I tried the following regex:
const static char * regex_string = "([a-zA-Z0-9]+).*";
void find_first(const std::string str);
int main(int argc, char ** argv)
{
find_first("0s7fg9078dfg09d78fg097dsfg7sdg\r\nfdfgdfg");
}
void find_first(const std::string str)
{
std::cout << str << std::endl;
std::regex rgx(regex_string);
std::smatch matcher;
if(std::regex_match(str, matcher, rgx))
{
std::cout << "Found : " << matcher.str(0) << std::endl;
} else {
std::cout << "Not found" << std::endl;
}
}
DEMO
I expected the regex will be completely correct and the group will be found. But it wasn't. Why? How can I match the line-break in c++ regex? In Java it works fine.
The dot in regex usually matches any character other than a newline std::ECMAScript syntax.
. not newline any character except line terminators (LF, CR, LS, PS).
0s7fg9078dfg09d78fg097dsfg7sdg\r\nfdfgdfg
[a-zA-Z0-9]+ matches until \r ↑___↑ .* would match from here
In many regex flavors there is a dotall flag available to make the dot also match newlines.
If not, there are workarounds in different languages such as [^] not nothing or [\S\s] any whitespace or non-whitespace together in a class wich results in any character including \n
regex_string = "([a-zA-Z0-9]+)[\\S\\s]*";
Or use optional line breaks: ([a-zA-Z0-9]+).*(?:\\r?\\n.*)* or ([a-zA-Z0-9]+)(?:.|\\r?\\n)*
See your updated demo
Update - Another idea worth mentioning: std::regex::extended
A <period> ( '.' ), when used outside a bracket expression, is an ERE that shall match any character in the supported character set except NUL.
std::regex rgx(regex_string, std::regex::extended);
See this demo at tio.run
You may try const static char * regex_string = "((.|\r\n)*)";
I hope It will help you.
I am using CTest and PROPERTIES PASS_REGULAR_EXPRESSION.
[\S\s]* did not work, but (.|\r|\n)* did.
This regex:
Function registered for ID 2 was called(.|\r|\n)*PASS
Matches:
Running test function: RegisterThreeDiffItemsTest04
ID 2 registered for callback
ID 4 registered for callback
ID 11 registered for callback
Function registered for ID 2 was called
ID 2 callback deregistered
ID 4 callback deregistered
ID 11 callback deregistered
Setup: PASS
Note: CMakeLists.txt needs to escape the backslashes:
SET (ANDPASS "(.|\\r|\\n)*PASS")
Related
I am trying to replace a single occurrence of a character '1' in a String with a different character.
This same character can occur multiple times in the String which I am not interested in.
For example, in the below string I want to replace the single occurrence of 1 with 2.
input:-0001011101
output:-0002011102
I tried the below regex but it is giving be wrong results
regex b1("(1){1}");
S1=regex_replace( S,
b1, "2");
Any help would be greatly appreciated.
If you used boost::regex, Boost regex library, you could simply use a lookaround-based solution like
(?<!1)1(?!1)
And then replace with 2.
With std::regex, you cannot use lookbehinds, but you can use a regex that captures either start of string or any one char other than your char, then matches your char, and then makes sure your char does not occur immediately on the right.
Then, you may replace with $01 backreference to Group 1 (the 0 is necessary since the $12 replacement pattern would be parsed as Group 12, an empty string here since there is no Group 12 in the match structure):
regex reg("([^1]|^)1(?!1)");
S1=std::regex_replace(S, regex, "$012");
See the C++ demo online:
#include <iostream>
#include <regex>
int main() {
std::string S = "-0001011101";
std::regex reg("([^1]|^)1(?!1)");
std::cout << std::regex_replace(S, reg, "$012") << std::endl;
return 0;
}
// => -0002011102
Details:
([^1]|^) - Capturing group 1: any char other than 1 ([^...] is a negated character class) or start of string (^ is a start of string anchor)
1 - a 1 char
(?!1) - a negative lookahead that fails the match if there is a 1 char immediately to the right of the current location.
Use a negative lookahead in the regexp to match a 1 that isn't followed by another 1:
regex b1("1(?!1)");
I have a RE2 regex as following
const re2::RE2 numRegex("(([0-9]+),)+([0-9])+");
std::string inputStr;
inputStr="apple with make,up things $312,412,3.00");
RE2::Replace(&inputStr, numRegex, "$1$3");
cout << inputStr;
Expected
apple with make,up,things $3124123.00
I was trying to remove the , in the recognized number, $1 would only match 312 but not 412 part. Wondering how to extract the recursive pattern in the group.
Note that RE2 doesn't support lookahead (see Using positive-lookahead (?=regex) with re2) and the solutions I found all use lookaheads.
RE2 based solution
As RE2 does not support lookarounds, there is no pure single-pass regex solution.
You can have a workaround (as usual, when no solution is available): replace the string twice with (\d),(\d) regex and $1$2 substitution:
const re2::RE2 numRegex(R"((\d),(\d))");
std::string inputStr("apple with make,up things $312,412,3.00");
RE2::Replace(&inputStr, numRegex, "$1$2");
RE2::Replace(&inputStr, numRegex, "$1$2"); // <- Second pass to remove commas in 1,2,3,4 like strings
std::cout << inputStr;
C++ std::regex based solution:
You can remove the commas between digits using
std::string inputStr("apple with make,up things $312,412,3.00");
std::regex numRegex(R"((\d),(?=\d))");
std::cout << regex_replace(inputStr, numRegex, "$1") << "\n";
// => apple with make,up things $3124123.00
See the C++ demo. Also, see the regex demo here.
Details:
(\d) - Capturing group 1 ($1): a digit
, - a comma
(?=\d) - a positive lookahead that requires a digit immediately to the right of the current location.
In the pattern that you tried, you are repeating the outer group (([0-9]+),)+ which will then contain the value of the last iteration where it can match a 1+ digits and a comma.
The last iteration will capture 412, and 312, will only be matched.
You are using regex, but as an alternative if you have boost available, you could make use of the \G anchor which can get iterative matches asserting the position at the end of the previous match and replace with an empty string.
(?:\$|\G(?!^))\d+\K,(?=\d)
The pattern matches:
(?: Non capture group
\$ match $
| Or
\G(?!^) Assert the position at the end of the previous match, not at the start
) Close non capture group
\d+\K Match 1+ digits and forget what is matched so far
,(?=\d) Match a comma and assert a digit directly to the right
Regex demo
#include<iostream>
#include <string>
#include <boost/regex.hpp>
using namespace std;
int main()
{
std::string inputStr = "apple with make,up things $312,412,3.00";
boost::regex numRegex("(?:\\$|\\G(?!^))\\d+\\K,(?=\\d)");
std::string result = boost::regex_replace(inputStr, numRegex, "");
std::cout << result << std::endl;
}
Output
apple with make,up things $3124123.00
I am trying to find substrings that are not surrounded by other a-zA-Z0-9 symbols.
For example: I want to find substring hello, so it won't match hello1 or hellow but will match Hello and heLLo!##$%.
And I have such sample below.
std::string s = "1mySymbol1, /_mySymbol_ mysymbol";
const std::string sub = "mysymbol";
std::regex rgx("[^a-zA-Z0-9]*" + sub + "[^a-zA-Z0-9]*", std::regex::icase);
std::smatch match;
while (std::regex_search(s, match, rgx)) {
std::cout << match.size() << "match: " << match[0] << '\n';
s = match.suffix();
}
The result is:
1match: mySymbol
1match: , /_mySymbol_
1match: mysymbol
But I don't understand why first occurance 1mySymbol1 also matches my regex?
How to create a proper regex that will ignore such strings?
UDP
If I do like this
std::string s = "mySymbol, /_mySymbol_ mysymbol";
const std::string sub = "mysymbol";
std::regex rgx("[^a-zA-Z0-9]+" + sub + "[^a-zA-Z0-9]+", std::regex::icase);
then I find only substring in the middle
1match: , /_mySymbol_
And don't find substrings at the beggining and at the end.
The regex [^a-zA-Z0-9]* will match 0 or more characters, so it's perfectly valid for [^a-zA-Z0-9]*mysymbol[^a-zA-Z0-9]* to match mysymbol in 1mySymbol1 (allowing for case insensitivity). As you saw, this is fixed when you use [^a-zA-Z0-9]+ (matching 1 or more characters) instead.
With your update, you see that this doesn't match strings at the beginning or end. That's because [^a-zA-Z0-9]+ has to match 1 or more characters (which don't exist at the beginning or end of the string).
You have a few options:
Use beginning/end anchors: (?:[^a-zA-Z0-9]+|^)mysymbol(?:[^a-zA-Z0-9]+|$) (non-alphanumeric OR beginning of string, followed by mysymbol, followed by non-alphanumeric OR end of string).
Use negative lookahead and negative lookbehind: (?<![a-zA-Z0-9])mysymbol(?![a-zA-Z0-9]) (match mysymbol which doesn't have an alphanumeric character before or after it). Note that using this the match won't include the characters before/after mysymbol.
I recommend using https://regex101.com/ to play around with regular expressions. It lists all the different constructs you can use.
Give some solution to this following example,
Scenario-1:
My String : Password={my_pswd}}123}
I want to select the value enclosed within the {} brackets(Example: I want to select the complete password key value {my_pswd}123} not {my_pswd})
If I'm using this regex \{(.*?)\} , this will select {my_pswd} not {my_pswd}}123}. So how to get complete word even if the word has } in between? Give me some suggestions by using regex or any other way.
Scenario-2:
I am using this regex ^\{|\}$ . If my string have both { bracket and } bracket like this {{my_password}} then only it want to select first and last bracket. If my string like this {{my_password, it don't want to select that starting bracket. Its like AND condition in Regex. I referred many posts they did with look up but I can't get clear idea. Give me some suggestion.
Thanks.
It seems that the {...} substrings you want to match must be followed with ; or end of string.
This will not work for cases when a } inside the values can also be followed with ;.
You may solve the first issue by adding a (?![^;]) lookaround:
\{(.*?)\}(?![^;])
See the regex demo.
Details
\{ - a { char
(.*?) - Group 1: any 0+ chars as few as possible
\} - a } char
(?![^;]) - no char other than ; is allowed right after the current position
See the C++ demo:
#include <iostream>
#include <vector>
#include <regex>
int main() {
const std::regex reg("\\{(.*?)\\}(?![^;])");
std::smatch match;
std::string s = "Username={My_{}user};Password={my_pswd}}123}}}kk};Password={my_pswd}}123}";
std::vector<std::string> results(
std::sregex_token_iterator(s.begin(), s.end(), reg, 1), // See 1, it extracts Group 1 value
std::sregex_token_iterator());
for (auto result : results)
{
std::cout << result << std::endl;
}
return 0;
}
Output:
My_{}user
my_pswd}}123}}}kk
my_pswd}}123
As for the second scenario, you may use
std::regex reg("^\\{([^]*)\\}$");
std::string s = "{My_{}user}";
std::cout << regex_replace(s, reg, "$1") << std::endl; // => My_{}user
See another C++ demo.
The \{([^]*)\}$ pattern matches the { at the start (^) of the string, then matches and captures into Group 1 (later referenced with the help of $1 in the replacement pattern) any 0+ chars, as many as possible, and then matches a } at the end of the string ($).
I have been trying to search for a clear answer on this one, but not been able to find it.
So lets say I have the string (where \n could be \r\n - I want to handle both - not sure if that is relevant or not)
"4345t435\ng54t a_file_123.xml rk\ngreg a_file_j34.xml fger 43t54"
Then I want to get matches:
a_file_123.xml
a_file_j34.xml
Here is my test code:
const str::string s = "4345t435\ng54t a_file_123.xml rk\ngreg a_file_j34.xml fger 43t54";
std::smatch matches;
if (std::regex_search(s, matches, std::regex("a_file_(.*)\\.xml")))
{
std::cout << "total: " << matches.size() << std::endl;
for (unsigned int i = 0; i < matches.size(); i++)
{
std::cout << "match: " << matches[i] << std::endl;
}
}
Output is:
total: 2
match: a_file_123.xml
match: 123
I don't quite understand why match 2 is just "123"...
You only have one match, not two, as the regex_search method returns a single match. What you printed is two group values, Group 0 (the whole match, a_file_123.xml here) and Group 1 (the capturing group value, here, 123 that is a substring captured with a capturing group you defined as (.*) in the pattern).
If you want to match multiple strings, you need to use the regex iterator, not just a regex_search that only returns the first match.
Besides, .* is too greedy and will return weird results if you have more than 1 match on the same line. It seems you want to match letter or digits, so .* can be replaced with \w+. Well, if there can really be anything, just use .*?.
Use
const std::string s = "4345t435\ng54t a_file_123.xml rk\ngreg a_file_j34.xml fger 43t54";
const std::regex rx("a_file_\\w+\\.xml");
std::vector<std::string> results(std::sregex_token_iterator(s.begin(), s.end(), rx),
std::sregex_token_iterator());
std::cout << "Number of matches: " << results.size() << std::endl;
for (auto result : results)
{
std::cout << result << std::endl;
}
See the C++ demo yielding
Number of matches: 2
a_file_123.xml
a_file_j34.xml
Notes on regex
a_file_ - a literal substring
\\w+ - 1+ word chars (letters, digits, _) (note you may use [^.]*? here instead of \\w+ if you want to match any char, 0 or more repetitions, as few as possible, up to the first .xml)
\\. - a dot (if you do not escape it, it will match any char except line break chars)
xml - a literal substring.
See the regex demo