I have a string, which represents a version:
abc-5.18.0.0_10
I'm trying to determine the regex to extract this subString:
abc-5.18.0
This in format terms is:
We want any characters up until the "-".
We also want between the "-" and the first '." which is the major version 1 to nn.
We also want between the first "." and the second "." which is the minor version 0 to nn.
We also want the subminor version between the second and third ".", 0-nn.
xxx-nn.nn.nn
I'm trying this:
.split("//|\\.")
And I'm getting this:
abc-5.18
What am I doing wrong? Should it be "///|\."?
I think the regex you want is stuff, period, stuff, period, stuff up to the third period.
This uses a group () to capture what you want. It looks confusing because . represents any character and \. represents a period.
s = 'abc-5.18.0.0_10'
re.match(r"(.*\..*\..*)\.", s).group(1)
Out: 'abc-5.18.0'
You can use restrictive regular expression like:
^([^-]+-\d+\.\d+\.\d+)(\..*)?$
You can apply it and extract the value in the following way in Groovy:
import java.util.regex.Pattern
final String str = 'abc-5.18.0.0_10'
final Pattern pattern = Pattern.compile(/^([^-]+-\d+\.\d+\.\d+)(\..*)?$/)
final String result = (str =~ pattern).replaceFirst('$1')
println result
Output:
abc-5.18.0
Related
Is there a way to replace the matched pattern substring using a single re.sub() line?.
What I would like to avoid is using a string replace method to the current re.sub() output.
Input = "/J&L/LK/Tac1_1/shareloc.pdf"
Current output using re.sub("[^0-9_]", "", input): "1_1"
Desired output in a single re.sub use: "1.1"
According to the documentation, re.sub is defined as
re.sub(pattern, repl, string, count=0, flags=0)
If repl is a function, it is called for every non-overlapping occurrence of pattern.
This said, if you pass a lambda function, you can remain the code in one line. Furthermore, remember that the matched characters can be accessed easier to an individual group by: x[0].
I removed _ from the regex to reach the desired output.
txt = "/J&L/LK/Tac1_1/shareloc.pdf"
x = re.sub("[^0-9]", lambda x: '.' if x[0] is '_' else '', txt)
print(x)
There is no way to use a string replacement pattern in Python re.sub to replace with two possible strings, as there is no conditional replacement construct support in Python re.sub. So, using a callable as the replacement argument or use other work-arounds.
It looks like you only expect one match of <DIGITS>_<DIGITS> in the input string. In this case, you can use
import re
text = "/J&L/LK/Tac1_1/shareloc.pdf"
print( re.sub(r'^.*?(\d+)_(\d+).*', r'\1.\2', text, flags=re.S) )
# => 1.1
See the Python demo. See the regex demo. Details:
^ - start of string
.*? - zero or more chars as few as possible
(\d+) - Group 1: one or more digits
_ - a _ char
(\d+) - Group 2: one or more digits
.* - zero or more chars as many as possible.
I am searching for a way to use a formatter to put a space between two characters. i thought it would be easy with a string formatter.
here is what i am trying to accomplish:
given: "AB" it will produce "A B"
Here is what i have tried so far:
"AB".format("%#s")
but this keep returning "AB" i want "A B". i thought the number sign could be used for space.
i also tried this:
"26".format("%#d") but its still prints "26"
is there anyway to do this with string.formatter.
It is kind of possible with the string formatter although not directly with a pattern.
jshell> String.format("%1$c %2$c", "AB".chars().boxed().toArray())
$10 ==> "A B"
We need to turn the string into an object array so it can be passed in as varargs and the formatter pattern can extract characters based on index (1$ and 2$) and format them as characters (c).
A much simpler regex solution is the following which scales to any number of characters:
jshell> "ABC^&*123".replaceAll(".", "$0 ").trim()
$3 ==> "A B C ^ & * 1 2 3"
All single characters are replaced with them-self ($0) followed by a space. Then the last extra space is removed with the trim() call.
I could not find way to do this using String#format. But here is a way to accomplish this using regex replacement:
String input = "AB";
String output = input.replaceAll("(?<=[A-Z])(?=[A-Z])", " ");
System.out.println(output);
The regex pattern (?<=[A-Z])(?=[A-Z]) will match every position in between two capital letters, and interpolate a space at that point. The above script prints:
A B
I am trying to split the string 'id#namespace' by #.
There is this special case that the id has a format name#gmail which makes the string I am trying to split look like name#gmail#namespace.
Is there any way to achieve that only split by the last # which will give me name#gmail and namespace?
If it is always the last index. Use lastIndex to find the last index of the character and then use substring
Something like this
int idx = string.lastIndexOf("#");
String[] splitStrings = {string.substring(0, idx), string.substring(idx)};
You could match the last # using a negative lookahead (?! to assert that there are no more # following:
#(?!.*#)
Regex demo
System.out.println("name#gmail#namespace".split("#(?!.*#)")[0]); //name#gmail
I am a beginner in programming. I have a string for example "test:1" and "test:2". And I want to remove ":1" and ":2" (including :). How can I do it using regular expression?
Hi andrew it's pretty easy. Think of a string as if it is an array of chars (letters) cause it actually IS. If the part of the string you want to delete is allways at the end of the string and allways the same length it goes like this:
var exampleString = 'test:1';
exampleString.length -= 2;
Thats it you just deleted the last two values(letters) of the string(charArray)
If you cant be shure it's allways at the end or the amount of chars to delete you'd to use the version of szymon
There are at least a few ways to do it with Groovy. If you want to stick to regular expression, you can apply expression ^([^:]+) (which means all characters from the beginning of the string until reaching :) to a StringGroovyMethods.find(regexp) method, e.g.
def str = "test:1".find(/^([^:]+)/)
assert str == 'test'
Alternatively you can use good old String.split(String delimiter) method:
def str = "test:1".split(':')[0]
assert str == 'test'
I have a problem with splitting string into two parts on special character.
For example:
12345#data
or
1234567#data
I have 5-7 characters in first part separated with "#" from second part, where are another data (characters,numbers, doesn't matter what)
I need to store two parts on each side of # in two variables:
x = 12345
y = data
without "#" character.
I was looking for some Lua string function like splitOn("#") or substring until character, but I haven't found that.
Use string.match and captures.
Try this:
s = "12345#data"
a,b = s:match("(.+)#(.+)")
print(a,b)
See this documentation:
First of all, although Lua does not have a split function is its standard library, it does have string.gmatch, which can be used instead of a split function in many cases. Unlike a split function, string.gmatch takes a pattern to match the non-delimiter text, instead of the delimiters themselves
It is easily achievable with the help of a negated character class with string.gmatch:
local example = "12345#data"
for i in string.gmatch(example, "[^#]+") do
print(i)
end
See IDEONE demo
The [^#]+ pattern matches one or more characters other than # (so, it "splits" a string with 1 character).