I really need help with coming up with the pattern matching solution...
If the string is <6>[ 84.982642] Killing the process
How can I extract them into three separate strings...
I need one for 6, 84.982642, and Killing the process..
I've tried many things but these brackets and blank spaces are really confusing me and I keep getting the error message
"WARNING: Use of uninitialized value $bracket in pattern match..."
Is there anyway I can somehow write in this way
($num_1, $num_2, $name_process) = split(/[\-,. :;!?()[\]{}]+/);
Not sure how to extract these..
Help Please?
Thank you so much
Assuming the input is in $_
($num_1, $num_2, $name_process) = /^<(\d+)>\[([^\]]+)\]\s+(.*)$/;
This assumes the first token in the angle brackets is always a number. For a little more generality use
($num_1, $num_2, $name_process) = /^<([^>]+)>\[([^\]]+)\]\s+(.*)$/;
Explanation:
<([^>]+)> - a left-angle-bracket followed one or more characters that are not a right angle-bracket, followed by a right-angle bracket.
\[([^\]]+)\] - a left-bracket followed by one or more characters that are not a right bracket, followed by a right bracket
\s+(.*) - one or more spaces, then capture everything starting with the first non-blank after that.
Related
I'm currently working with postgresql, I learned about this function btrim, I checked many websites for explanation, but I don't really understand.
Here they mention this example:
btrim('xyxtrimyyx', 'xyz')
It gives trim.
When I try this example:
btrim('xyxtrimyyx', 'yzz')
or
btrim('xyxtrimyyx', 'y')
I get this: xyxtrimyyx
I don't understand this. Why didn't it remove the y?
From the docs you point to, the definition says:
Remove the longest string consisting only of characters in characters
(a space by default) from the start and end of string
The reason your example doesn't work is because the function tries to strip the text from Both sides of the text, consisting only of the characters specified
Lets take a look at the first example (from the docs):
btrim('xyxtrimyyx', 'xyz')
This returns trim, because it goes through xyxtrimyyx and gets up to the t and doesn't see that letter in xyz, so that is where the function stops stripping from the front.
We are now left with trimyyx
Now we do the same, but from the end of the string.
While one of xyz is the last letter, remove that letter.
We do this until m, so we are left with trim.
Note: I have never worked with any form of sql. I could be wrong about the exact way that postgresql does this, But I am fairly certain from the docs that this is how it is done.
I'm typing up a table with org mode, where the equal sign(=) if the first character in the cell and it want to start a formula. how do I get it to display the symbol without it being a formula, of a way to use formulas to display it. I get errors when I use single quotes, and I see the Unicode decimal value when using double quotes.
I have tried the following
='=+'
="=+"
they give
#ERROR
[61, 43]
Use an escaped entity, \equal{} and it should display as you wish. See the variable org-entities for others you can use.
I'm a bit late :D
There may be a better way, but you can try with :='(format "=+")
Source: https://emacs.stackexchange.com/questions/19183/can-i-use-formula-to-manipulate-text-in-org-mode-table
When I ran into this problem just now, I found that I was able to get around it by replacing the equals sign with some other similar-looking character. Two which come to mind are ꞊ ‘U+A78A MODIFIER LETTER SHORT EQUALS SIGN’ and ⹀ ‘U+2E40 DOUBLE HYPHEN’.
I asked a question earlier today and got a really quick answer from llbrink. I really should have asked that question before I spent several hours trying to find an answer.
So - here's another question that I have never found an answer for (although I have created a work-around which seems very cludgy).
My AHK program asks the user for a login name. The program then compares the login name with an existing list of names in a file.
The login name in the file may contain spaces, but there are never spaces at the beginning of the name. When the user enters the name, he may include spaces at the beginning. This means that when my program compares the name with those in the file, it can not find a match (because of the extra spaces).
I want to find a way of stripping the spaces from the beginning of the input.
My work-round has been to split the input string into an array (which does ignore leading spaces) and then use the first element of the array. This is my code :
name := DoStrip(name)
DoStrip(xyz) ; strip leading and trailing spaces from string
{
StringSplit, out, xyz, `,, %A_Space%
Return out1
}
This seems to be a very laboured way to do it - is there a better way ?
I don't see a problem with your example if it works on all cases.
There is a much simpler way; just use Autotrim which works like this.
AutoTrim, On ; not required it is on by default
my_variable = %my_variable%
There are also many other different ways to trim string in autohotkey,
which you can combine into something useful.
You can also use #LTrim and #RTrim to remove white spaces at the beginning and at the end of the string.
I have an NSArray of lines (objective-c iphone), and I'm trying to find the line which starts with a number, followed by a dot and a space, but can have any number of spaces (including none) before it, and have any text following it eg:
1. random text
2. text random
3.
what regular expression would I use to get this? (I'm trying to learn it, and I needed the above expression anyway, so I thought I'd use it as an example)
With C#:
#"^ *[0-9]+\. "
It doesn't check for the presence of something after the ., so this is legal:
1.(space)
If you delete the # and escape the \ it should work with other languages (it is pretty "down-to-earth" as RegExpes go)
I may suggest (Perl-compatible regexp):
^\s*\d+\.\s
At the beginning of a line:
Any number (0-n) of spaces
One or more digits
A dot
A space
Something like
^\s*\d+\.
But it depends on the language.
/^\s*[0-9]+\.\s+/
would be my guess providing you don't have any space before the number
I'm using regular expression lib icucore via RegKit on the iPhone to
replace a pattern in a large string.
The Pattern i'm looking for looks some thing like this
| hello world (P1)|
I'm matching this pattern with the following regular expression
\|((\w*|.| )+)\((\w\d+)\)\|
This transforms the input string into 3 groups when a match is found, of which group 1(string) and group 3(string in parentheses) are of interest to me.
I'm converting these formated strings into html links so the above would be transformed into
Hello world
My problem is the trailing space in the third group. Which when the link is highlighted and underlined, results with the line extending beyond the printed characters.
While i know i could extract all the matches and process them manually, using the search and replace feature of the icu lib is a much cleaner solution, and i would rather not do that as a result.
Many thanks as always
Would the following work as an alternate regular expression?
\|((\w*|.| )+)\s+\((\w\d+)\)\| Where inserting the extra \s+ pulls the space outside the 1st grouping.
Though, given your example & regex, I'm not sure why you don't just do:
\|(.+)\s+\((\w\d+)\)\|
Which will have the same effect. However, both your original regex and my simpler one would both fail, however on:
| hello world (P1)| and on the same line | howdy world (P1)|
where it would roll it up into 1 match.
\|\s*([\w ,.-]+)\s+\((\w\d+)\)\|
will put the trailing space(s) outside the capturing group. This will of course only work if there always is a space. Can you guarantee that?
If not, use
\|\s*([\w ,.-]+(?<!\s))\s*\((\w\d+)\)\|
This uses a lookbehind assertion to make sure the capturing group ends in a non-space character.