I came along this
__date__ = "$Date: 2011/06$"
and found this in the docs
$$ is an escape; it is replaced with a single $.
$identifier names a substitution placeholder matching a mapping key of "identifier". By default, "identifier" must spell a Python identifier. The first non-identifier character after the $ character terminates this placeholder specification.
${identifier} is equivalent to $identifier. It is required when valid identifier characters follow the placeholder but are not part of the placeholder, such as "${noun}ification".
but I don't understand it.
Could someone explain in plain english what's the $ for and give some examples preferably?
To Python, those dollar signs mean nothing at all. Just like the 'D' or 'a' that follow, the dollar sign is merely a character in a string.
To your source-code control system, the dollar signs indicate a substitution command. When you check out a new copy of your source code, that string is replaced with the timestamp of the last committed change to the file.
Reference:
http://svnbook.red-bean.com/en/1.6/svn.advanced.props.special.keywords.html
http://www.badgertronics.com/writings/cvs/keywords.html
This has been used in the context of string replace. For ex, if you have scenario with a variable which takes different value in same string, you can use this as follows:
import string
mytext = "$dog is an animal"
replaceDogtoCat = {"dog":"cat"}
mytemplate = string.Template(mytext)
print mytemplate.substitute(replaceDogtoCat) #output: cat is an animal
replaceDogtoGoat = {"dog":"goat"}
print mytemplate.substitute(replaceDogtoGoat) #output: goat is an animal
$dog is a variable which would get replaced when substitute gets executed
Related
Groovy How to replace the exact match word in a String.
I wanted to replace the exact matched word in a given string in Groovy. and when i tried the below am not getting the exact matched word
def str="My Name is Richards and Richardson"
log.info(str)
str=str.replace("Richards","Praveen")
log.info("After"+str)
Output after executing the above
My Name is Richards and Richardson
AfterMy Name is Praveen and Praveenon
Am Looking for the output like : AfterMy Name is Praveen and Richardson
I tried the boundaries \b
str=str.replace("\bRichards\b","Praveen")
which is in Java and its not working. Looks \b is ba backslash escape sequence in the Groovy
can someone help
def str="My Name is Richards and Richardson"
log.info(str)
str=str.replace("Richards","Praveen")
log.info("After"+str)
expecting:AfterMy Name is Praveen and Richardson
Using boundaries (/b) will not work with String::replace because the method argument does not accept a regular expression pattern but a simple string literal.
You have two options to get the expected outcome:
Instead of using String::replace you can use String::replaceFirst. As the method name suggests it will replace only the first occurrence of the Richards substring leaving the Richardson as is.
str = str.replaceFirst("Richards", "Praveen")
Instead of using String::replace you can use String::replaceAll, in opposite to String::replace it supports regular expressions so you can use word boundaries tokens
str = str.replaceAll("\\bRichards\\b","Praveen")
Mind the double slashes!
Also, according to the String::replaceAll documentation:
Note that backslashes (\) and dollar signs ($) in the replacement string may cause the results to be different than if it were being treated as a literal replacement string; see Matcher.replaceAll. Use Matcher.quoteReplacement(java.lang.String) to suppress the special meaning of these characters, if desired.
I am searching for a way to use a formatter to put a space between two characters. i thought it would be easy with a string formatter.
here is what i am trying to accomplish:
given: "AB" it will produce "A B"
Here is what i have tried so far:
"AB".format("%#s")
but this keep returning "AB" i want "A B". i thought the number sign could be used for space.
i also tried this:
"26".format("%#d") but its still prints "26"
is there anyway to do this with string.formatter.
It is kind of possible with the string formatter although not directly with a pattern.
jshell> String.format("%1$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
In Swift I can create a String variable such as this:
let s = "Hello\nMy name is Jack!"
And if I use s, the output will be:
Hello
My name is Jack!
(because the \n is a linefeed)
But what if I want to programmatically obtain the raw characters in the s variable? As in if I want to actually do something like:
let sRaw = s.raw
I made the .raw up, but something like this. So that the literal value of sRaw would be:
Hello\nMy name is Jack!
and it would literally print the string, complete with literal "\n"
Thank you!
The newline is the "raw character" contained in the string.
How exactly you formed the string (in this case from a string literal with an escape sequence in source code) is not retained (it is only available in the source code, but not preserved in the resulting program). It would look exactly the same if you read it from a file, a database, the concatenation of multiple literals, a multi-line literal, a numeric escape sequence, etc.
If you want to print newline as \n you have to convert it back (by doing text replacement) -- but again, you don't know if the string was really created from such a literal.
You can do this with escaped characters such as \n:
let secondaryString = "really"
let s = "Hello\nMy name is \(secondaryString) Jack!"
let find = Character("\n")
let r = String(s.characters.split(find).joinWithSeparator(["\\","n"]))
print(r) // -> "Hello\nMy name is really Jack!"
However, once the string s is generated the \(secondaryString) has already been interpolated to "really" and there is no trace of it other than the replaced word. I suppose if you already know the interpolated string you could search for it and replace it with "\\(secondaryString)" to get the result you want. Otherwise it's gone.
I'd like to calculate the length of a replace string used in a substitution. That is, "bar" in :s/foo/bar. Suppose I have access to this command string, I can run and undo it, and may separate the parts marked by / with split(). How would I get the string length of the replace string if it contains special characters like \1, \2 etc or ~?
For instance if I have
:s/\v(foo)|(bars)/\2\rreplace/
the replace length would be strlen("bars\rreplace") = 12.
EDIT: Just to be clear, I hope to use this to move the cursor past the text that was affected by a substitute operation. I'd appreciate alternative solutions as well.
You have to use :help sub-replace-expression. In it, you use submatch(2) instead of \2. If the expression is a custom function, you can as a side effect store the original length in a variable, and access that later:
function! Replace()
let g:replaceLength = strlen(submatch(0))
" Equivalent of \2\rreplace
return submatch(2) . "\r" . 'replace'
endfunction
:s/\v(foo)|(bars)/\=Replace()/
How do I remove lines from a string begins with another string in Lua ? For instance i want to remove all line from string result begins with the word <Table. This is the code I've written so far:
for line in result:gmatch"<Table [^\n]*" do line = "" end
string.gmtach is used to get all occurrences of a pattern. For replacing certain pattern, you need to use string.gsub.
Another problem is your pattern <Table [^\n]* will match all line containing the word <Table, not just begins with it.
Lua pattern doesn't support beginning of line anchor, this almost works:
local str = result:gsub("\n<Table [^\n]*", "")
except that it will miss on the first line. My solution is using a second run to test the first line:
local str1 = result:gsub("\n<Table [^\n]*", "")
local str2 = str1:gsub("^<Table [^\n]*\n", "")
The LPEG library is perfect
for this kind of task.
Just write a function to create custom line strippers:
local mk_striplines
do
local lpeg = require "lpeg"
local P = lpeg.P
local Cs = lpeg.Cs
local lpegmatch = lpeg.match
local eol = P"\n\r" + P"\r\n" + P"\n" + P"\t"
local eof = P(-1)
local linerest = (1 - eol)^1 * (eol + eof) + eol
mk_striplines = function (pat)
pat = P (pat)
local matchline = pat * linerest
local striplines = Cs (((matchline / "") + linerest)^1)
return function (str)
return lpegmatch (striplines, str)
end
end
end
Note that the argument to mk_striplines() may be a string or a
pattern.
Thus the result is very flexible:
mk_striplines (P"<Table" + P"</Table>") would create a stripper
that drops lines with two different patterns.
mk_striplines (P"x" * P"y"^0) drops each line starting with an
x followed by any number of y’s -- you get the idea.
Usage example:
local linestripper = mk_striplines "foo"
local test = [[
foo lorem ipsum
bar baz
buzz
foo bar
xyzzy
]]
print (linestripper (test))
The other answers provide good solutions to actually stripping lines from a string, but don't address why your code is failing to do that.
Reformatting for clarity, you wrote:
for line in result:gmatch"<Table [^\n]*" do
line = ""
end
The first part is a reasonable way to iterate over result and extract all spans of text that begin with <Table and continue up to but not including the next newline character. The iterator returned by gmatch returns a copy of the matching text on each call, and the local variable line holds that copy for the body of the for loop.
Since the matching text is copied to line, changes made to line are not and cannot modifying the actual text stored in result.
This is due to a more fundamental property of Lua strings. All strings in Lua are immutable. Once stored, they cannot be changed. Variables holding strings are actually holding a pointer into the internal table of reference counted immutable strings, which permits only two operations: internalization of a new string, and deletion of an internalized string with no remaining references.
So any approach to editing the content of the string stored in result is going to require the creation of an entirely new string. Where string.gmatch provides an iteration over the content but cannot allow it to be changed, string.gsub provides for creation of a new string where all text matching a pattern has been replaced by something new. But even string.gsub is not changing the immutable source text; it is creating a new immutable string that is a copy of the old with substitutions made.
Using gsub could be as simple as this:
result = result:gsub("<Table [^\n]*", "")
but that will disclose other defects in the pattern itself. First, and most obviously, nothing requires that the pattern match at only the beginning of the line. Second, the pattern does not include the newline, so it will leave the line present but empty.
All of that can be refined by careful and clever use of the pattern library. But it doesn't change the fact that you are starting with XML text and are not handling it with XML aware tools. In that case, any approach based on pattern matching or even regular expressions is likely to end in tears.
result = result:gsub('%f[^\n%z]<Table [^\n]*', '')
The start of this pattern, '%f[^\n%z], is a frontier pattern which will match any transition from either a newline or zero character to another character, and for frontier patterns the pre-first character counts as a zero character. In other words, using that prefix allows the rest of the pattern to match at either the first line or any other start-of-line.
Reference: the Lua 5.3 manual, section 6.4.1 on string patterns