help needed to match full message in a Lookbehind.
Lets say i have the following simplified string:
1 hostname Here is some Text
at the beginning i could have 1 or 2 digits followed by space, which i would ignore.
then i need the first word captured as "host"
and then i would like to look behind to the first space, so that capture group "message" has everything starting after the first 2 digits and space. i.e. "hostname Here is some Text"
my regex is:
^[1-9]\d{0,2}\s(?<host>[\w][\w\d\.#-]*)\s(?<message>(?<=\s).*$)
this gives me:
host = "hostname"
message = "Here is some Text"
I can't figure out how my lookbehind needs to look like.
Thanks for your help.
ok, i found it. What needs to be done is to put message as the first group, and everything else, including the other groups inside the message group:
^[1-9]\d{0,2}\s(?<message>(?<host>[\w][\w\d\.#-]*)\s.*$)
Related
i have a column in openrefine, which has cells with content like:
This dog is a great dog.
This cat is a great cat,
i would like to remove the words dog, cat from the end of each cell (if punctuation could be removed also, it would be great).
i have tried with
\bdog\s*$
but i get errors, or no replacement done
I am using openrefine 3.3.
value.replace(\bdog|\bcat\s*$,'')
error i get:
Parsing error at offset 14: Missing number, string, identifier, regex, or parenthesized expression
desired output:
This dog is a great
This cat is a great
also, it would be great if i could also remove all characters in the end like " : , . (actually i am looking for a regex to cluster publishers -librarian data) so if you could suggest words i should remove from the end of the cells i would be grateful
I combined Ettore answer with the split() function value.split(' ')[-1] that select the last part word of a string.
The results is :
replace(value,value.split(' ')[-1],'') + value.split(' ')[-1].replace(/cat|dog/,'')
where
replace(value,value.split(' ')[-1],'') select your string expect the last work
value.split(' ')[-1].replace(/cat|dog/,'') replace the last word with nothing if it contains cat or dog.
Note that the expression is working because of the punctuation at the end of the string. Not a perfect solution but you may be able to build something from here.
I have a long method, after I type it in, it looks like this:
[someObj action1:param1 action2:param2 action3:param3 action4:param4];
But I want it to become like this... :
[someObj action1:param1
action2:param2
action3:param3
action4:param4];
...automatically after I type in the last semicolon.
I just saw a video that did this, so how do I do that?
(It is a paid video so can't give link here)
For the question text:
[someObj action1:param1 action2:param2 action3:param3 action4:param4];
if you have already entered that on one line just select the space between parameters and return
That will tend to alligh the colon ":" characters on multiple lines.
To auto-indent already entered code select the text that you would like auto-indented and then control i and that selection will be indented to Xcode rules.
I use that all the time.
If you just want to move a selected block of code to the left command [, to the right command ]
Have a look at https://github.com/travisjeffery/ClangFormat-Xcode.
It will reformat your code to adhere to style rules, like Chromium's for example, where the line length is 80 chars max.
In this case, the method will be wrapped as you mentioned if it's more than 80 chars.
You can either reformat on a particular key combination, or on each save.
I'm using a manual verification point - it works well - but when it appears in the log its name -or title- appears empty, what should i do to make the manual VP's title appears in the log, thanks in advance
Abed.
The syntax is
vpManual("Your VP name here", expectedData, actualData).performTest();
The VP name should not contain spaces, dots or other "strange" characters, keep on letters, digits and underscore _
If you use a string variable for the name, check it is not null or empty.
Also note the name must be unique in the script. If you provide a little example of your problem I can try answering with more details.
Take a look at the RFT docs
I'm trying to create a Regex to check for 6-12 characters, one being a digit, the rest being any characters, no spaces. Can Regex do this? I'm trying to do this in objective-c and I'm not familiar with Regex at all. I've been reading a couple tutorials, but most are for matching simple cases of a number, or a set of numbers, but not exactly what i'm looking for. I can do it with methods, but I was wondering if it that would be too slow and I figured I could try learning something new.
asdfg1 == ok
asdfg 1 != ok
asdfgh != ok
123456 != ok
asdfasgdasgdasdfasdf != ok
use this regex ^(?=.*\d)(?=.*[a-zA-Z])[^ ]{6,12}$
It seems that you mean "letter" when you say "character", right? And (thanks to burning_LEGION for pointing that out) there may be only one digit?
In that case, use
^(?=\D*\d\D*$)[^\W_]{6,12}$
Explanation:
^ # Start of string
(?=\D*\d\D*$) # Assert that there is exactly one digit in the string
[^\W_] # Match a letter or digit (explanation below)
{6,12} # 6-12 times
$ # End of string
[^\W_] might look a little odd. How does it work? Well, \w matches any letter, digit or underscore. \W matches anything that \w doesn't match. So [^\W] (meaning "match any character that is not not alphanumeric/underscore") is essentially the same as \w, but by adding _ to this character class, we can remove the underscore from the list of allowed characters.
i didn't try though, but i think here is the answer
(^[^\d\x20]*\d[^\d\x20]*$){6,12}
This is for one digit: ^[^\d\x20]{0,11}\d{1}[^\d\x20]{0,11}$ but I can`t get limited to 6-12 length, you can use other function to check length first and if it from 6 to 12 check with this regex witch I wrote.
I have this code
Dim parts As New List(Of String)(Regex.Split(RichTextBox2.Text, "~\d"))
That splits lines in this format into parts:
~1Hello~2~3Bye~4~5Morning~6
So if I do MsgBox(parts(5)), it will show me "Morning".
I want to do the exact same thing, but now my line is arranged like this:
Hello, Bye, Morning,
Change "~\d" to ", ?". The question mark after the space means that the space is optional.
Alternatively, assuming that you are only looking for single words, instead of Regex.Split you could use Regex.Matches with the regular expression "\w+".