I am trying to test string like {"count":0} in Scala using matches. Since Integer part can be different I am trying to do something like this:
assert(response.matches(s"^\\{\"count\":${notificationCount}\\}$"), s"Actual response: $response")
But I am getting wrong string literal in second $ sign which indicate end of string in regular expression.
Any suggestion?
When using string interpolation in Scala, you can escape $ by using a double $$:
val foo = 5
s"$foo${foo + 1}$$" //56$
You might also consider using a triple-quoted raw string to help with the escaped brace and quote characters:
s"""{"count":${foo}}$$""" //{"count":5}$
Related
I'm having trouble escaping double dollars from a string to be used with regex functions pattern/matcher.
This is part of the String:
WHERE oid_2 = $$test$$ || oid_2 = $$test2$$
and this is the closest code I've tried to get near the solution:
List<String> strList = new ArrayList<String>();
Pattern pattern = Pattern.compile("\$\$.*?\$\$");
log.debug("PATTERN: "+pattern)
Matcher matcher = pattern.matcher(queryText);
while (matcher.find()) {
strList.add(matcher.group());
}
log.debug(strList)
This is the debug output i get
- PATTERN: $$.*?$$
- []
So the pattern is actually right, but the placeholders are not found in the string.
As a test I've tried to replace "$$test$$" with "XXtestXX" and everything works perfectly. What am I missing? I've tried "/$" strings, "\\" but still have no solution.
Note that a $ in regex matches the end of the string. To use it as a literal $ symbol, you need to escape it with a literal backslash.
You used "\$\$.*?\$\$" that got translated into a literal string like $$.*?$$ that matches 2 end of string positions, any 0+ chars as few as possible and then again 2 end of strings, which has little sense. You actually would need a backslash to first escape the $ that is used in Groovy to inject variables into a double quoted string literal, and then use 2 backslashes to define a literal backslash - "\\\$\\\$.*?\\\$\\\$".
However, when you work with regex, slashy strings are quite helpful since all you need to escape a special char is a single backslash.
Here is a sample code extracting all matches from the string you have in Groovy:
def regex = /\$\$.*?\$\$/;
def s = 'WHERE oid_2 = $$test$$ || oid_2 = $$test2$$'
def m = s =~ regex
(0..<m.count).each { print m[it] + '\n' }
See the online demo.
Anyone who gets here might like to know another answer to this, if you want to use Groovy slashy strings:
myComparisonString ==~ /.*something costs [$]stuff.*/
I couldn't find another way of putting a $ in a slashy string, at least if the $ is to be followed by text. If, conversely, it is followed by a number (or presumably any non-letter), this will work:
myComparisonString ==~ /.*something costs \$100.*/
... the trouble being, of course, that the GString "compiler" (if that's its name) would recognise "$stuff" as an interpolated variable.
So I am experiencing some inconsistent behavior in my regex
My regex:
(?<=test\\\\)(.*)(?=\",)
The input string:
"test.exe /c echo teststring > \\\\.\\test\\teststring",
When I run this in https://Regex101.com
I get the value teststring however when I run this in F#
Regex.Match(inputString, "(?<=test\\\\)(.*)(?=\",)")
I get \teststring back. My goal is to get just teststring. I'm not sure what I'm doing wrong.
I had success using triple quoted strings. Then only the regex escapes need be considered, and not the F# string escapes.
let inputString = """test.exe /c echo teststring > \\\\.\\test\\teststring","""
let x = Regex.Match(inputString, """(?<=test\\\\)(.*)(?=\",)""")
"teststring" comes out
The string in your source comes out as
(?<=test\\)(.*)(?=",)
If you don't want to use triple quotes or verbatim, you will have to write this in F# :
"(?<=test\\\\\\\\)(.*)(?=\\\",)"
This string in F# uses backslashes to escape backslashes and a quote character. There are eight backslashes in a row in one place, and this then becomes four actual backslashes in the string value. There is also this:
\\\"
which translates to one actual \ and one actual " in the actual string value.
So then we end up with a string value of
(?<=test\\\\)(.*)(?=\",)
This then is the actual string value fed to the regex engine. The regex engine, like the F# compiler, also uses the backslash to escape characters. That's why any actual backslash had to be doubled and then doubled again.
I'm trying to verify if a string correspons to a regex with the stl like this :
regex rgx("#^\([ \r\t\n\f]*([-]?[0-9]+)[ \r\t\n\f]*,[ \r\t\n\f]*([-]?[0-9]+)[ \r\t\n\f]*\)$#");
bool test = regex_search("(12,3)", rgx);
The string is supposed to match, but test = false after that !
You need to wrap your literal to Raw literal:
regex rgx(R"#(^\([ \r\t\n\f]*([-]?[0-9]+)[ \r\t\n\f]*,[ \r\t\n\f]*([-]?[0-9]+)[ \r\t\n\f]*\)$)#");
BTW, I believe the following regexp is identical to yours but much simpler:
regex rgx(R"#(^\(\s*(-?\d+)\s*,\s*(-?\d+)\s*\)$)#");
Or without raw literals:
regex rgx("^\\(\\s*(-?\\d+)\\s*,\\s*(-?\\d+)\\s*\\)$")
Note: I had to simplify my actual use case to spare SO a lot of backstory. So if your first reaction to this question is: why would you ever do this, trust me, I just need to.
I'm trying to write a Groovy expression that replaces double-quotes (""") that appear in a string with single-quotes ("'").
// BEFORE: Replace my "double" quotes with 'single' quotes.
String toReplace = "Replace my \"double-quotes\" with 'single' quotes.";
// Wrong: compiler error
String replacerExpression = "toReplace.replace(""", "'");";
Binding binding = new Binding();
binding.setVariable("toReplace", toReplace);
GroovyShell shell = new GroovyShell(binding);
// AFTER: Replace my 'double' quotes with 'single' quotes.
String replacedString = (String)shell.evaluate(replacerExpression);
The problem is, I'm getting a compile error on the line where I assign replacerExpression:
Syntax error on token ""toReplace.replace("", { expected
I think it's because I need to escape the string that contains the double-quote character (""") but since it's a string-inside-a-string, I'm not sure how to properly escape it here. Any ideas?
You need to escape the quote within quotes in this line:
String replacerExpression = "toReplace.replace(""", "'");";
The string will be evaluated twice: once as a string literal, and once as a script. This means you have to escape it with a backslash, and escape the backslash too. Also, with the embedded quotes, it'll be much more readable if you use triple quotes.
Try this (in groovy):
String replacerExpression = """toReplace.replace("\\"", "'");""";
In Java, you're stuck with using backslashes to escape all the quotes and the embedded backslash:
String replacerExpression = "toReplace.replace(\"\\\"\", \"\'\");";
Triple-quotes work well, but one can also use single-quoted string to specify a double-quote, and a double-quoted string for a single-quote.
Consider this:
String toReplace = "Replace my \"double-quotes\" with 'single' quotes."
// key line:
String replacerExpression = """toReplace.replace('"', "'");"""
Binding binding = new Binding(); binding.setVariable("toReplace", toReplace)
GroovyShell shell = new GroovyShell(binding)
String replacedString = (String)shell.evaluate(replacerExpression)
That is, after the string literal evaluation, this is evaluated in the Groovy shell:
toReplace.replace('"', "'");
If that is too hard on the eyes, replace the "key line" above with another style (using slashy strings):
String ESC_DOUBLE_QUOTE = /'"'/
String ESC_SINGLE_QUOTE = /"'"/
String replacerExpression = """toReplace.replace(${ESC_DOUBLE_QUOTE}, ${ESC_SINGLE_QUOTE});"""
Please try to use regular expressions to solve this kind of problems, instead of messing your head to tackle the escaping of quotes.
I have put up a solution using groovy console. Please see if that helps.
I want to write a function, to replace all the numbers in a string with literal \d. My code is:
val r = """\d""".r
val s = r.replaceAllIn("123abc", """\d""")
println(s)
I expect the result is \d\d\dabc, but get:
dddabc
Then I change my code (line 2) to:
val s = r.replaceAllIn("123abc", """\\d""")
The result is correct now: \d\d\dabc
But I don't understand why the method replaceAllIn converts the string, not use it directly?
There was a toList in my previous code, that now what I want. I have just update the question. Thanks to everyone.
Just remove the toList.
val r = """\d""".r
val list = r.replaceAllIn("123abc", """\\d""")
println(list)
Strings are (implicitly, via WrappedString, convertible to) Seq[Char]. If you invoke toList, you will have a List[Char].
Scala's Regex uses java.util.regex underneath (at least on the JVM). Now, if you look up replaceAll on Java docs, you'll see this:
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. Dollar
signs may be treated as references to
captured subsequences as described
above, and backslashes are used to
escape literal characters in the
replacement string.