regex_search trying to match a string containing '[' - regex

Being relatively new to regular expressions I am having trouble figuring out the correct syntax. I am trying to match a string with the following pattern: String[string]!='string'. I want to divide it into three matches as it follows:
String[string] // can contain numbers
!= // operator that can also be: =,> and < or a combination of these
'string' // can contain _ and numbers
So far I have managed to match a string with the following pattern: string='string'. Using this code:
const string strExpression = "TypeValue!='Set_Site'";
regex regex("([a-zA-Z0-9]+)([=><!]+)(['a-zA-Z0-9_]+)");
smatch match;
if (regex_search(strExpression.begin(), strExpression.end(), match, regex))
{
string indicator(match[1]);
string op(match[2]);
string value(match[3]);
}
However when I try to add '[' and ']' to the regex syntax I don't get any matches. I have modified the code like this:
const string strExpression = "Type[Value]!='Set_Site'";
regex regex("([]a-zA-Z0-9[]+)([=><!]+)(['a-zA-Z0-9_]+)");
smatch match;
if (regex_search(strExpression.begin(), strExpression.end(), match, regex))
{
string indicator(match[1]);
string op(match[2]);
string value(match[3]);
}
Аccording to the documentation I am reading the right-square-bracket ( ']' ) will lose its special meaning (terminating the bracket expression) and represent itself in a bracket expression if it occurs first in the list. For the left-square-bracket( '[' ) it says that it will lose its special meaning within a bracket expression. So following this rules and definitions I cannot identify why i am not getting any matches
Can someone give me some guidelines what I am doing wrong?
Thank you.

Related

Regular expresion with a specific character and without another

I'm trying to implement the escape character functionality in a macro generator I'm writing in Dart. For example, I would like the program to grab all the occurrences of '&param' in my string and replace it with 'John', unless the '&' character is preceded with the escape character '\'. Example: "My name is &param and my parameter is called \&param." -> "My name is John and my parameter is called &param". What would be the regular expression to catch all the substrings that contain the '&', then my parameter's name, and without the preceding '\'?
It's possible to match that, even avoiding escapes of backslashes, as:
var re = RegExp(r"(?<!(?:^|[^\\])(?:\\{2})*\\)&\w+");
This uses negative lookbehind to find a & followed by word-characters, and not preceded by an odd number of backslashes.
More likely, you want to also recognize double-backslashes and convert them to single-backslashes. That's actually easier if you try to find all matches, because then you know all preceding double-backslashes are part of an earlier match:
var re = RegExp(r"\\\\|(?<!\\)&\w+");
This, when used as re.allMatches will find all occurrences of \\ and &word where the latter is not preceded by an odd number of backslashes.
var _re = RegExp(r"\\\\|(?<!\\)&(\w+)");
String template(String input, Map<String, String> values) {
return input.replaceAllMapped(_re, (m) {
var match = m[0]!;
if (match == r"\\") return r"\";
var replacement = values[m[1]!];
if (replacement != null) return replacement;
// do nothing for undefined words.
return match;
});
}
(You might also want to allow something like &{foo} if parameters can occur next to other characters, like &{amount)USD).
To keep the character before &param when it matches a non-backslash character you need to use so called capturing groups. These are are subexpressions of a regular expression inside parentheses. To use capturing groups in Dard you need to use the method replaceAllMapped. We also have the case when the template starts with &param and in this case we match at the beginning of the string instead.
Try this:
void main() {
final template = 'My name is &param and my parameter is called \\&param.';
final populatedTemplate = template.replaceAllMapped(RegExp(r'(^|[^\\])&param\b'), (match) {
return '${match.group(1)}John';
});
final result = populatedTemplate.replaceAll(RegExp(r'\\&param\b'), 'John');
print(result);
}

java.util.regex.PatternSyntaxException: Dangling meta character '+' for the string +91

I am using below regular expression string to replace the supplied string with ****
String output=output.replaceAll("(?<!\\w)(?i)"+requesterView.getFirstname()+"(?!\\w)","****");
Above the supplied string +91
If it contains + then getting below exception
java.util.regex.PatternSyntaxException: Dangling meta character '+' near index 12
(?<!\w)(?i)(+91)(?!\w)
^
at java.util.regex.Pattern.error(Pattern.java:1955)
at java.util.regex.Pattern.sequence(Pattern.java:2123)
at java.util.regex.Pattern.expr(Pattern.java:1996)
at java.util.regex.Pattern.group0(Pattern.java:2905)
at java.util.regex.Pattern.sequence(Pattern.java:2051)
at java.util.regex.Pattern.expr(Pattern.java:1996)
at java.util.regex.Pattern.compile(Pattern.java:1696)
at java.util.regex.Pattern.<init>(Pattern.java:1351)
at java.util.regex.Pattern.compile(Pattern.java:1028)
How to resolve above exception ?
You need to escape regex meta-characters in your input String, which you can do with the Pattern.quote(String str) static method :
String output=output.replaceAll("(?<!\\w)(?i)"+Pattern.quote(requesterView.getFirstname())+"(?!\\w)","****");
Currently Java tries to parse the tokens of the input string (+91) as regex tokens and fails to make sense of the + meta-character in the context it's found in. Additionnally the parenthesis would have been understood as a capturing group.
Like Aaron mentioned you need to quote the regular expression.
This can be achieved either with Pattern.quote or using \Q together with \E. Here is an example:
public static String transformRegex(String input, String testStr) {
return input.replaceAll("(?<!\\w)(?i)\\Q" + testStr + "\\E(?!\\w)", "****");
}
Here is a test of the method above:
String output = transformRegex("+91 123123123", "+91");
System.out.println(output);
This prints:
**** 123123123

How to match a string with an opening brace { in C++ regex

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.

C++11 regex to tokenize Mathematical Expression

I have the following code to tokenize a string of the format: (1+2)/((8))-(100*34):
I'd like to throw an error to the user if they use an operator or character that isn't part of my regex.
e.g if user enters 3^4 or x-6
Is there a way to negate my regex, search for it and if it is true throw the error?
Can the regex expression be improved?
//Using c++11 regex to tokenize input string
//[0-9]+ = 1 or many digits
//Or [\\-\\+\\\\\(\\)\\/\\*] = "-" or "+" or "/" or "*" or "(" or ")"
std::regex e ( "[0-9]+|[\\-\\+\\\\\(\\)\\/\\*]");
std::sregex_iterator rend;
std::sregex_iterator a( infixExpression.begin(), infixExpression.end(), e );
queue<string> infixQueue;
while (a!=rend) {
infixQueue.push(a->str());
++a;
}
return infixQueue;
-Thanks
You can run a search on the string using the search expression [^0-9()+\-*/] defined as C++ string as "[^0-9()+\\-*/]" which finds any character which is NOT a digit, a round bracket, a plus or minus sign (in real hyphen), an asterisk or a slash.
The search with this regular expression search string should not return anything otherwise the string contains a not supported character like ^ or x.
[...] is a positive character class which means find a character being one of the characters in the square brackets.
[^...] is a negative character class which means find a character NOT being one of the characters in the square brackets.
The only characters which must be escaped within square brackets to be interpreted as literal character are ], \ and - whereby - must not be escaped if being first or last character in the list of characters within the square brackets. But it is nevertheless better to escape - always within square brackets as this makes it easier for the regular expression engine / function to detect that the hyphen character should be interpreted as literal character and not with meaning "FROM x to z".
Of course this expression does not check for missing closing round brackets. But formula parsers do often not require that there is always a closing parenthesis for every opening parenthesis in comparison to a compiler or script interpreter simply because not needed to calculate the value based on entered formula.
Answer is given already but perhaps someone might need this
[0-9]?([0-9]*[.])?[0-9]+|[\\-\\+\\\\\(\\)\\/\\*]
This regex separates floats, integers and arithmetic operators
Heres the trick:
[0-9]?([0-9]*[.])?[0-9]+ -> if its a digit and has a point, then grab the digits with the point and the digits that follows it, if not, just grab the digits.
Sorry if my answer isn't clear, i just learned regex and found this solution by my own by just trial and errors.
Heres the code (it takes a mathematical expression and split all digits and operators into a vector)
NOTE: I don't know if it accepts whitespaces, meaning that the mathematical expression that i worked with had no whitespaces. Example: 4+2*(3+1) and would separate everything nicely, but i havent tried with whitespaces.
/* Separate every int or float or operator into a single string using regular expression and store it in untokenize vector */
string infix; //The string to be parse (the arithmetic operation if you will)
vector<string> untokenize;
std::regex words_regex("[0-9]?([0-9]*[.])?[0-9]+|[\\-\\+\\\\\(\\)\\/\\*]");
auto words_begin = std::sregex_iterator(infix.begin(), infix.end(), words_regex);
auto words_end = std::sregex_iterator();
for (std::sregex_iterator i = words_begin; i != words_end; ++i) {
cout << (*i).str() << endl;
untokenize.push_back((*i).str());
}
Output:
(<br/>
1<br/>
+<br/>
2<br/>
)<br/>
/<br/>
(<br/>
(<br/>
8<br/>
)<br/>
)<br/>
-<br/>
(<br/>
100<br/>
*<br/>
34<br/>
)<br/>

Ignore String containing special words (Months)

I am trying to find alphanumeric strings by using the following regular expression:
^(?=.*\d)(?=.*[a-zA-Z]).{3,90}$
Alphanumeric string: an alphanumeric string is any string that contains at least a number and a letter plus any other special characters it can be # - _ [] () {} ç _ \ ù %
I want to add an extra constraint to ignore all alphanumerical strings containing the following month formats :
JANVIER|FEVRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AOUT|SEPTEMBRE|OCTOBRE|NOVEMBRE|DECEMBRE|Jan|Feb|Mar|Apr|May|Jun|JUN|Jul|Aug|Sep|Oct|Nov|Dec|[jJ]anvier|[fF][ée]vrier|[mM]ars|[aA]vril|[mM]ai|[jJ]uin|[jJ]uillet|[aA]o[éû]t|aout|[sS]eptembre|[oO]ctobre|[nN]ovembre|[dD][eé]cembre
One solution is to actually match an alphanumerical string. Then check if this string contains one of these names by using the following function:
vector<string> findString(string s)
{
vector<string> vec;
boost::regex rgx("JANVIER|FEVRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AOUT|SEPTEMBRE|OCTOBRE|NOVEMBRE|DECEMBRE|Jan|Feb|Mar|Apr|May|Jun|JUN|Jul|Aug|Sep|Oct|Nov|Dec|[jJ]anvier|[fF][ée]vrier|[mM]ars|[aA]vril|[mM]ai|[jJ]uin|[jJ]uillet|[aA]o[éû]t|aout|[sS]eptembre|[oO]ctobre|[nN]ovembre|[dD][eé]cembre
");
boost::smatch match;
boost::sregex_iterator begin {s.begin(), s.end(), rgx},
end {};
for (boost::sregex_iterator& i = begin; i != end; ++i)
{
boost::smatch m = *i;
vec.push_back(m.str());
}
return vec;
}
Question: How can I add this constraint directly into the regular expression instead of using this function.
One solution is to use negative lookahead as mentioned in How to ignore words in string using Regular Expressions.
I used it as follows:
String : 2-hello-001
Regular expression : ^(?=.*\d)(?=.*[a-zA-Z]^(?!Jan|Feb|Mar)).{3,90}$
Result: no match
Test website: http://regexlib.com/
The edit provided by #Robin and #RyanCarlson : ^[][\w#_(){}ç\\ù%-]{3,90}$ works perfectly in detecting alphanumeric strings with special characters. It's just the negative lookahead part that isn't working.
You can use negative look ahead, the same way you're using positive lookahead:
(?=.*\d)(?=.*[a-zA-Z])
(?!.*(?:JANVIER|FEVRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AOUT|SEPTEMBRE|OCTOBRE|NOVEMBRE|DECEMBRE|Jan|Feb|Mar|Apr|May|Jun|JUN|Jul|Aug|Sep|Oct|Nov|Dec|[jJ]anvier|[fF][ée]vrier|[mM]ars|[aA]vril|[mM]ai|[jJ]uin|[jJ]uillet|[aA]o[éû]t|aout|[sS]eptembre|[oO]ctobre|[nN]ovembre|[dD][eé]cembre)).{3,90}$
Also you regex is pretty unclear. If you want alphanumerical strings with a length between 3 and 90, you can just do:
/^(?!.*(?:JANVIER|F[Eé]VRIER|MARS|AVRIL|MAI|JUIN|JUILLET|AO[Uù]T|SEPTEMBRE|OCTOBRE|NOVEMBRE|D[Eé]CEMBRE|Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec))
[][\w#_(){}ç\\ù%-]{3,90}$/i
the i flag means it will match upper and lower case (so you can reduce your forbidden list), \w is a shortcut for [0-9a-zA-Z_] (careful if you copy-paste, there's a linebreak here for readability between (?! ) and [ ]). Just add in the final [...] whatever special characters you wanna match.