I want to extract number from string. This is the string
#all/30
All I want is 30. How can I extract?
I try to use :
echo "#all/30" | sed 's/.*\/([^0-9])\..*//'
But nothing happen.
How should I write for the regular expression?
Sorry for bad english.
You may consider using grep to extract the numbers from a simple string like this.
echo "#all/30" | grep -o '[0-9]\+'
-o option shows only the matching part that matches the pattern.
You could try the below sed command,
$ echo "#all/30" | sed 's/[^0-9]*\([0-9]\+\)[^0-9]*/\1/'
30
[^0-9]* [^...] is a negated character class. It matches any character but not the one inside the negated character class. [^0-9]* matches zero or more non-digit characters.
\([0-9]\+\) Captures one or more digit characters.
[^0-9]* Matches zero or more non-digit characters.
Replacing the matched characters with the chars inside group 1 will give you the number 30
echo "all/30" | sed 's/[^0-9]*\/\([0-9][0-9]*\)/\1/'
Avoid writing '.*' as it consumes entire string. Default matches are always greedy.
echo "all/30" | sed 's/[^0-9]*//g'
# OR
echo "all/30" | sed 's#.*/##'
# OR
echo "all/30" | sed 's#.*\([0-9]*\)#\1#'
without more info about possible input string we can only assume that structure is #all/ followed by the number (only)
Related
I have versions like:
v1.0.3-preview2
v1.0.3-sometext
v1.0.3
v1.0.2
v1.0.1
I am trying to get the latest version that is not preview (doesn't have text after version number) , so result should be:
v1.0.3
I used this grep: grep -m1 "[v\d+\.\d+.\d+$]"
but it still outputs: v1.0.3-preview2
what I could be missing here?
To return first match for pattern v<num>.<num>.<num>, use:
grep -m1 -E '^v[0-9]+(\.[0-9]+){2}$' file
v1.0.3
If you input file is unsorted then use grep | sort -V | head as:
grep -E '^v[0-9]+(\.[0-9]+){2}$' f | sort -rV | head -1
When you use ^ or $ inside [...] they are treated a literal character not the anchors.
RegEx Details:
^: Start
v: Match v
[0-9]+: Match 1+ digits
(\.[0-9]+){2}: Match a dot followed by 1+ dots. Repeat this group 2 times
$: End
To match the digits with grep, you can use
grep -m1 "v[[:digit:]]\+\.[[:digit:]]\+\.[[:digit:]]\+$" file
Note that you don't need the [ and ] in your pattern, and to escape the dot to match it literally.
With awk you could try following awk code.
awk 'match($0,/^v[0-9]+(\.[0-9]+){2}$/){print;exit}' Input_file
Explanation of awk code: Simple explanation of awk program would be, using match function of awk to match regex to match version, once match is found print the matched value and exit from program.
Regular expressions match substrings, not whole strings. You need to explicitly match the start (^) and end ($) of the pattern.
Keep in mind that $ has special meaning in double quoted strings in shell scripts and needs to be escaped.
The boundary characters need to be outside of any group ([]).
Can you please tell me how to get the token value correctly? At the moment I am getting: "1jdq_dnkjKJNdo829n4-xnkwe",258],["FbtResult
echo '{"facebookdotcom":true,"messengerdotcom":false,"workplacedotcom":false},827],["DTSGInitialData",[],{"token":"aaaaaaa"},258],["FbtResult' | sed -n 's/.*"token":\([^}]*\)\}/\1/p'
You need to match the full string, and to get rid of double quotes, you need to match a " before the token and use a negated bracket expression [^"] instead of [^}]:
sed -n 's/.*"token":"\([^"]*\).*/\1/p'
Details:
.* - any zero or more chars
"token":" - a literal "token":" string
\([^"]*\) - Group 1 (\1 refers to this value): any zero or more chars other than "
.* - any zero or more chars.
This replacement works:
echo '{"facebookdotcom":true,"messengerdotcom":false,"workplacedotcom":false},827],["DTSGInitialData",[],{"token":"aaaaaaa"},258],["FbtResult'
| sed -n 's/.*"token":"\([a-z]*\)"\}.*/\1/p'
Key capture after "token" found between quotes via \([a-z]*\), followed by a closing brace \} and remaining characters after that as .* (you were missing this part before, which caused the replacement to include the text after keyword as well).
Output:
aaaaaaa
A grep solution:
echo '{"facebookdotcom":true,"messengerdotcom":false,"workplacedotcom":false},827],["DTSGInitialData",[],{"token":"aaaaaaa"},258],["FbtResult' | grep -Po '(?<="token":")[^"]+'
yields
aaaaaaa
The -P option to grep enables the Perl-compatible regex (PCRE).
The -o option tells grep to print only the matched substring, not the entire line.
The regex (?<="token":") is a PCRE-specific feature called a zero-width positive lookbehind assertion. The expression (?<=pattern) matches a pattern without including it in the matched result.
I have the following rule:
https://regex101.com/r/noX9lj/4
I want to make this work in a script so I'm using grep like this:
echo "\$this->table('test')" | grep -Po "qr/\$this->table\(\'(test)\'\);/"
The output should be "test"
It's not working, not sure why..
You may use
echo "\$this->table('test');" | grep -oP "\\\$this->table\\('\\K[^']+(?='\\);)"
Or, if you feed a file path to grep:
grep -oP "\\\$this->table\\('\\K[^']+(?='\\);)" file
See the online grep demo
To match $, you need to escape it with a literal backslash, and inside a double quoted string, you need to escape $ itself with one backslash char in order to stop variable expansion, and then you need to add two more backslashes to regex-escape the literal $ char, hence is the "\\\$" in the pattern.
To match any text between two single quotes, you may use [^']+ - 1 or more chars other than '.
See the regex demo
Pattern details
\$this->table\(' - $this->table(' string
\K - match reset operator that discards the text matched so far from the overall match buffer
[^']+ - one or more chars other than '
(?='\);) - a positive lookahead that requires '); string to be present immediately to the right of the current position.
There were multiple issues:
had to use "cat" instead of echo for some reason
used this rule instead:
grep -oP "this->table\('\K\w+(?='\);)"
I know a colon: should be literal, so I'm not clear why a grep matches all lines. Here's a file called "test":
cat test
123|4444
4546|4444
666666|5678
7777777|7890675::1
I need to match the line with::1. Of course, the real case is more complicated, so I can't simply search for "::1". I tried many iterations, like
grep -E '^[0-9]|[0-9]:' test
grep -E '^[0-9]|[0-9]::1' test
But they return all lines:
123|4444
4546|4444
666666|5678
7777777|7890675::1
I am expecting to match just the last line. Any idea why that is?
This is GNU/Linux bash.
The pipe needs to be escaped and you need to allow repeated digits:
grep -E '^[0-9]+\|[0-9]+:' test
Otherwise ^[0-9] is all that needs to match for a line to be retained by the grep.
Given:
$ echo "$txt"
123|4444
4546|4444
666666|5678
7777777|7890675::1
Use repetition (+ means 'one or more') and character classes:
$ echo "$txt" | grep -E '^[[:digit:]]+[|][[:digit:]]+[:]+'
7777777|7890675::1
Since | is a regex meta character, it has to be either escaped (\|) or in a character class.
There are two issues:
The regex [0-9] matches any single digit. Since you have multiple digits, you need to replace those parts with [0-9]+, which matches one or more digits. If you want to allow an empty sequence with no digits, replace the + with a *, which means “zero or more”.
The pipe character | means “alternative”s in regex. What you provided will match either a digit at the start of the line, or a digit followed by a colon. Since every line has at least one of those, you match every line. To get a literal | character, you can use either [|] or \|; the second option is usually preferred in most styles.
Applying both of these, you get ^[0-9]+\|[0-9]+::1.
Another approach is to use a tool like awk that can process the fields of each line, and match lines where the 2nd field ends with "::1"
awk -F'|' '$2 ~ /::1$/' test
example file:
blahblah 123.a.site.com some-junk
yoyoyoyo 456.a.site.com more-junk
hihohiho 123.a.site.org junk-in-the-trunk
lalalala 456.a.site.org monkey-junk
I want to grep out all those domains in the middle of each line, they all have a common part a.site with which I can grep for, but I can't work out how to do it without returning the whole line?
Maybe sed or a regex is need here as a simple grep isn't enough?
You can do:
grep -o '[^ ]*a\.site[^ ]*' input
or
awk '{print $2}' input
or
sed -e 's/.*\([^ ]*a\.site[^ ]*\).*/\1/g' input
Try this to find anything in that position
$ sed -r "s/.* ([0-9]*)\.(.*)\.(.*)/\2/g"
[0-9]* - For match number zero or more time.
.* - Match anything zero or more time.
\. - Match the exact dot.
() - Which contain the value particular expression in parenthesis, it can be printed using \1,\2..\9. It contain only 1 to 9 buffer space. \0 means it contain all the expressed pattern in the expression.