I am working on string manipulation using LUA and having trouble with the following problem.
Using this as an example of the original data I am given -
"[0;1;36m(Web): You say, "Text here."[0;37m"
I want to keep the string intact except for removing the ANSI codes.
I have been pointed toward using gsub with the LUA pattern matching but I cannot seem to get the pattern correct. I am also unsure how to reference exactly the escape character sent.
text:gsub("[\27\[([\d\;]+)m]", "")
or
text:gsub("%x%[[%d+;+]m", "")
If successful, all I want to be left with, using the above example, would be:
(Web): You say, "Text here."
Your string example is missing the escape character, ASCII 27.
Here's one way:
s = '\x1b[0;1;36m(Web): You say, "Text here."\x1b[0;37m'
s = s:gsub('\x1b%[%d+;%d+;%d+;%d+;%d+m','')
:gsub('\x1b%[%d+;%d+;%d+;%d+m','')
:gsub('\x1b%[%d+;%d+;%d+m','')
:gsub('\x1b%[%d+;%d+m','')
:gsub('\x1b%[%d+m','')
print(s)
Related
I have used GetText on an area of screen text and it has returned values separated by "\n". This is perfectly fine as there are line-breaks in the text and actually is exactly what I wanted, however, I want to convert the string into a list, splitting on the "\n".
My issue is that I get the following error from this line of foce:
put text split by "\\n" into itemList
Sensetalk compiler exception: syntax error - cant understand "n" at line.....
I initially thought there was a delimiting issue, having originally tried to split on "\n" so I switched to "\n" but the same error occurs..
How can you use the split function when there are escape sequences in the string?
many thanks
Depending on how OCR is reading this in, one of the predefined variables should be equivalent to your \n. I would expect this to work:
put text split by newline into itemList
You can find out more about predefined variables in SenseTalk here: https://docs.eggplantsoftware.com/studio/stk-restricted-words/#predefined-variables
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.
How to print all special characters without inserting escape sign before every of them? I have very large textiles with many special characters and I'm looking for something like # in c# which prints string literally as it is
What you're referring to, is called a verbatim string literal in C# and that concept does not translate exactly to Swift.
However, with the introduction of multiline string Literals in Swift 4, you can get close.
let multilineString = """
Here you can use \ and newline characters.
Also single " or double "" are allowed.
"""
For reference, find the grammar of a Swift String literal here.
I have a requirement to escape a string with url information but also some special characters such as '<'.
Using cl_http_utility=>escape_url this translates to '%3c'. However due to our backend webserver, it is unable to recognize this as special character and takes the value literally. What it does recognize as special character is '%3C' (C is upper case). Also if one checks http://www.w3schools.com/tags/ref_urlencode.asp it shows the value with all caps as the proper encoding.
I guess my question is is there an alternative to cl_http_utility=>escape_url that does essentially the same thing except outputs the value in upper case?
Thanks.
Use the string function.
l_escaped = escape( val = l_unescaped
format = cl_abap_format=>e_url ).
Other possible formats are e_url_full, e_uri, e_uri_full, and a bunch of xml/json stuff too. The string function escape is documented pretty well, demo programs and all.
Could anybody help me make a proper regular expression from a bunch of text in Ruby. I tried a lot but I don't know how to handle variable length titles.
The string will be of format <sometext>title:"<actual_title>"<sometext>. I want to extract actual_title from this string.
I tried /title:"."/ but it doesnt find any matches as it expects a closing quotation after one variable from opening quotation. I couldn't figure how to make it check for variable length of string. Any help is appreciated. Thanks.
. matches any single character. Putting + after a character will match one or more of those characters. So .+ will match one or more characters of any sort. Also, you should put a question mark after it so that it matches the first closing-quotation mark it comes across. So:
/title:"(.+?)"/
The parentheses are necessary if you want to extract the title text that it matched out of there.
/title:"([^"]*)"/
The parentheses create a capturing group. Inside is first a character class. The ^ means it's negated, so it matches any character that's not a ". The * means 0 or more. You can change it to one or more by using + instead of *.
I like /title:"(.+?)"/ because of it's use of lazy matching to stop the .+ consuming all text until the last " on the line is found.
It won't work if the string wraps lines or includes escaped quotes.
In programming languages where you want to be able to include the string deliminator inside a string you usually provide an 'escape' character or sequence.
If your escape character was \ then you could write something like this...
/title:"((?:\\"|[^"])+)"/
This is a railroad diagram. Railroad diagrams show you what order things are parsed... imagine you are a train starting at the left. You consume title:" then \" if you can.. if you can't then you consume not a ". The > means this path is preferred... so you try to loop... if you can't you have to consume a '"' to finish.
I made this with https://regexper.com/#%2Ftitle%3A%22((%3F%3A%5C%5C%22%7C%5B%5E%22%5D)%2B)%22%2F
but there is now a plugin for Atom text editor too that does this.