I want to separate string and number in a file to get a specific number in bash script, such as:
Branches executed:75.38% of 1190
I want to only get number
75.38
. I have try like the code below
$new_value=value | sed -r 's/.*_([0-9]*)\..*/\1/g'
but it was incorrect and it was failed.
How should it works? Thank you before for your help.
You can use the following regex to extract the first number in a line:
^[^0-9]*\([0-9.]*\).*$
Usage:
% echo 'Branches executed:75.38% of 1190' | sed 's/^[^0-9]*\([0-9.]*\).*$/\1/'
75.38
Give this a try:
value=$(sed "s/^Branches executed:\([0-9][.0-9]*[0-9]*\)%.*$/\1/" afile)
It is assumed that the line appears only once in afile.
The value is stored in the value variable.
There are several things here that we could improve. One is that you need to escape the parentheses in sed: \(...\)
Another one is that it would be good to have a full specification of the input strings as well as a good script that can help us to play with this.
Anyway, this is my first attempt:
Update: I added a little more bash around this regex so it'll be more easy to play with it:
value='Branches executed:75.38% of 1190'
new_value=`echo $value | sed -e 's/[^0-9]*\([0-9]*\.[0-9]*\).*/\1/g'`
echo $new_value
Update 2: as john pointed out, it will match only numbers that contain a decimal dot. We can fix it with an optional group: \(\.[0-9]\+\)?.
An explanation for the optional group:
\(...\) is a group.
\(...\)? Is a group that appears zero or one times (mind the question mark).
\.[0-9]\+ is the pattern for a dot and one or more digits.
Putting all together:
value='Branches executed:75.38% of 1190'
new_value=`echo $value | sed -e 's/[^0-9]*\([0-9]\+\(\.[0-9]\+\)\?\).*/\1/g'`
echo $new_value
I have spent some time trying to find the answer to this question but with no luck...
I am writing a script which takes a string as input, then searches through a cvs log for a match. The output should be the filename prior to the matching string. I can write a regex to match the filename without any trouble, however I am unsure how to look back for the last occurrence of a regex after a match is made...
There are obviously many files in the cvs log that I will be parsing and the number of lines between the filename and the input string is unknown so cannot do anything like 'grep -B4' etc...
EXAMPLE BELOW (cvs log with irrelevant bits removed) - Need to match a string (which will be given as input to script (e.g. 120233)) and retrieve the filename associated with it, which here will be (Aliases.xml). As I said there are many files so although the regex will match many filenames, i am only interested in the one prior to my matched string.
This is my first time posting here and I am new to programming so I hope this makes sense.
----------
RCS file: /user1/cvs/Aliases.xml
revision 1.18
date: 2015/03/13 16:21:07;
FIX - 217427 - fixed error....
revision 1.17
date: 2013/03/27 08:03:36;
IMPROVEMENT - 120233 - some improvement
You can use awk command for this:
cvs log | awk -F '[:, ]+' -v s='120233' '/RCS file:/{f=$3} f && $0 ~ s{print f; f=""}'
Explanation:
-F '[:, ]+' # make one or more of colon, comma or <space> a field saprator
-v s='120233' # pass search string to awk in variable s
/RCS file:/ # search for string "RCS file:"
{f=$3} # store filename in variable f
f && $0 ~ s # if f is NOT_EMPTY and line matches variable s
print f; f="" # print filename from variable f and set f to EMPTY
I have price strings formatted as
$25.00
in various html files. I would like to use the Linux command line (BASH, presumably with awk or sed) to increase each price by a certain dollar amount ($3 in this case).
In short, I need to find $nn.00 and replace it with $(n+3)n.00
Started to put it together but I don't know how to add 3 sed -r 's/([^$][0-9][0-9][.]00). ????' file.html
Thanks!
Sample data:
$ cat prices_file.html
<p>$25.00</p><p>$78.00</p>
<p>$2.00</p>
<p>$101.00</p>
Solution with Perl:
$ perl -pi.bak -e 's/\$(\d+\.\d+)/sprintf("\$%.2f", $1 + 3)/eg' prices_file.html
After:
$ cat prices_file.html
<p>$28.00</p><p>$81.00</p>
<p>$5.00</p>
<p>$104.00</p>
Above example is one of most common perl use cases with substitution.
It will also backup your original file (in prices_file.html.bak) in case you do something unwanted to it.
What is maybe not so common is evaluation modifier (s///e) which allows you to execute arbitrary perl code in substitution.
Global modifier (s///g) tells perl to replace all occurrences (here in a context of line, if you remove g modifier if would only replace first price in 1st line of given sample data).
In sprintf("\$%.2f", $1 + 3) replacement, $1 refers to matched group [(\d+\.\d+)].
I wanted to organize my mp3 files and rename them using the pattern: artist - song.
I need a regular expression that selects all the words before the first dash, and then the last dash and all characters proceeding after.
In the example, [09] System Of A Down -Toxicity - 03 - Chop Suey.mp3:
all the word items before the first dash: System of a down
the last dash: -
everything else after the last dash: Chop Suey.mp3
How do I do this?
There is a linux command rename which is exactly what you want.
For example, to rename all files matching "*.bak" to strip the extension, you might say
rename 's/\.bak$//' *.bak
To translate uppercase names to lower, you'd use
rename 'y/A-Z/a-z/' *
To rename [09] System Of A Down -Toxicity - 03 - Chop Suey.mp3 to Toxicity - Chop Suey.mp3
# you should rewrite regex to meet your requirement
rename 's/.*-(.*)-.*-(.*)/$1 - $2/' *.mp3
These is a Windows music player foobar2000 which can do the job very well.
Properties -> Format from other fields...
If you are using Windows, I would highly recommend using the freeware program: MP3Tag. It will easily do the renaming you request (and a whole lot more).
You can find the "middle pattern" by doing (-.*-) and replace with -. To remove the numbers you can use (\[[0-9]*\])
May be this could help -
sed 's/[[0-9]\+] \([A-Za-z ]*[^-]\) -.*[^-]- \(.*\)/\1-\2/'
Test:
[jaypal:~/temp] echo "[09] System Of A Down -Toxicity - 03 - Chop Suey.mp3" |
sed 's/[[0-9]\+] \([A-Za-z ]*[^-]\) -.*[^-]- \(.*\)/\1 - \2/'
System Of A Down - Chop Suey.mp3
OR
awk -F"-" '{print $1"-"$4}' | sed 's/[[0-9]\+] //g'
Test:
[jaypal:~/temp] echo "[09] System Of A Down -Toxicity - 03 - Chop Suey.mp3" |
awk -F"-" '{print $1"-"$4}' | sed 's/[[0-9]\+] //g'
System Of A Down - Chop Suey.mp3
I see lots of examples and man pages on how to do things like search-and-replace using sed, awk, or gawk.
But in my case, I have a regular expression that I want to run against a text file to extract a specific value. I don't want to do search-and-replace. This is being called from bash. Let's use an example:
Example regular expression:
.*abc([0-9]+)xyz.*
Example input file:
a
b
c
abc12345xyz
a
b
c
As simple as this sounds, I cannot figure out how to call sed/awk/gawk correctly. What I was hoping to do, is from within my bash script have:
myvalue=$( sed <...something...> input.txt )
Things I've tried include:
sed -e 's/.*([0-9]).*/\\1/g' example.txt # extracts the entire input file
sed -n 's/.*([0-9]).*/\\1/g' example.txt # extracts nothing
My sed (Mac OS X) didn't work with +. I tried * instead and I added p tag for printing match:
sed -n 's/^.*abc\([0-9]*\)xyz.*$/\1/p' example.txt
For matching at least one numeric character without +, I would use:
sed -n 's/^.*abc\([0-9][0-9]*\)xyz.*$/\1/p' example.txt
You can use sed to do this
sed -rn 's/.*abc([0-9]+)xyz.*/\1/gp'
-n don't print the resulting line
-r this makes it so you don't have the escape the capture group parens().
\1 the capture group match
/g global match
/p print the result
I wrote a tool for myself that makes this easier
rip 'abc(\d+)xyz' '$1'
I use perl to make this easier for myself. e.g.
perl -ne 'print $1 if /.*abc([0-9]+)xyz.*/'
This runs Perl, the -n option instructs Perl to read in one line at a time from STDIN and execute the code. The -e option specifies the instruction to run.
The instruction runs a regexp on the line read, and if it matches prints out the contents of the first set of bracks ($1).
You can do this will multiple file names on the end also. e.g.
perl -ne 'print $1 if /.*abc([0-9]+)xyz.*/' example1.txt example2.txt
If your version of grep supports it you could use the -o option to print only the portion of any line that matches your regexp.
If not then here's the best sed I could come up with:
sed -e '/[0-9]/!d' -e 's/^[^0-9]*//' -e 's/[^0-9]*$//'
... which deletes/skips with no digits and, for the remaining lines, removes all leading and trailing non-digit characters. (I'm only guessing that your intention is to extract the number from each line that contains one).
The problem with something like:
sed -e 's/.*\([0-9]*\).*/&/'
.... or
sed -e 's/.*\([0-9]*\).*/\1/'
... is that sed only supports "greedy" match ... so the first .* will match the rest of the line. Unless we can use a negated character class to achieve a non-greedy match ... or a version of sed with Perl-compatible or other extensions to its regexes, we can't extract a precise pattern match from with the pattern space (a line).
You can use awk with match() to access the captured group:
$ awk 'match($0, /abc([0-9]+)xyz/, matches) {print matches[1]}' file
12345
This tries to match the pattern abc[0-9]+xyz. If it does so, it stores its slices in the array matches, whose first item is the block [0-9]+. Since match() returns the character position, or index, of where that substring begins (1, if it starts at the beginning of string), it triggers the print action.
With grep you can use a look-behind and look-ahead:
$ grep -oP '(?<=abc)[0-9]+(?=xyz)' file
12345
$ grep -oP 'abc\K[0-9]+(?=xyz)' file
12345
This checks the pattern [0-9]+ when it occurs within abc and xyz and just prints the digits.
perl is the cleanest syntax, but if you don't have perl (not always there, I understand), then the only way to use gawk and components of a regex is to use the gensub feature.
gawk '/abc[0-9]+xyz/ { print gensub(/.*([0-9]+).*/,"\\1","g"); }' < file
output of the sample input file will be
12345
Note: gensub replaces the entire regex (between the //), so you need to put the .* before and after the ([0-9]+) to get rid of text before and after the number in the substitution.
If you want to select lines then strip out the bits you don't want:
egrep 'abc[0-9]+xyz' inputFile | sed -e 's/^.*abc//' -e 's/xyz.*$//'
It basically selects the lines you want with egrep and then uses sed to strip off the bits before and after the number.
You can see this in action here:
pax> echo 'a
b
c
abc12345xyz
a
b
c' | egrep 'abc[0-9]+xyz' | sed -e 's/^.*abc//' -e 's/xyz.*$//'
12345
pax>
Update: obviously if you actual situation is more complex, the REs will need to me modified. For example if you always had a single number buried within zero or more non-numerics at the start and end:
egrep '[^0-9]*[0-9]+[^0-9]*$' inputFile | sed -e 's/^[^0-9]*//' -e 's/[^0-9]*$//'
The OP's case doesn't specify that there can be multiple matches on a single line, but for the Google traffic, I'll add an example for that too.
Since the OP's need is to extract a group from a pattern, using grep -o will require 2 passes. But, I still find this the most intuitive way to get the job done.
$ cat > example.txt <<TXT
a
b
c
abc12345xyz
a
abc23451xyz asdf abc34512xyz
c
TXT
$ cat example.txt | grep -oE 'abc([0-9]+)xyz'
abc12345xyz
abc23451xyz
abc34512xyz
$ cat example.txt | grep -oE 'abc([0-9]+)xyz' | grep -oE '[0-9]+'
12345
23451
34512
Since processor time is basically free but human readability is priceless, I tend to refactor my code based on the question, "a year from now, what am I going to think this does?" In fact, for code that I intend to share publicly or with my team, I'll even open man grep to figure out what the long options are and substitute those. Like so: grep --only-matching --extended-regexp
why even need match group
gawk/mawk/mawk2 'BEGIN{ FS="(^.*abc|xyz.*$)" } ($2 ~ /^[0-9]+$/) {print $2}'
Let FS collect away both ends of the line.
If $2, the leftover not swallowed by FS, doesn't contain non-numeric characters, that's your answer to print out.
If you're extra cautious, confirm length of $1 and $3 both being zero.
** edited answer after realizing zero length $2 will trip up my previous solution
there's a standard piece of code from awk channel called "FindAllMatches" but it's still very manual, literally, just long loops of while(), match(), substr(), more substr(), then rinse and repeat.
If you're looking for ideas on how to obtain just the matched pieces, but upon a complex regex that matches multiple times each line, or none at all, try this :
mawk/mawk2/gawk 'BEGIN { srand(); for(x = 0; x < 128; x++ ) {
alnumstr = sprintf("%s%c", alnumstr , x)
};
gsub(/[^[:alnum:]_=]+|[AEIOUaeiou]+/, "", alnumstr)
# resulting str should be 44-chars long :
# all digits, non-vowels, equal sign =, and underscore _
x = 10; do { nonceFS = nonceFS substr(alnumstr, 1 + int(44*rand()), 1)
} while ( --x ); # you can pick any level of precision you need.
# 10 chars randomly among the set is approx. 54-bits
#
# i prefer this set over all ASCII being these
# just about never require escaping
# feel free to skip the _ or = or r/t/b/v/f/0 if you're concerned.
#
# now you've made a random nonce that can be
# inserted right in the middle of just about ANYTHING
# -- ASCII, Unicode, binary data -- (1) which will always fully
# print out, (2) has extremely low chance of actually
# appearing inside any real word data, and (3) even lower chance
# it accidentally alters the meaning of the underlying data.
# (so intentionally leaving them in there and
# passing it along unix pipes remains quite harmless)
#
# this is essentially the lazy man's approach to making nonces
# that kinda-sorta have some resemblance to base64
# encoded, without having to write such a module (unless u have
# one for awk handy)
regex1 = (..); # build whatever regex you want here
FS = OFS = nonceFS;
} $0 ~ regex1 {
gsub(regex1, nonceFS "&" nonceFS); $0 = $0;
# now you've essentially replicated what gawk patsplit( ) does,
# or gawk's split(..., seps) tracking 2 arrays one for the data
# in between, and one for the seps.
#
# via this method, that can all be done upon the entire $0,
# without any of the hassle (and slow downs) of
# reading from associatively-hashed arrays,
#
# simply print out all your even numbered columns
# those will be the parts of "just the match"
if you also run another OFS = ""; $1 = $1; , now instead of needing 4-argument split() or patsplit(), both of which being gawk specific to see what the regex seps were, now the entire $0's fields are in data1-sep1-data2-sep2-.... pattern, ..... all while $0 will look EXACTLY the same as when you first read in the line. a straight up print will be byte-for-byte identical to immediately printing upon reading.
Once i tested it to the extreme using a regex that represents valid UTF8 characters on this. Took maybe 30 seconds or so for mawk2 to process a 167MB text file with plenty of CJK unicode all over, all read in at once into $0, and crank this split logic, resulting in NF of around 175,000,000, and each field being 1-single character of either ASCII or multi-byte UTF8 Unicode.
you can do it with the shell
while read -r line
do
case "$line" in
*abc*[0-9]*xyz* )
t="${line##abc}"
echo "num is ${t%%xyz}";;
esac
done <"file"
For awk. I would use the following script:
/.*abc([0-9]+)xyz.*/ {
print $0;
next;
}
{
/* default, do nothing */
}
gawk '/.*abc([0-9]+)xyz.*/' file