Here is my sample string.
[echo] The SampleProject solution currently has 85% code coverage.
My desired output should be.
The SampleProject solution currently has 85% code coverage.
Btw, I had this out because I'm getting through the logs in my CI using Jenkins.
Any help? Thanks..
You can try substText parameter in BUILD_LOG_REGEX token to substitute the text matching your regex
New optional arg: ${BUILD_LOG_REGEX, regex, linesBefore, linesAfter, maxMatches, showTruncatedLines, substText} which allows substituting text for the matched regex. This is particularly useful when the text contains references to capture groups (i.e. $1, $2, etc.)
Using below will remove prefix [echo] from all of your logs ,
${BUILD_LOG_REGEX, regex="^\[echo] (.*)$", maxMatches=0, showTruncatedLines=false, substText="$1"}
\[[^\]]*\] will match the bit you want to remove. Just use a string replace function to replace that bit with an empty string.
Andrew has the right idea, but with Perl-style regex syntaxes (which includes Java's built-in regex engine), you can do even better:
str.replaceAll("\\[.*?\\]", "");
(i.e., use the matching expression \[.*?\]. The ? specifies minimal match: so it will finish matching upon the first ] found.)
Related
I can't figure out how to get Lua to return ALL matches for a particular pattern match.
I have the following regex which works and is so basic:
.*\n
This just splits a long string per line.
The equivelent of this in Lua is:
.-\n
If you run the above in a regex website against the following text it will find three matches (if using the global flag).
Hello
my name is
Someone
If you do not use the global flag it will return only the first match. This is the behaviour of LUA; it's as if it does not have a global switch and will only ever return the first match.
The exact code I have is:
local test = {string.match(string_variable_here, ".-\n")}
If I run it on the above test for example, test will be a table with only one item (the first row). I even tried using capture groups but the result is the same.
I cannot find a way to make it return all occurrences of a match, does anyone know if this is possible in LUA?
Thanks,
You can use string.gmatch(s, pattern) / s:gmatch(pattern):
This returns a pattern finding iterator. The iterator will search through the string passed looking for instances of the pattern you passed.
See the online Lua demo:
local a = "Hello\nmy name is\nSomeone\n"
for i in string.gmatch(a, ".*\n") do
print(i)
end
Note that .*\n regex is equivalent to .*\n Lua pattern. - in Lua patterns is the equivalent of *? non-greedy ("lazy") quantifier.
I am attempting to define a syntax to parse data definitions in COBOL and had a particular definition for picture clauses like this:
syntax PictureClause = pic: "PIC" PictureStringType PictureStringLen ("VALUE"|"VALUES") ValueSpec
My matching ADT for this syntax was as so:
data PictureClause = pic(str pictype, PictureStringLen plen, str valuespec);
However, I noticed that it seems as if the implode function was attempting to match the parenthesized statement to the second str parameter, instead of ignoring it like the "PIC" string literal. However, this syntax definition worked as expected:
syntax PictureClause = pic: "PIC" PictureStringType PictureStringLen "VALUE" ValueSpec
|pic: "PIC" PictureStringType PictureStringLen "VALUES" ValueSpec;
As the title states, how can I define alternatives in a single statement for literals that I do not want in my ADT in a syntax definition? I can see that alternatives are possible, but I'm wondering if there is a more concise way of defining it, in the spirit of my first attempt
I seem to recall that the current version of implode treats alternatives as nodes and does not flatten them, even if the alternatives are merely literals. Your definition is perfect, nevertheless.
It's a relatively simple feature request imho, if you have time to register it on GitHub.
Another option is to not implode at all and use concrete syntax
I'm new to GREP in BBEdit. I need to find a string inside an XML file. Such string is enclosed in quotes. I need to replace only what's inside the quotes.
The problem is that the replacement string starts with a number thus confuses BBEdit when I put together the replacement pattern. Example:
Original string in XML looks like this:
What I need to replace it with:
01 new file name.png
My grep search and replace patterns:
Using the replacement pattern above, BBEdit wrongly thinks that the first backreference is "\101" when what I really need it understand is that I mean "\01".
TIA for any help.
Your example is highly artificial because in fact there is no need for your \1 or \3 as you know their value: it is " and you can just type that directly to get the desired result.
"01 new file name.png"
However, just for the sake of completeness, the answer to your actual question (how to write a replacement group number followed by a number) is that you write this:
\0101 new file name.png\3
The reason that works is that there can only be 99 capture groups, so \0101 is parsed as \01 (the first capture group) followed by literal 01.
Hi I've been struggling with this for the last hour and am no closer. How exactly do I strip everything except numbers, commas and decimal points from a rails string? The closest I have so far is:-
rate = rate.gsub!(/[^0-9]/i, '')
This strips everything but the numbers. When I try add commas to the expression, everything is getting stripped. I got the aboves from somewhere else and as far as I can gather:
^ = not
Everything to the left of the comma gets replaced by what's in the '' on the right
No idea what the /i does
I'm very new to gsub. Does anyone know of a good tutorial on building expressions?
Thanks
Try:
rate = rate.gsub(/[^0-9,\.]/, '')
Basically, you know the ^ means not when inside the character class brackets [] which you are using, and then you can just add the comma to the list. The decimal needs to be escaped with a backslash because in regular expressions they are a special character that means "match anything".
Also, be aware of whether you are using gsub or gsub!
gsub! has the bang, so it edits the instance of the string you're passing in, rather than returning another one.
So if using gsub! it would be:
rate.gsub!(/[^0-9,\.]/, '')
And rate would be altered.
If you do not want to alter the original variable, then you can use the version without the bang (and assign it to a different var):
cleaned_rate = rate.gsub!(/[^0-9,\.]/, '')
I'd just google for tutorials. I haven't used one. Regexes are a LOT of time and trial and error (and table-flipping).
This is a cool tool to use with a mini cheat-sheet on it for ruby that allows you to quickly edit and test your expression:
http://rubular.com/
You can just add the comma and period in the square-bracketed expression:
rate.gsub(/[^0-9,.]/, '')
You don't need the i for case-insensitivity for numbers and symbols.
There's lots of info on regular expressions, regex, etc. Maybe search for those instead of gsub.
You can use this:
rate = rate.gsub!(/[^0-9\.\,]/g,'')
Also check this out to learn more about regular expressions:
http://www.regexr.com/
I'm trying to look at a string and reject anything that has seq= or app= in the string. Where it gets tricky is I need elements with q=something or p=something.
The seq= part of the string is always preceded an & and app= is always preceded by a ?
I have absolutely no idea where to start. I've been using http://www.rubular.com/ to try and figure it out but to no avail.
Any help would be hugely appreciated.
Based on your question, I believe you could just reject any strings that match the following expression:
[\?&](?:seq|app)=
This will match any string that contains a ? or & followed by either app= or seq=. The ?: inside the parentheses just tells the regular expression not to bother to capture matching groups as sub-matches. They're not really necessary, but what the heck.
Here's a Rubular link with some samples.