Groovy 2.4 here. I am trying to build a regex that will filter out all the following characters:
`,./;[]-&<>?:"()|
Here's my best attempt:
static void main(String[] args) {
// `,./;[]-&<>?:"()|
String regex = "`,./;[]-&<>?:\"()|"
String test = "ooekrofkrofor ` oxkeoe , wdkeodeko / kodek ] woekoedk \" swjiej ' wsjwdjeiji :"
println test.replaceAll(regex, "")
}
However this produces a compile error on the regex string definition, complaining:
illegal character range (to < from)
Not sure if this is a Java or Groovy thing, but I can't figure out how to define the regex properly so that it quiets the error and correctly strips these "illegal characters" out of my string. Any ideas?
It seems to me you want to remove all the characters listed in your regex variable. The problem is that you declared a sequence while you need a character class (enclose the characters with []).
See Groovy demo:
String regex = "[`,./;\\[\\]&<>?:\"()|-]+"
^ ^^^^^^ ^ ^
String test = "ooekrofkrofor ` oxkeoe , wdkeodeko / kodek ] woekoedk \" swjiej ' wsjwdjeiji :"
println test.replaceAll(regex, "")
Output: ooekrofkrofor oxkeoe wdkeodeko kodek woekoedk swjiej ' wsjwdjeiji
The pattern now contains a character class matching any of the characters defined inside it - [`,./;\[\]&<>?:\"()|-] - one or more times due to the + quantifier. Note that inside the character class, ] and [ must always be escaped, and the - can be left unescaped when placed at the start/end of the character class.
You need to escape a few special characters in your pattern:
String regex = "[`,./;\\[]\\-&<>?:\"\\(\\)|]+"
Note using double \\ to turn them into a single \ in the string, so when the pattern is parsed, the next character is escaped.
Related
I'm trying to isolate a " character when (simultaneously):
it's not in the beginning of the line
it's not followed by the character ";"
it's not preceded by the character ";"
E.g.:
Line: "Best Before - NO MATCH
Line: Best Before"; - NO MATCH
Line: ;"Best "Before - NO MATCH
Line: Best "Before - MATCH
My best solution is (?<![;])([^^])(")(?![;]) but it's not working correctly.
I also tried (?<![;])(")(?![;]), but it's only partial (missing the "not at the beginning" part)
I don't understand why I'm spelling the "AND not at the beginning" wrong.
Where am I missing it?
If you want to allow partial matches, you can extend the lookbehind with an alternation not asserting the start of the string to the left.
The semi colon [;] does not have to be between square brackets.
(?<!;|^)"(?!;)
Regex demo
if you want to match the " when there is no occurrence of '" to the left and right, and a infinite quantifier in a lookbehind assertion is allowed:
(?<!^.*;(?=").*|^)"(?!;|.*;")
Regex demo
In notepad++ you can use
^.*(?:;"|";).*$(*SKIP)(*F)|(?<!^)"
Regex demo
You can use the fact that not preceded by ; means that it's also not the first character on the line to simplify things
[^;]"(?:[^;]|$)
This gives you
Match a character that's not a ; (so there must be a character and thus the next character can't be the start of the line)
Match a "
Match a character that's not a ; or the end of the line
I know you are asking for a regex solution, but, almost always, strings can also be filtered using string methods in whatever language you are working in.
For the sake of completeness, to show that regex is not your only available tool here, here is a short javascript using the string methods:
myString.charAt()
myString.includes()
Working Example:
const checkLine = (line) => {
switch (true) {
// DOUBLE QUOTES AT THE BEGINNING
case(line.charAt(0) === '"') :
return console.log(line, '// NO MATCH');
// DOUBLE QUOTES IMMEDIATELY FOLLOWED BY SEMI-COLON
case(line.includes('";')) :
return console.log(line, '// NO MATCH');
// DOUBLE QUOTES IMMEDIATELY PRECEDED BY SEMI-COLON
case(line.includes(';"')) :
return console.log(line, '// NO MATCH');
default:
return console.log(line, '// MATCH');
}
}
checkLine('"Best Before');
checkLine('Best Before";');
checkLine(';"Best "Before');
checkLine('Best "Before');
Further Reading:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/charAt
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/includes
I need to replace two characters {, } with {\n, \n}.
But they must be not surrounded in '' or "".
I tried this code to achieve that
text = 'hello(){imagine{myString("HELLO, {WORLD}!")}}'
replaced = re.sub(r'{', "{\n", text)
Ellipsis...
Naturally, This code replaces curly brackets that are surrounded in quote marks.
What are the negative statements like ! or not that can be used in regular expressions?
And the following is what I wanted.
hello(){
imagine{
puts("{HELLO}")
}
}
In a nutshell - what I want to do is
Search { and }.
If that is not enclosed in '' or ""
replace { or } to {\n or \n}
In the opposite case, I can solve it with (?P<a>\".*){(?P<b>.*?\").
But I have no clue how I can solve it in my case.
First replace all { characters with {\n. You will also be replacing {" with {\n". Now, you can replace back all {\n" characters with {".
text = 'hello(){imagine{puts("{HELLO}")}}'
replaced = text.replace('{', '{\n').replace('{\n"','{"')
You may match single and double quoted (C-style) string literals (those that support escape entities with backslashes) and then match { and } in any other context that you may replace with your desired values.
See Python demo:
import re
text = 'hello(){imagine{puts("{HELLO}")}}'
dblq = r'(?<!\\)(?:\\{2})*"[^"\\]*(?:\\.[^"\\]*)*"'
snlq = r"(?<!\\)(?:\\{2})*'[^'\\]*(?:\\.[^'\\]*)*'"
rx = re.compile(r'({}|{})|[{{}}]'.format(dblq, snlq))
print(rx.pattern)
def repl(m):
if m.group(1):
return m.group(1)
elif m.group() == '{':
return '{\n'
else:
return '\n}'
# Examples
print(rx.sub(repl, text))
print(rx.sub(repl, r'hello(){imagine{puts("Nice, Mr. \"Know-all\"")}}'))
print(rx.sub(repl, "hello(){imagine{puts('MORE {HELLO} HERE ')}}"))
The pattern that is generated in the code above is
((?<!\\)(?:\\{2})*"[^"\\]*(?:\\.[^"\\]*)*"|(?<!\\)(?:\\{2})*'[^'\\]*(?:\\.[^'\\]*)*')|[{}]
It can actually be reduced to
(?<!\\)((?:\\{2})*(?:"[^"\\]*(?:\\.[^"\\]*)*"|'[^'\\]*(?:\\.[^'\\]*)*'))|[{}]
See the regex demo.
Details:
The pattern matches 2 main alternatives. The first one matches single- and double-quoted string literals.
(?<!\\) - no \ immediately to the left is allowed
((?:\\{2})*(?:"[^"\\]*(?:\\.[^"\\]*)*"|'[^'\\]*(?:\\.[^'\\]*)*')) - Group 1:
(?:\\{2})* - 0+ repetitions of two consecutive backslashes
(?: - a non-capturing group:
"[^"\\]*(?:\\.[^"\\]*)*" - a double quoted string literal
| - or
'[^'\\]*(?:\\.[^'\\]*)*' - a single quoted string literal
) - end of the non-capturing group
| - or
[{}] - a { or }.
In the repl method, Group 1 is checked for a match. If it matched, the single- or double-quoted string literal is matched, it must be put back where it was. Else, if the match value is {, it is replaced with {\n, else, with \n}.
Replace { with {\n:
text.replace('{', '{\n')
Replace } with \n}:
text.replace('}', '\n}')
Now to fix the braces that were quoted:
text.replace('"{\n','"{')
and
text.replace('\n}"', '}"')
Combined together:
replaced = text.replace('{', '{\n').replace('}', '\n}').replace('"{\n','"{').replace('\n}"', '}"')
Output
hello(){
imagine{
puts("{HELLO}")
}
}
You can check the similarities with the input and try to match them.
text = 'hello(){imagine{puts("{HELLO}")}}'
replaced = text.replace('){', '){\n').replace('{puts', '{\nputs').replace('}}', '\n}\n}')
print(replaced)
output:
hello(){
imagine{
puts("{HELLO}")
}
}
UPDATE
try this: https://regex101.com/r/DBgkrb/1
I have about writing regexes in C++. I have 2 regexes which work fine in java. But these throws an error namely
one of * + was not preceded by a valid regular expression C++
These regexes are as follows:
regex r1("^[\s]*{[\s]*\n"); //Space followed by '{' then followed by spaces and '\n'
regex r2("^[\s]*{[\s]*\/\/.*\n") // Space followed by '{' then by '//' and '\n'
Can someone help me how to fix this error or re-write these regex in C++?
See basic_regex reference:
By default, regex patterns follow the ECMAScript syntax.
ECMAScript syntax reference states:
characters:
\character
description: character
matches: the character character as it is, without interpreting its special meaning within a regex expression.
Any character can be escaped except those which form any of the special character sequences above.
Needed for: ^ $ \ . * + ? ( ) [ ] { } |
So, you need to escape { to get the code working:
std::string s("\r\n { \r\nSome text here");
regex r1(R"(^\s*\{\s*\n)");
regex r2(R"(^\s*\{\s*//.*\n)");
std::string newtext = std::regex_replace( s, r1, "" );
std::cout << newtext << std::endl;
See IDEONE demo
Also, note how the R"(pattern_here_with_single_escaping_backslashes)" raw string literal syntax simplifies a regex declaration.
I want to find the number between [/ and ] (12345 in this case).
I have written such code:
float num;
string line = "A111[/12345]";
boost::regex e ("[/([0-9]{5})]");
boost::smatch match;
if (boost::regex_search(line, match, e))
{
std::string s1(match[1].first, match[1].second);
num = boost::lexical_cast<float>(s1); //convert to float
cout << num << endl;
}
However, I get this error: The error occurred while parsing the regular expression fragment: '/([0-9]{5}>>>HERE>>>)]'.
You need to double escape the [ and ] that special characters in regex denoting character classes. The correct regex declaration will be
boost::regex e ("\\[/([0-9]{5})\\]");
This is necessary because C++ compiler also uses a backslash to escape entities like \n, and regex engine uses the backslash to escape special characters so that they are treated like literals. Thus, backslash gets doubled. When you need to match a literal backslash, you will have to use 4 of them (i.e. \\\\).
Use the following (escape [ and ] because they are special characters in regex meaning a character class):
\\[/([0-9]{5})\\]
^^ ^^
I am looking for a perl regex which will validate a string containing only the letters ACGT. For example "AACGGGTTA" should be valid while "AAYYGGTTA" should be invalid, since the second string has "YY" which is not one of A,C,G,T letters. I have the following code, but it validates both the above strings
if($userinput =~/[A|C|G|T]/i)
{
$validEntry = 1;
print "Valid\n";
}
Thanks
Use a character class, and make sure you check the whole string by using the start of string token, \A, and end of string token, \z.
You should also use * or + to indicate how many characters you want to match -- * means "zero or more" and + means "one or more."
Thus, the regex below is saying "between the start and the end of the (case insensitive) string, there should be one or more of the following characters only: a, c, g, t"
if($userinput =~ /\A[acgt]+\z/i)
{
$validEntry = 1;
print "Valid\n";
}
Using the character-counting tr operator:
if( $userinput !~ tr/ACGT//c )
{
$validEntry = 1;
print "Valid\n";
}
tr/characterset// counts how many characters in the string are in characterset; with the /c flag, it counts how many are not in the characterset. Using !~ instead of =~ negates the result, so it will be true if there are no characters not in characterset or false if there are characters not in characterset.
Your character class [A|C|G|T] contains |. | does not stand for alternation in a character class, it only stands for itself. Therefore, the character class would include the | character, which is not what you want.
Your pattern is not anchored. The pattern /[ACGT]+/ would match any string that contains one or more of any of those characters. Instead, you need to anchor your pattern, so that only strings that contain just those characters from beginning to end are matched.
$ can match a newline. To avoid that, use \z to anchor at the end. \A anchors at the beginning (although it doesn't make a difference whether you use that or ^ in this case, using \A provides a nice symmetry.
So, you check should be written:
if ($userinput =~ /\A [ACGT]+ \z/ix)
{
$validEntry = 1;
print "Valid\n";
}