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 $new_value
Related
I have an URL in the format like https://foo.bar.whoo.dum.io, for which I like to replace the foo string with something else. Of course, the foo part is unknown and can be anything.
I tried with a simple regex like (.+?)\.(.+), but it seems that regex in Bash is always greedy (or?).
My best attempt is to split the string by . and then join it back with the first part left out, but I was wondering, whether there is a more intuitive, different solution.
Thank you
There are a lot of ways of getting the desired output.
If you're sure the url will always start with https://, we can use parameter expansion to remove everything before the first . and then add the replacement you need:
input="https://foo.bar.whoo.dum.io"
echo "https://new.${input#*.}"
Will output
https://new.bar.whoo.dum.io
Try it online!
You can use sed:
url='https://foo.bar.whoo.dum.io'
url=$(sed 's,\(.*://\)[^/.]*,\1new_value,' <<< "$url")
Here, the sed command means:
\(.*://\) - Capturing group 1: any text and then ://
[^/.]* - zero or more chars other than / and .
\1new_value - replaces the match with the Group 1 and new_value is appended to this group value.
See the online demo:
url='https://foo.bar.whoo.dum.io'
sed 's,\(.*://\)[^/.]*,\1new_value,' <<< "$url"
# => https://new_value.bar.whoo.dum.io
1st solution: Using Parameter expansion capability of bash here, adding this solution. Where newValue is variable with new value which you want to have in your url.
url='https://foo.bar.whoo.dum.io'
newValue="newValue"
echo "${url%//*}//$newValue.${url#*.}"
2nd solution: With your shown samples, please try following sed code here. Where variable url has your shown sample url value in it.
echo "$url" | sed 's/:\/\/[^.]*/:\/\/new_value/'
Explanation: Simple explanation would be, printing shell variable named url value by echo command and sending it as a standard input to sed command. Then in sed command using its capability of substitution here. Where substituting :// just before1st occurrence of . with ://new_value as per requirement.
From the following URL:
https://console.developers.google.com/storage/browser/test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/]
I need to extract the following part:
test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/
I'm pretty bad at regex. I came up with the following but it doesn't work:
sed -n "s/^.*browser\(test-lab.*/.*/\).*$/\1/p"
Can anyone help with what I'm doing wrong?
Could you please try with awk solution also and let me know if this helps you.
echo "https://console.developers.google.com/storage/browser/test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/" | awk '{sub(/.*browser\//,"");sub(/\/$/,"");print}'
Explanation: Simply, substituting everything till browser/ then substituting last / with NULL.
EDIT1: Adding a sed solution here too.
sed 's/\(.[^//]*\)\/\/\(.[^/]*\)\(.[^/]*\)\(.[^/]*\)\/\(.*\)/\5/' Input_file
Output will be as follows.
test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/
Explanation of sed command: Dividing the whole line into parts and using sed's ability to keep the matched regex into memory so here are the dividers I used.
(.[^//]):* Which will have the value till https: in it and if anyone wants to print it you could use \1 for it because this is very first buffer for sed.
//: Now as per URL // comes to mentioning them now.
(.[^/]):* Now comes the 2nd part for sed's buffer which will have value console.developers.google.com in it, because REGEX looks for very first occurrence of / and stops matching there itself.
(.[^/]) && (.[^/]) && /(.):* These next 3 occurrences works on same method of storing buffers like they will look for first occurrence of / and keep the value from last matched letter's next occurrence to till 1st / comes.
/\5/: Now I am substituting everything with \5 means 5th buffer which contains values as per OP's instructions.
Use a different sed delimiter and don't forget to escape the braces.
avinash:~/Desktop$ echo 'https://console.developers.google.com/storage/browser/test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/]' | sed 's~.*/browser/\([^/]*/[^/]*/\).*~\1~'
test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/
OR
Use grep with oP parameters.
avinash:~/Desktop$ echo 'https://console.developers.google.com/storage/browser/test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/]' | grep -oP '/browser/\K[^/]*/[^/]*/'
test-lab-acteghe53j0sf-jrf3f8u8p12n4/2017-09-27_15:23:07.566833_MPoy/
I have an issue with string manipulation in bash. I have a list of names, each name being composed of two parts, chars and numbers: for example
abcdef01234
I want to cut the last character before the numeric part starts, in this case
f
I think there is a regular expression to help me with this but just can't figure it out. AWK/sed solutions are accepted too. Hope someone can help.
Thank you.
In bash it can be done with parameter expansion with substring removal and string indexes, e.g.,
a=abcdef01234 # your string
tmp=${a%%[0-9]*} # remove all numbers from right
echo ${tmp:(-1)} # output last of remaining chars
Output: f
You can use a regexp like [a-zA-Z]+([a-zA-Z])[0-9]+. If you know how to use sed is pretty easy.
Check https://regex101.com/r/XCkKM5/1
The match will be the letter you want.
^\w+([a-zA-Z])\d+$
As a sed command (on OSX) this will be :
echo "abcdef12345" | sed -E "s#^[a-zA-Z]+([a-zA-Z])[0-9]+\$#\1#"
try following too once.
echo "abcdef01234" | awk '{match($0,/[a-zA-Z]+/);print substr($0,RLENGTH,1)}'
I have a list of names I assume is a file, file. Using grep's PCRE and (positive) lookahead:
$ grep -oP "[a-z](?=[^a-z])" file
f
It prints out the first (lowercase) letter followed by a non-(lowercase)-letter.
I am looking to extract everything form a string but ignore everything after encountering numbers after a slash(alphanumeric allowed)
Examples:
http://www.test.com/products/cards/product_code100/12345/something_else
http://www.test.com/products/123abc/45678/
Desired output -
http://www.test.com/products/cards/product_code100/
http://www.test.com/products/123abc/
The following regex gives me everything in backreferences but it'll be great if I could get rid of numbers after a slash-
^(.*:)//([a-z\-.]+)(:[0-9]+)?(.*)
Additional Information - Languauge independent regex needed.
Many Thanks
this should work with most languages and should produce the desired output
(http://.*)(?=/\d+(?!\w+))
It takes every character until it finds (lookahead) \ followed by a number.
If you'd try to match
http://www.test.com/products/123abc/
or
http://www.test.com/products/123abc
it just would not find a match and you could be sure the string checked doesnt encounter a pure number after a slash
Example in Perl:
echo "http://...." | perl -pe 's/(.*\/)\d+\/.*/$1/'
or:
echo "http://...." | perl -ne 'print "$1\n" if /(.*\/)\d+\/.*/'
Edit: It's true what #creinig noted in his comment - there is no such thing as generic regex. Nonetheless, Perl is widely used, so it's an option.
I want to print only the lines that meet the criteria : "worde:" and "wordo;"
I got this far:
sed -n '/\([a-z]*\)\1e:\1o;/p;'
But it doesn't quite work.
Can someone please perfect it and tell me exactly how its a fixed version/what was wrong with mine?
(Please note there are no capital letters ever, hence why I didn't bother including that within my initial character range)
Thanks heaps,
This will handle lines where "worde:wordo;" (nothing between the words) appears:
sed -n '/\([a-z]*\)e:\1o;/p;'
If you need to allow for characters BETWEEN the words, you'll need something like this:
sed -n '/\([a-z]*\)e:.*\1o;/p;'
My interpretation of your question is that you want to match lines which contain both worde: and wordo;
sed -n '/worde:/{/wordo;/p}' infile
The -n parameter prevents sed from printing the pattern space (infile), the first regex matches, then control flows into the block, if the regex isn't matched, then the line is ignored. Inside the block, the if the second regex is matched, the line is printed.
One way using alternation:
sed -n '/word\(e:\|o;\)/ p' infile
Is it a requirement to use capture groups? I went without them.
$ sed -n '/[\w]*[oe][:;]/p'
[\w]* - Match any word character. (if you really want only [a-z], swap
that back in)
[oe] - Those word characters must end in an e or
o
[:;] - And then have a : or ;
This might work for you:
sed '/^\(.*\)[eE]:\s*\1[oO];/!d' file