I have a file that has a line stating
version = "12.0.08-SNAPSHOT"
The word version and quoted strings can occur on multiple lines in that file.
I am looking for a single line bash statement that can output the following string:
12.0.08-SNAPSHOT
The version can have RELEASE tag too instead of SNAPSHOT.
So to summarize, given
version = "12.0.08-SNAPSHOT"
expected output: 12.0.08-SNAPSHOT
And given
version = "12.0.08-RELEASE"
expected output: 12.0.08-RELEASE
The following command prints strings enquoted in version = "...":
grep -Po '\bversion\s*=\s*"\K.*?(?=")' yourFile
-P enables perl regexes, which allow us to use features like \K and so on.
-o only prints matched parts instead of the whole lines.
\b ensures that version starts at a word boundary and we do not match things like abcversion.
\s stands for any kind of whitespace.
\K lets grep forget, that it matched the part before \K. The forgotten part will not be printed.
.*? matches as few chararacters as possible (the matching part will be printed) ...
(?=") ... until we see a ", which won't be included in the match either (this is called a lookahead).
Not all grep implementations support the -P option. Alternatively, you can use perl, as described in this answer:
perl -nle 'print $& if m{\bversion\s*=\s*"\K.*?(?=")}' yourFile
Seems like a job for cut:
$ echo 'version = "12.0.08-SNAPSHOT"' | cut -d'"' -f2
12.0.08-SNAPSHOT
$ echo 'version = "12.0.08-RELEASE"' | cut -d'"' -f2
12.0.08-RELEASE
Portable solution:
$ echo 'version = "12.0.08-RELEASE"' |sed -E 's/.*"(.*)"/\1/g'
12.0.08-RELEASE
or even:
$ perl -pe 's/.*"(.*)"/\1/g'.
$ awk -F"\"" '{print $2}'
Related
I am trying to extract the version number from a string. I am unable to find the exact regex to find what I need.
For eg -
1012-EPS-Test-OF-Something-1.3
I need sed to only extract 1.3 from the above line.
I have tried quite a few things until now something like but it is clearly not working out
sed 's/[^0-9.0-9]*//')
With your shown samples, easiest way could be. Simply print value of shell variable into awk program as input and then setting field separator as - and printing the last field value in it.
echo "$string" | awk -F'-' '{print $NF}'
2nd solution: In case you could have anything else also apart from version number in last field of your value(where - is field delimiter) then use match function of awk.
echo "$var" |
awk -F'-' 'match($NF,/[0-9]+(\.[0-9]+)*/){print substr($NF,RSTART,RLENGTH)}'
3rd solution: Using GNU grep try following once. Using \K option for GNU grep here. This will match everything till - and then mentioning \K will forget OR wouldn't consider that matched value for printing and will print all further matched value(with further mentioned regex).
echo "$var" | grep -oP '.*-\K\d+(\.\d+)*'
This should work in any grep:
s='1012-EPS-Test-OF-Something-1.3'
grep -Eo '[0-9]+(\.[0-9]+)+' <<< "$s"
1.3
This might work for you (GNU sed):
sed -n 's/.*[^0-9.]//p' file
The regexp is greedy and swallows the whole line .* then steps back a character at a time till the first match of [^0-9.], removes the front portion and prints the remainder.
You can use string manipulation to get the last part after -:
s='1012-EPS-Test-OF-Something-1.3'
s="${s##*-}"
See this online demo:
#!/bin/bash
s='1012-EPS-Test-OF-Something-1.3'
s="${s##*-}"
echo "$s"
# => 1.3
See 10.1. Manipulating Strings:
${string##substring}
Deletes longest match of $substring from front of $string.
$ acpi
Battery 0: Charging, 18%, 01:37:09 until charged
How to grep the battery level value without percentage character (18)?
This should do it but I'm getting an empty result:
acpi | grep -e '(?<=, )(.*)(?=%)'
Your regex is correct but will work with experimental -P or perl mode regex option in gnu grep. You will also need -o to show only matching text.
Correct command would be:
grep -oP '(?<=, )\d+(?=%)'
However, if you don't have gnu grep then you can also use sed like this:
sed -nE 's/.*, ([0-9]+)%.*/\1/p' file
18
Could you please try following, written and tested in link https://ideone.com/nzSGKs
your_command | awk 'match($0,/Charging, [0-9]+%/){print substr($0,RSTART+10,RLENGTH-11)}'
Explanation: Adding detailed explanation for above only for explanation purposes.
your_command | ##Running OP command and passing its output to awk as standrd input here.
awk ' ##Starting awk program from here.
match($0,/Charging, [0-9]+%/){ ##Using match function to match regex Charging, [0-9]+% in line here.
print substr($0,RSTART+10,RLENGTH-11) ##Printing sub string and printing from 11th character from starting and leaving last 11 chars here in matched regex of current line.
}'
Using awk:
awk -F"," '{print $2+0}'
Using GNU sed:
sed -rn 's/.*\, *([0-9]+)\%\,.*/\1/p'
You can use sed:
$ acpi | sed -nE 's/.*Charging, ([[:digit:]]*)%.*/\1/p'
18
Or, if Charging is not always in the string, you can look for the ,:
$ acpi | sed -nE 's/[^,]*, ([[:digit:]]*)%.*/\1/p'
Using bash:
s='Battery 0: Charging, 18%, 01:37:09 until charged'
res="${s#*, }"
res="${res%%%*}"
echo "$res"
Result: 18.
res="${s#*, }" removes text from the beginning to the first comma+space and "${res%%%*}" removes all text from end till (and including) the last occurrence of %.
I am running a linux bash script. From stout lines like: /gpx/trk/name=MyTrack1, I want to keep only the end of line after =.
I am struggling to understand why the following sed command is not working as I expect:
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
(I also tried)
echo "/gpx/trk/name=MyTrack1" | sed -e "s/^*\=//"
The return is always /gpx/trk/name=MyTrack1 and not MyTrack1
An even simpler way if this is the only structure you are concerned about:
echo "/gpx/trk/name=MyTrack1" | cut -d = -f 2
Simply try:
echo "/gpx/trk/name=MyTrack1" | sed 's/.*=//'
Solution 2nd: With another sed.
echo "/gpx/trk/name=MyTrack1" | sed 's/\(.*=\)\(.*\)/\2/'
Explanation: As per OP's request adding explanation for this code here:
s: Means telling sed to do substitution operation.
\(.*=\): Creating first place in memory to keep this regex's value which tells sed to keep everything in 1st place of memory from starting to till = so text /gpx/trk/name= will be in 1 place.
\(.*\): Creating 2nd place in memory for sed telling it to keep everything now(after the match of 1st one, so this will start after =) and have value in it as MyTrack1
/\2/: Now telling sed to substitute complete line with only 2nd memory place holder which is MyTrack1
Solution 3rd: Or with awk considering that your Input_file is same as shown samples.
echo "/gpx/trk/name=MyTrack1" | awk -F'=' '{print $2}'
Solution 4th: With awk's match.
echo "/gpx/trk/name=MyTrack1" | awk 'match($0,/=.*$/){print substr($0,RSTART+1,RLENGTH-1)}'
$ echo "/gpx/trk/name=MyTrack1" | sed -e "s/^.*=//"
MyTrack1
The regular expression ^.*= matches anything up to and including the last = in the string.
Your regular expression ^*= would match the literal string *= at the start of a string, e.g.
$ echo "*=/gpx/trk/name=MyTrack1" | sed -e "s/^*=//"
/gpx/trk/name=MyTrack1
The * character in a regular expression usually modifies the immediately previous expression so that zero or more of it may be matched. When * occurs at the start of an expression on the other hand, it matches the character *.
Not to take you off the sed track, but this is easy with Bash alone:
$ echo "$s"
/gpx/trk/name=MyTrack1
$ echo "${s##*=}"
MyTrack1
The ##*= pattern removes the maximal pattern from the beginning of the string to the last =:
$ s="1=2=3=the rest"
$ echo "${s##*=}"
the rest
The equivalent in sed would be:
$ echo "$s" | sed -E 's/^.*=(.*)/\1/'
the rest
Where #*= would remove the minimal pattern:
$ echo "${s#*=}"
2=3=the rest
And in sed:
$ echo "$s" | sed -E 's/^[^=]*=(.*)/\1/'
2=3=the rest
Note the difference in * in Bash string functions vs a sed regex:
The * in Bash (in this context) is glob like - itself means 'any character'
The * in a regex refers to the previous pattern and for 'any character' you need .*
Bash has extensive string manipulation functions. You can read about Bash string patterns in BashFAQ.
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 have a file with strings similar to this:
abcd u'current_count': u'2', u'total_count': u'3', u'order_id': u'90'
I have to find current_count and total_count for each line of file. I am trying below command but its not working. Please help.
grep current_count file | sed "s/.*\('current_count': u'\d+'\).*/\1/"
It is outputting the whole line but I want something like this:
'current_count': u'3', 'total_count': u'3'
It's printing the whole line because the pattern in the s command doesn't match, so no substitution happens.
sed regexes don't support \d for digits, or x+ for xx*. GNU sed has a -r option to enable extended-regex support so + will be a meta-character, but \d still doesn't work. GNU sed also allows \+ as a meta-character in basic regex mode, but that's not POSIX standard.
So anyway, this will work:
echo -e "foo\nabcd u'current_count': u'2', u'total_count': u'3', u'order_id': u'90'" |
sed -nr "s/.*('current_count': u'[0-9]+').*/\1/p"
# output: 'current_count': u'2'
Notice that I skip the grep by using sed -n s///p. I could also have used /current_count/ as an address:
sed -r -e '/current_count/!d' -e "s/.*('current_count': u'[0-9]+').*/\1/"
Or with just grep printing only the matching part of the pattern, instead of the whole line:
grep -E -o "'current_count': u'[[:digit:]]+'
(or egrep instead of grep -E). I forget if grep -o is POSIX-required behaviour.
For me this looks like some sort of serialized Python data. Basically I would try to find out the origin of that data and parse it properly.
However, while being hackish, sed can also being used here:
sed "s/.*current_count': [a-z]'\([0-9]\+\).*/\1/" input.txt
sed "s/.*total_count': [a-z]'\([0-9]\+\).*/\1/" input.txt