your Input_file is same as example shown.
awk -F'[= >]' '{print $12}' file
8
awk -F= '{split($8,a,">");print a[1]}' file
8
awk -F= '$8=="8>; rel"{print substr($8,1,1)}' file
8
The fact that a greedy regex is needed here (only the last occurrence of &page= should be matched) enables a simple sed solution:
sed -E 's/^.*&page=([0-9]+).*$/\1/' file
^.*&page= matches everything up to the last occurrence of &page on the line.
([0-9]+) matches one or more digits, and - thanks to enclosure in (...) stores the match in the 1st (and only) capture group, which the replacement string then reference as \1.
.*$ matches any remaining character on the line.
By virtue of the regex having matched the entire line, \1 therefore results in just the captured number as the output.
The above works with both GNU and BSD/macOS sed and takes advantage of modern extended regular expressions (-E), but in case you need a POSIX-compliant solution (which must use basic regular expressions and is therefore more cumbersome):
sed 's/^.*&page=\([0-9]\{1,\}\).*$/\1/' file
With GNU grep (on Linux, as requested), a single-pass grep -Po solution is also possible; like the sed solution, it relies on greedily matching up to the last &page=:
grep -Po "^.*&page=\K[0-9]+" file
-P activates support for PRCEs (Perl-compatible Regular Expressions).
-o only outputs the matching part of the line.
\K drops everything matched so far, so that what [0-9]+ matches - one or more digits - is the only output.
I have some file xxx.conf in text format. I have some text "disablelog = 1" in this file.
When I use
grep -r "disablelog" oscam.conf
output is
disablelog = 1
But i need only value 1.
Do you have some idea please?
one way is to use awk to print just the value
grep -r "disablelog" oscam.conf | awk '{print $3}'
you could also use sed to replace diablelog = with empty
grep -r 'disablelog' oscam.conf | sed -e 's/disablelog = //'
If you also want to get the lines with or without space before and after = use
grep -r 'disablelog' oscam.conf | sed 's/disablelog\s*=\s*//'
above command will also match
disablelog=1
Assuming you need it as a var in a script:
#!/bin/bash
DISABLELOG=$(awk -F= '/^.*disablelog/{gsub(/ /,"",$2);print $2}' /path/to/oscam.conf)
echo $DISABLELOG
When calling this script, the output should be 1.
Edit: No matter wether there is whitespace or not between the equals sign and the value, the above will handle that. The regex should be anchored in either way to improve performance.
Try:
grep -r "disablelog" oscam.conf | awk -F= '{print $2}'
Just for fun a solution without awk
grep -r disablelog | cut -d= -f2 | xargs
xargs is used here to trim the whitespace
I have a file that has the following as the last three lines. I want to retrieve the penultimate line, i.e. 100.000;8438; 06:46:12.
.
.
.
99.900; 8423; 06:44:41
100.000;8438; 06:46:12
Number of patterns: 8438
I don't know the line number. How can I retrieve it using a shell script? Thanks in advance for your help.
Try this:
tail -2 yourfile | head -1
A short sed one-liner inspired by https://stackoverflow.com/a/7671772/5287901
sed -n 'x;$p'
Explanation:
-n quiet mode: dont automatically print the pattern space
x: exchange the pattern space and the hold space (hold space now store the current line, and pattern space the previous line, if any)
$: on the last line, p: print the pattern space (the previous line, which in this case is the penultimate line).
Use this
tail -2 <filename> | head -1
ed and sed can do it as well.
str='
99.900; 8423; 06:44:41
100.000;8438; 06:46:12
Number of patterns: 8438
'
printf '%s' "$str" | sed -n -e '${x;1!p;};h' # print last line but one
printf '%s\n' H '$-1p' q | ed -s <(printf '%s' "$str") # same
printf '%s\n' H '$-2,$-1p' q | ed -s <(printf '%s' "$str") # print last line but two
From: Useful sed one-liners by Eric Pement
# print the next-to-the-last line of a file
sed -e '$!{h;d;}' -e x # for 1-line files, print blank line
sed -e '1{$q;}' -e '$!{h;d;}' -e x # for 1-line files, print the line
sed -e '1{$d;}' -e '$!{h;d;}' -e x # for 1-line files, print nothing
You don't need all of them, just pick one.
tail +2 <filename>
This prints from second line to last line.
To clarify what has already been said:
ec2thisandthat | sort -k 5 | grep 2012- | awk '{print $2}' | tail -2 | head -1
snap-e8317883
snap-9c7227f7
snap-5402553f
snap-3e7b2c55
snap-246b3c4f
snap-546a3d3f
snap-2ad48241
snap-d00150bb
returns
snap-2ad48241
tac <file> | sed -n '2p'
I am using grep recursive to search files for a string, and all the matched files and the lines containing that string are print on the terminal. But is it possible to get the line numbers of those lines too??
ex: presently what I get is /var/www/file.php: $options = "this.target", but what I am trying to get is /var/www/file.php: 1142 $options = "this.target";, well where 1142 would be the line number containing that string.
Syntax I am using to grep recursively is sudo grep -r 'pattern' '/var/www/file.php'
One more question is, how do we get results for not equal to a pattern. Like all the files but not the ones having a certain string?
grep -n SEARCHTERM file1 file2 ...
Line numbers are printed with grep -n:
grep -n pattern file.txt
To get only the line number (without the matching line), one may use cut:
grep -n pattern file.txt | cut -d : -f 1
Lines not containing a pattern are printed with grep -v:
grep -v pattern file.txt
If you want only the line number do this:
grep -n Pattern file.ext | gawk '{print $1}' FS=":"
Example:
$ grep -n 9780545460262 EXT20130410.txt | gawk '{print $1}' FS=":"
48793
52285
54023
grep -A20 -B20 pattern file.txt
Search pattern and show 20 lines after and before pattern
grep -nr "search string" directory
This gives you the line with the line number.
In order to display the results with the line numbers, you might try this
grep -nr "word to search for" /path/to/file/file
The result should be something like this:
linenumber: other data "word to search for" other data
When working with vim you can place
function grepn() {
grep -n $# /dev/null | awk -F $':' '{t = $1; $1 = $2; $2 = t; print; }' OFS=$':' | sed 's/^/vim +/' | sed '/:/s// /' | sed '/:/s// : /'
}
in your .bashrc and then
grepn SEARCHTERM file1 file2 ...
results in
vim +123 file1 : xxxxxxSEARCHTERMxxxxxxxxxx
vim +234 file2 : xxxxxxSEARCHTERMxxxxxxxxxx
Now, you can open vim on the correspondending line (for example line 123) by simply copying
vim +123 file1
to your shell.