i need a regex that replaces everything except the content between the first " and the last ".
I need it like this:
Input String:["Key:"Value""]
And after the regex i only need this:
Output String:Key:"Value"
Thanks!
You can try something like this.
patern:
^.*?"(.*)".*$
Substion:
$1
On Regex101
Explination:
the first part ^.*?" matches as few characters as possible that are between the start of the string and a double quote
the second part(.*)" makes the largest match it can that ends in a double quote, and stuffs it all in a capture group
the last part .*$ grabs what ever is left and includes it in the match
Finally you replace the entire match with the contents of the first capture group
Can you say why you need a RegExp?
A function like:
String unquote(String input) {
int start = input.indexOf('"');
if (start < 0) return input; // or throw.
int end = input.lastIndexOf('"');
if (start == end) return input; // or throw
return input.substring(start + 1, end);
}
is going to be faster and easier to understand than a RegExp.
Anyway, for the challenge, let's say we do want a RegExp that replaces the part up to the first " and from the last " with nothing. That's two replaces, so you can do an
input.replaceAll(RegExp(r'^[^"]*"|"[^"]*$'), "")`
or you can use a capturing group and a computed replacement like:
input.replaceFirstMapped(RegExp(r'^[^"]*"([^]*)"[^"]*$'), (m) => m[1])
Alternatively, you can use the capturing group to select the text between the two and extract it in code, instead of doing string replacement:
String unquote(String input) {
var re = RegExp(r'^[^"]*"([^]*)"[^"]$');
var match = re.firstMatch(input);
if (match == null) return input; // or throw.
return match[1];
}
Related
I have a scenario where i want to extract some substring based on following condition.
search for any pattern myvalue=123& , extract myvalue=123
If the "myvalue" present at end of the line without "&", extract myvalue=123
for ex:
The string is abcdmyvalue=123&xyz => the it should return myvalue=123
The string is abcdmyvalue=123 => the it should return myvalue=123
for first scenario it is working for me with following regex - myvalue=(.?(?=[&,""]))
I am looking for how to modify this regex to include my second scenario as well. I am using https://regex101.com/ to test this.
Thanks in Advace!
Some notes about the pattern that you tried
if you want to only match, you can omit the capture group
e* matches 0+ times an e char
the part .*?(?=[&,""]) matches as least chars until it can assert eiter & , or " to the right, so the positive lookahead expects a single char to the right to be present
You could shorten the pattern to a match only, using a negated character class that matches 0+ times any character except a whitespace char or &
myvalue=[^&\s]*
Regex demo
function regex(data) {
var test = data.match(/=(.*)&/);
if (test === null) {
return data.split('=')[1]
} else {
return test[1]
}
}
console.log(regex('abcdmyvalue=123&3e')); //123
console.log(regex('abcdmyvalue=123')); //123
here is your working code if there is no & at end of string it will have null and will go else block there we can simply split the string and get the value, If & is present at the end of string then regex will simply extract the value between = and &
if you want to use existing regex then you can do it like that
var test = data1.match(/=(.*)&|=(.*)/)
const result = test[1] ? test[1] : test[2];
console.log(result);
I have a string that sometimes contains a certain substring at the end and sometimes does not. When the string is present I want to update its value. When it is absent I want to add it at the end of the existing string.
For example:
int _newCount = 7;
_myString = 'The count is: COUNT=1;'
_myString2 = 'The count is: '
_rRuleString.replaceAllMapped(RegExp('COUNT=(.*?)\;'), (match) {
//if there is a match (like in _myString) update the count to value of _newCount
//if there is no match (like in _myString2) add COUNT=1; to the string
}
I have tried using a return of:
return "${match.group(1).isEmpty ? _myString + ;COUNT=1;' : 'COUNT=$_newCount;'}";
But it is not working.
Note that replaceAllMatched will only perform a replacement if there is a match, else, there will be no replacement (insertion is still a replacement of an empty string with some string).
Your expected matches are always at the end of the string, and you may leverage this in your current code. You need a regex that optionally matches COUNT= and then some text up to the first ; including the char and then checks if the current position is the end of string.
Then, just follow the logic: if Group 1 is matched, set the new count value, else, add the COUNT=1; string:
The regex is
(COUNT=[^;]*;)?$
See the regex demo.
Details
(COUNT=[^;]*;)? - an optional group 1: COUNT=, any 0 or more chars other than ; and then a ;
$ - end of string.
Dart code:
_myString.replaceFirstMapped(RegExp(r'(COUNT=[^;]*;)?$'), (match) {
return match.group(0).isEmpty ? "COUNT=1;" : "COUNT=$_newCount;" ; }
)
Note the use of replaceFirstMatched, you need to replace only the first match.
I am using regex for CSV processing where data can be in Quotes, or no quotes. But if there is just a comma at the starting column, it skips it.
Here is the regex I am using:
(?:,"|^")(""|[\w\W]*?)(?=",|"$)|(?:,(?!")|^(?!"))([^,]*?|)(?=$|,)
Now the example data I am using is:
,"data",moredata,"Data"
Which should have 4 matches ["","data","moredata","Data"], but it always skips the first comma. It is fine if there is quotes on the first column, or it is not blank, but if it is empty with no quotes, it ignores it.
Here is a sample code I am using for testing purposes, it is written in Dart:
void main() {
String delimiter = ",";
String rawRow = ',,"data",moredata,"Data"';
RegExp exp = new RegExp(r'(?:'+ delimiter + r'"|^")(^,|""|[\w\W]*?)(?="'+ delimiter + r'|"$)|(?:'+ delimiter + '(?!")|^(?!"))([^'+ delimiter + r']*?)(?=$|'+ delimiter + r')');
Iterable<Match> matches = exp.allMatches(rawRow.replaceAll("\n","").replaceAll("\r","").trim());
List<String> row = new List();
matches.forEach((Match m) {
//This checks to see which match group it found the item in.
String cellValue;
if (m.group(2) != null) {
//Data found without speech marks
cellValue = m.group(2);
} else if (m.group(1) != null) {
//Data found with speech marks (so it removes escaped quotes)
cellValue = m.group(1).replaceAll('""', '"');
} else {
//Anything left
cellValue = m.group(0).replaceAll('""', '"');
}
row.add(cellValue);
});
print(row.toString());
}
Investigating your expression
(,"|^")
(""|[\w\W]*?)
(?=",|"$)
|
(,(?!")|^(?!"))
([^,]*?|)
(?=$|,)
(,"|^")(""|[\w\W]*?)(?=",|"$) This part is to match quoted strings, that seem to work for you
Going through this part (,(?!")|^(?!"))([^,]*?|)(?=$|,)
(,(?!")|^(?!")) start with comma not followed by " OR start of line not followed by "
([^,]*?|) Start of line or comma zero or more non greedy and |, why |
(?=$|,) end of line or , .
In CSV this ,,,3,4,5 line should give 6 matches but the above only gets 5
You could add (^(?=,)) at the begining of second part, the part that matches non quoted sections.
Second group with match of start and also added non capture to groups
(?:^(?=,))|(?:,(?!")|^(?!"))(?:[^,]*?)(?=$|,)
Complete: (?:,"|^")(?:""|[\w\W]*?)(?=",|"$)|(?:^(?=,))|(?:,(?!")|^(?!"))(?:[^,]*?)(?=$|,)
Here is another that might work
(?:(?:"(?:[^"]|"")*"|(?<=,)[^,]*(?=,))|^[^,]+|^(?=,)|[^,]+$|(?<=,)$)
How that works i described here: Build CSV parser using regex
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 $1 in the replacement pattern) any 0+ chars, as many as possible, and then matches a } at the end of the string ($).
As stated in the title I have an program in golang where I have a string with a reoccurring pattern. I have a beginning and end delimiters for this pattern, and I would like to extract them from the string. The following is pseudo code:
string := "... This is preceding text
PATTERN BEGINS HERE (
pattern can continue for any number of lines...
);
this is trailing text that is not part of the pattern"
In short what I am attempting to do is from the example above is extract all occurrences of of the pattern that begins with "PATTERN BEGINS HERE" and ends with ");" And I need help in figuring out what the regex for this looks like.
Please let me know if any additional info or context is needed.
The regex is:
(?s)PATTERN BEGINS HERE.*?\);
where (?s) is a flag to let .* match multiple lines (see Go regex syntax).
See demo
Not regex, but works
func findInString(str, start, end string) ([]byte, error) {
var match []byte
index := strings.Index(str, start)
if index == -1 {
return match, errors.New("Not found")
}
index += len(start)
for {
char := str[index]
if strings.HasPrefix(str[index:index+len(match)], end) {
break
}
match = append(match, char)
index++
}
return match, nil
}
EDIT: Best to handle individual character as bytes and return a byte array