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)");
Related
I'm sorry I found it difficult to express this question with my poor English. So, let's go directly to a simple example.
Assume we have a subject string "apple:banana:cherry:durian". We want to match the subject and have $1, $2, $3 and $4 become "apple", "banana", "cherry" and "durian", respectively. The pattern I'm using is ^(\w+)(?::(.*?))*$, and $1 will be "apple" as expected. However, $2 will be "durian" instead of "banana".
Because the subject string to match doesn't need to be 4 items, for example, it could be "one:two:three", and $1 and $2 will be "one" and "three" respectively. Again, the middle item is missing.
What is the correct pattern to use in this case? By the way, I'm going to use PCRE2 in C++ codes, so there is no split, a Perl built-in function. Thanks.
If the input contains strictly items of interest separated by :, like item1:item2:item3, as the attempt in the question indicates, then you can use the regex pattern
[^:]+
which matches consecutive characters which are not :, so a substring up to the first :. That may need to capture as well, ([^:]+), depending on the overall approach. How to use this to get all such matches depends on the language.†
In C++ there are different ways to approach this. Using std::regex_iterator
#include <string>
#include <vector>
#include <iterator>
#include <regex>
#include <iostream>
int main()
{
std::string str{R"(one:two:three)"};
std::regex r{R"([^:]+)"};
std::vector<std::string> result{};
auto it = std::sregex_iterator(str.begin(), str.end(), r);
auto end = std::sregex_iterator();
for(; it != end; ++it) {
auto match = *it;
result.push_back(match[0].str());
}
std::cout << "Input string: " << str << '\n';
for(auto i : result)
std::cout << i << '\n';
}
Prints as expected.
One can also use std::regex_search, even as it returns at first match -- by iterating over the string to move the search start after every match
#include <string>
#include <regex>
#include <iostream>
int main()
{
std::string str{"one:two:three"};
std::regex r{"[^:]+"};
std::smatch res;
std::string::const_iterator search_beg( str.cbegin() );
while ( regex_search( search_beg, str.cend(), res, r ) )
{
std::cout << res[0] << '\n';
search_beg = res.suffix().first;
}
std::cout << '\n';
}
(With this string and regex we don't need the raw string literal so I've removed them here.)
† This question was initially tagged with perl (with no c++), also with an explicit mention of it in text (still there), and the original version of this answer referred to Perl with
/([^:]+)/g
The /g "modifier" is for "global," to find all matches. The // are pattern delimiters.
When this expression is bound (=~) to a variable with a target string then the whole expression returns a list of matches when used in a context in which a list is expected, which can thus be directly assigned to an array variable.
my #captures = $string =~ /[^:]+/g;
(when this is used literally as shown then the capturing () aren't needed)
Assigning to an array provides this "list context." If the matching is used in a "scalar context," in which a single value is expected, like in the condition for an if test or being assigned to a scalar variable, then a single true/false is returned (usually 1 or '', empty string).
Repeating a capture group will only capture the value of the last iteration. Instead, you might make use of the \G anchor to get consecutive matches.
If the whole string can only contain word characters separated by colons:
(?:^(?=\w+(?::\w+)+$)|\G(?!^):)\K\w+
The pattern matches:
(?: Non capture group
^ Assert start of string
(?=\w+(?::\w+)+$) Assert from the current position 1+ word characters and 1+ repetitions of : and 1+ word characters till the end of the string
| Or
\G(?!^): Assert the position at the end of the previous match, not at the start and match :
) Close non capture group
\K\w+ Forget what is matched so far, and match 1+ word characters
Regex demo
To allow only words as well from the start of the string, and allow other chars after the word chars:
\G:?\K\w+
Regex demo
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 $1 in the replacement pattern) any 0+ chars, as many as possible, and then matches a } at the end of the string ($).
Look at this example:
string str = "January 19934";
The Outcome should be
Jan 1993
I think I have created the right RegEx ([A-z]{3}).*([\d]{4}) to use in this case but I do not know what I should do now?
How can I extract what I am looking for, using RegEx? Is there a way like receiving 2 variables, the first one being the result of the first RegEx bracket: ([A-z]{3}) and the second result being 2nd bracket:[[\d]{4}]?
Your regex contains a common typo: [A-z] matches more than just ASCII letters. Also, the .* will grab all the string up to its end, and backtracking will force \d{4} match the last 4 digits. You need to use lazy quantifier with the dot, *?.
Then, use regex_search and concat the 2 group values:
#include <regex>
#include <string>
#include <iostream>
using namespace std;
int main() {
regex r("([A-Za-z]{3}).*?([0-9]{4})");
string s("January 19934");
smatch match;
std::stringstream res("");
if (regex_search(s, match, r)) {
res << match.str(1) << " " << match.str(2);
}
cout << res.str(); // => Jan 1993
return 0;
}
See the C++ demo
Pattern explanation:
([A-Za-z]{3}) - Group 1: three ASCII letters
.*? - any 0+ chars other than line break symbols as few as possible
([0-9]{4}) - Group 2: 4 digits
This could work.
([A-Za-z]{3})([a-z ])+([\d]{4})
Note the space after a-z is important to catch space.