I need to format the price string in dart.
String can be: ₹ 2,19,990.00
String can be: $1,114.99
String can be: $14.99
What I tried:
void main() {
String str = "₹ 2,19,990.00";
RegExp regexp = RegExp("(\\d+[,.]?[\\d]*)");
RegExpMatch? match = regexp.firstMatch(str);
str = match!.group(1)!;
print(str);
}
What my output is: 2,19
What my output is: 1,114
What my output is: 14.99
Expected output: 219990.00
Expected output: 1114.99
Expected output: 14.99 (This one is correct because there is no comma)
The simplest solution would be to replace all non-digit/non-dot characters with nothing.
The most efficient way to do that is:
final re = RegExp(r"[^\d.]+");
String sanitizeCurrency(String input) => input.replaceAll(re, "");
You can't do it by matching because a match is always contiguous in the source string, and you want to omit the embedded ,s.
You can use this regex for search:
^\D+|(?<=\d),(?=\d)
And replace with an empty string i.e. "".
RegEx Details:
^: Start
\D+: Match 1+ non-digit characters
|: OR
(?<=\d),(?=\d): Match a comma if it surrounded with digits on both sides
RegEx Demo
Code: Using replaceAll method:
str = str.replaceAll(RegExp(r'^\D+|(?<=\d),(?=\d)'), '');
Related
I defined a regular expression to check if the string only contains alphabetic characters and with length 5:
use regex::Regex;
fn main() {
let re = Regex::new("[a-zA-Z]{5}").unwrap();
println!("{}", re.is_match("this-shouldn't-return-true#"));
}
The text I use contains many illegal characters and is longer than 5 characters, so why does this return true?
You have to put it inside ^...$ to match the whole string and not just parts:
use regex::Regex;
fn main() {
let re = Regex::new("^[a-zA-Z]{5}$").unwrap();
println!("{}", re.is_match("this-shouldn't-return-true#"));
}
Playground.
As explained in the docs:
Notice the use of the ^ and $ anchors. In this crate, every expression is executed with an implicit .*? at the beginning and end, which allows it to match anywhere in the text. Anchors can be used to ensure that the full text matches an expression.
Your pattern returns true because it matches any consecutive 5 alpha chars, in your case it matches both 'shouldn't' and 'return'.
Change your regex to: ^[a-zA-Z]{5}$
^ start of string
[a-zA-Z]{5} matches 5 alpha chars
$ end of string
This will match a string only if the string has a length of 5 chars and all of the chars from start to end fall in range a-z and A-Z.
I would like to take a string like "My String 2022-01-07" extract the date part into a named capture.
I've tried the following regex, but it only works when there's an exact match:
# Does not work
iex> Regex.named_captures(~r/(?<date>\$?(\d{4}-\d{2}-\d{2})?)/, "My String 2021-01-01")
%{"date" => ""}
# Works
iex> Regex.named_captures(~r/(?<date>\$?(\d{4}-\d{2}-\d{2})?)/, "2021-01-01")
%{"date" => "2021-01-01"}
I've also tried this without luck:
iex> Regex.named_captures(~r/([a-zA-Z0-9 ]+?)(?<date>\$?(\d{4}-\d{2}-\d{2})?)/, "My String 2021-01-01")
%{"date" => ""}
Is there a way to use named captures to extract the date part of a string when you don't care about the characters surrounding the date?
I think I'm looking for a regex that will work like this:
iex> Regex.named_captures(REGEX???, "My String 2021-01-01 Other Parts")
%{"date" => "2021-01-01"}
You want
Regex.named_captures(~r/(?<date>\$?\d{4}-\d{2}-\d{2})/, "My String 2021-01-01")
Your regex - (?<date>\$?(\d{4}-\d{2}-\d{2})?) - represents a named capturing group with date as a name and a \$?(\d{4}-\d{2}-\d{2})? as a pattern. The \$?(\d{4}-\d{2}-\d{2})? pattern matches
\$? - an optional $ char
(\d{4}-\d{2}-\d{2})? - an optional sequence of four digits, -, two digits, -, two digits.
Since the pattern is not anchored (does not have to match the whole string) and both consecutive pattern parts are optional and thus can match an empty string, the ~r/(?<date>\$?(\d{4}-\d{2}-\d{2})?)/ regex **matches the first empty location (empty string) at the start of the "My String 2021-01-01" string.
Rule of thumb: If you do not want to match an empty string, make sure your pattern contains obligatory patterns, that must match at least one char.
Extract Date only:
void main() {
String inputString = "Your String 1/19/2023 9:29:11 AM";
RegExp dateRegex = new RegExp(r"(\d{1,2}\/\d{1,2}\/\d{4})");
Iterable<RegExpMatch> matches = dateRegex.allMatches(inputString);
for (RegExpMatch m in matches) {
print(m.group(0));
}
}
This will output:
1/19/2023
Extract Date and time:
void main() {
String inputString = "Your String 1/19/2023 9:29:11 AM";
RegExp dateTimeRegex = new RegExp(r"(\d{1,2}\/\d{1,2}\/\d{4} \d{1,2}:\d{2}:\d{2} [AP]M)");
Iterable<RegExpMatch> matches = dateTimeRegex.allMatches(inputString);
for (RegExpMatch m in matches) {
print(m.group(0));
}
}
This will output: 1/19/2023 9:29:11 AM
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);
This question already has an answer here:
How to put all regex matches into a string list
(1 answer)
Closed 1 year ago.
I want to return a pattern through regEx in flutter every time it' found, I tested using the Regex operation it worked on the same string, returning the match after that included match 'text:' to '}' letters, but it does not print the matches in the flutter application.
The code I am using:
String myString = '{boundingBox: 150,39,48,25, text: PM},';
RegExp exp = RegExp(r"text:(.+?(?=}))");
print("allMatches : "+exp.allMatches(myString).toString());
The output print statement is printing I/flutter ( 5287): allMatches : (Instance of '_RegExpMatch', Instance of '_RegExpMatch')
instead of text: PM
Following is the screenshot of how it is parsing on regexr.com
Instead of using a non greedy match with a lookahead, I would suggest using a negated character class matching any char except } in capture group 1, and match the } after the group to prevent some backtracking.
\b(text:[^}]+)}
You can loop the result from allMatches and print group 1:
String myString = '{boundingBox: 150,39,48,25, text: PM},';
RegExp exp = RegExp(r"\b(text:[^}]+)}");
for (var m in exp.allMatches(myString)) {
print(m[1]);
}
Output
text: PM
You need to use map method to retrieve the string from the matches:
String myString = '{boundingBox: 150,39,48,25, text: PM},';
RegExp exp = RegExp(r"text:(.+?(?=}))");
final matches = exp.allMatches(myString).map((m) => m.group(0)).toString();
print("allMatches : $matches");
I am trying to replace a certain group to "" by using regex.
I was searching and doing my best, but it's over my head.
What I want to do is,
string text = "(12je)apple(/)(jj92)banana(/)cat";
string resultIwant = {apple, banana, cat};
In the first square bracket, there must be 4 character including numbers.
and '(/)' will come to close.
Here's my code. (I was using matches function)
string text= #"(12dj)apple(/)(88j1)banana(/)cat";
string pattern = #"\(.{4}\)(?<value>.+?)\(/\)";
Regex rex = new Regex(pattern);
MatchCollection mc = rex.Matches(text);
if(mc.Count > 0)
{
foreach(Match str in mc)
{
print(str.Groups["value"].Value.ToString());
}
}
However, the result was
apple
banana
So I think I should use replace or something else instead of Matches.
The below regex would capture the word characters which are just after to ),
(?<=\))(\w+)
DEMO
Your c# code would be,
{
string str = "(12je)apple(/)(jj92)banana(/)cat";
Regex rgx = new Regex(#"(?<=\))(\w+)");
foreach (Match m in rgx.Matches(str))
Console.WriteLine(m.Groups[1].Value);
}
IDEONE
Explanation:
(?<=\)) Positive lookbehind is used here. It sets the matching marker just after to the ) symbol.
() capturing groups.
\w+ Then it captures all the following word characters. It won't capture the following ( symbol because it isn't a word character.