I'm trying to parse in Dart long strings containing hashtags, so far I tried various combinations with regexp but I cannot find the right use.
My code is
String mytestString = "#one #two, #three#FOur,#five";
RegExp regExp = new RegExp(r"/(^|\s)#\w+/g");
print(regExp.allMatches(mytestString).toString());
The desidered output would be a list of hahstags
#one #two #three #FOur #five
Thankyou in advance
You should not use a regex literal inside a string literal, or backslashes and flags will become part of the regex pattern. Also, omit the left-hand boundary pattern (that matches start of string or whitespace) if you need to match # followed with 1+ word chars in any context.
Use
String mytestString = "#one #two, #three#FOur,#five";
final regExp = new RegExp(r"#\w+");
Iterable<String> matches = regExp.allMatches(mytestString).map((m) => m[0]);
print(matches);
Output: (#one, #two, #three, #FOur, #five)
String mytestString = "#one #two, #three#FOur,#five";
RegExp regExp = new RegExp(r"/(#\w+)/g");
print(regExp.allMatches(mytestString).toString());
This should match all of the hashtags, placing them into capture groups for you to later use.
Related
The following should be matched:
AAA123
ABCDEFGH123
XXXX123
can I do: ".*123" ?
Yes, you can. That should work.
. = any char except newline
\. = the actual dot character
.? = .{0,1} = match any char except newline zero or one times
.* = .{0,} = match any char except newline zero or more times
.+ = .{1,} = match any char except newline one or more times
Yes that will work, though note that . will not match newlines unless you pass the DOTALL flag when compiling the expression:
Pattern pattern = Pattern.compile(".*123", Pattern.DOTALL);
Matcher matcher = pattern.matcher(inputStr);
boolean matchFound = matcher.matches();
Use the pattern . to match any character once, .* to match any character zero or more times, .+ to match any character one or more times.
The most common way I have seen to encode this is with a character class whose members form a partition of the set of all possible characters.
Usually people write that as [\s\S] (whitespace or non-whitespace), though [\w\W], [\d\D], etc. would all work.
.* and .+ are for any chars except for new lines.
Double Escaping
Just in case, you would wanted to include new lines, the following expressions might also work for those languages that double escaping is required such as Java or C++:
[\\s\\S]*
[\\d\\D]*
[\\w\\W]*
for zero or more times, or
[\\s\\S]+
[\\d\\D]+
[\\w\\W]+
for one or more times.
Single Escaping:
Double escaping is not required for some languages such as, C#, PHP, Ruby, PERL, Python, JavaScript:
[\s\S]*
[\d\D]*
[\w\W]*
[\s\S]+
[\d\D]+
[\w\W]+
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RegularExpression{
public static void main(String[] args){
final String regex_1 = "[\\s\\S]*";
final String regex_2 = "[\\d\\D]*";
final String regex_3 = "[\\w\\W]*";
final String string = "AAA123\n\t"
+ "ABCDEFGH123\n\t"
+ "XXXX123\n\t";
final Pattern pattern_1 = Pattern.compile(regex_1);
final Pattern pattern_2 = Pattern.compile(regex_2);
final Pattern pattern_3 = Pattern.compile(regex_3);
final Matcher matcher_1 = pattern_1.matcher(string);
final Matcher matcher_2 = pattern_2.matcher(string);
final Matcher matcher_3 = pattern_3.matcher(string);
if (matcher_1.find()) {
System.out.println("Full Match for Expression 1: " + matcher_1.group(0));
}
if (matcher_2.find()) {
System.out.println("Full Match for Expression 2: " + matcher_2.group(0));
}
if (matcher_3.find()) {
System.out.println("Full Match for Expression 3: " + matcher_3.group(0));
}
}
}
Output
Full Match for Expression 1: AAA123
ABCDEFGH123
XXXX123
Full Match for Expression 2: AAA123
ABCDEFGH123
XXXX123
Full Match for Expression 3: AAA123
ABCDEFGH123
XXXX123
If you wish to explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
There are lots of sophisticated regex testing and development tools, but if you just want a simple test harness in Java, here's one for you to play with:
String[] tests = {
"AAA123",
"ABCDEFGH123",
"XXXX123",
"XYZ123ABC",
"123123",
"X123",
"123",
};
for (String test : tests) {
System.out.println(test + " " +test.matches(".+123"));
}
Now you can easily add new testcases and try new patterns. Have fun exploring regex.
See also
regular-expressions.info/Tutorial
No, * will match zero-or-more characters. You should use +, which matches one-or-more instead.
This expression might work better for you: [A-Z]+123
Specific Solution to the example problem:-
Try [A-Z]*123$ will match 123, AAA123, ASDFRRF123. In case you need at least a character before 123 use [A-Z]+123$.
General Solution to the question (How to match "any character" in the regular expression):
If you are looking for anything including whitespace you can try [\w|\W]{min_char_to_match,}.
If you are trying to match anything except whitespace you can try [\S]{min_char_to_match,}.
Try the regex .{3,}. This will match all characters except a new line.
[^] should match any character, including newline. [^CHARS] matches all characters except for those in CHARS. If CHARS is empty, it matches all characters.
JavaScript example:
/a[^]*Z/.test("abcxyz \0\r\n\t012789ABCXYZ") // Returns ‘true’.
I like the following:
[!-~]
This matches all char codes including special characters and the normal A-Z, a-z, 0-9
https://www.w3schools.com/charsets/ref_html_ascii.asp
E.g. faker.internet.password(20, false, /[!-~]/)
Will generate a password like this: 0+>8*nZ\\*-mB7Ybbx,b>
I work this Not always dot is means any char. Exception when single line mode. \p{all} should be
String value = "|°¬<>!\"#$%&/()=?'\\¡¿/*-+_#[]^^{}";
String expression = "[a-zA-Z0-9\\p{all}]{0,50}";
if(value.matches(expression)){
System.out.println("true");
} else {
System.out.println("false");
}
For a project of mine, I want to create 'blocks' with Regex.
\xyz\yzx //wrong format
x\12 //wrong format
12\x //wrong format
\x12\x13\x14\x00\xff\xff //correct format
When using Regex101 to test my regular expressions, I came to this result:
([\\x(0-9A-Fa-f)])/gm
This leads to an incorrect output, because
12\x
Still gets detected as a correct string, though the order is wrong, it needs to be in the order specified below, and in no other order.
backslash x 0-9A-Fa-f 0-9A-Fa-f
Can anyone explain how that works and why it works in that way? Thanks in advance!
To match the \, folloed with x, followed with 2 hex chars, anywhere in the string, you need to use
\\x[0-9A-Fa-f]{2}
See the regex demo
To force it match all non-overlapping occurrences, use the specific modifiers (like /g in JavaScript/Perl) or specific functions in your programming language (Regex.Matches in .NET, or preg_match_all in PHP, etc.).
The ^(?:\\x[0-9A-Fa-f]{2})+$ regex validates a whole string that consists of the patterns like above. It happens due to the ^ (start of string) and $ (end of string) anchors. Note the (?:...)+ is a non-capturing group that can repeat in the string 1 or more times (due to + quantifier).
Some Java demo:
String s = "\\x12\\x13\\x14\\x00\\xff\\xff";
// Extract valid blocks
Pattern pattern = Pattern.compile("\\\\x[0-9A-Fa-f]{2}");
Matcher matcher = pattern.matcher(s);
List<String> res = new ArrayList<>();
while (matcher.find()){
res.add(matcher.group(0));
}
System.out.println(res); // => [\x12, \x13, \x14, \x00, \xff, \xff]
// Check if a string consists of valid "blocks" only
boolean isValid = s.matches("(?i)(?:\\\\x[a-f0-9]{2})+");
System.out.println(isValid); // => true
Note that we may shorten [a-zA-Z] to [a-z] if we add a case insensitive modifier (?i) to the start of the pattern, or just use \p{Alnum} that matches any alphanumeric char in a Java regex.
The String#matches method always anchors the regex by default, we do not need the leading ^ and trailing $ anchors when using the pattern inside it.
I'm having trouble escaping double dollars from a string to be used with regex functions pattern/matcher.
This is part of the String:
WHERE oid_2 = $$test$$ || oid_2 = $$test2$$
and this is the closest code I've tried to get near the solution:
List<String> strList = new ArrayList<String>();
Pattern pattern = Pattern.compile("\$\$.*?\$\$");
log.debug("PATTERN: "+pattern)
Matcher matcher = pattern.matcher(queryText);
while (matcher.find()) {
strList.add(matcher.group());
}
log.debug(strList)
This is the debug output i get
- PATTERN: $$.*?$$
- []
So the pattern is actually right, but the placeholders are not found in the string.
As a test I've tried to replace "$$test$$" with "XXtestXX" and everything works perfectly. What am I missing? I've tried "/$" strings, "\\" but still have no solution.
Note that a $ in regex matches the end of the string. To use it as a literal $ symbol, you need to escape it with a literal backslash.
You used "\$\$.*?\$\$" that got translated into a literal string like $$.*?$$ that matches 2 end of string positions, any 0+ chars as few as possible and then again 2 end of strings, which has little sense. You actually would need a backslash to first escape the $ that is used in Groovy to inject variables into a double quoted string literal, and then use 2 backslashes to define a literal backslash - "\\\$\\\$.*?\\\$\\\$".
However, when you work with regex, slashy strings are quite helpful since all you need to escape a special char is a single backslash.
Here is a sample code extracting all matches from the string you have in Groovy:
def regex = /\$\$.*?\$\$/;
def s = 'WHERE oid_2 = $$test$$ || oid_2 = $$test2$$'
def m = s =~ regex
(0..<m.count).each { print m[it] + '\n' }
See the online demo.
Anyone who gets here might like to know another answer to this, if you want to use Groovy slashy strings:
myComparisonString ==~ /.*something costs [$]stuff.*/
I couldn't find another way of putting a $ in a slashy string, at least if the $ is to be followed by text. If, conversely, it is followed by a number (or presumably any non-letter), this will work:
myComparisonString ==~ /.*something costs \$100.*/
... the trouble being, of course, that the GString "compiler" (if that's its name) would recognise "$stuff" as an interpolated variable.
Im new to regular expressions and Im trying to use RegExp on gwt Client side. I want to do a simple * matching. (say if user enters 006* , I want to match 006...). Im having trouble writing this. What I have is :
input = (006*)
input = input.replaceAll("\\*", "(" + "\\" + "\\" + "S\\*" + ")");
RegExp regExp = RegExp.compile(input).
It returns true with strings like BKLFD006* too. What am I doing wrong ?
Put a ^ at the start of the regex you're generating.
The ^ character means to match at the start of the source string only.
I think you are mixing two things here, namely replacement and matching.
Matching is used when you want to extract part of the input string that matches a specific pattern. In your case it seems that is what you want, and in order to get one or more digits that are followed by a star and not preceded by anything then you can use the following regex:
^[0-9]+(?=\*)
and here is a Java snippet:
String subjectString = "006*";
String ResultString = null;
Pattern regex = Pattern.compile("^[0-9]+(?=\\*)");
Matcher regexMatcher = regex.matcher(subjectString);
if (regexMatcher.find()) {
ResultString = regexMatcher.group();
}
On the other hand, replacement is used when you want to replace a re-occurring pattern from the input string with something else.
For example, if you want to replace all digits followed by a star with the same digits surrounded by parentheses then you can do it like this:
String input = "006*";
String result = input.replaceAll("^([0-9]+)\\*", "($1)");
Notice the use of $1 to reference the digits that where captured using the capture group ([0-9]+) in the regex pattern.
Excel returns a reference of the form
=Sheet1!R14C1R22C71junk
("junk" won't normally be there, but I want to be sure that there's no extraneous text.)
I would like to 'split' this into a VB array, where
a(0)="Sheet1"
a(1)="14"
a(2)="1"
a(3)="22"
a(4)="71"
a(5)="junk"
I'm sure it can be done easily with a regular expression, but I just can't get the hang of it.
Is there a kind soul who could help me?
Thanks
=([^!]+)!R(\d+)C(\d+)R(\d+)C(\d+)(.*)
should work.
[^!]+ matches a sequence of non-exclamation-point characters.
\d+ matches a sequence of digits.
.* matches anything.
So, in VB.NET:
Dim a As Match
a = Regex.Match(SubjectString, "=([^!]+)!R(\d+)C(\d+)R(\d+)C(\d+)(.*)")
If a.Success Then
' matched text: a.Value
' backreference n text: a.Groups(n).Value
Else
' Match attempt failed
End If
A straightforward String.Split would work, provided the "junk" text wasn't there:
Dim input As String = "=Sheet1!R14C1R22C71"
Dim result = input.Split(New Char() { "="c, "!"c, "R"c, "C"c }, StringSplitOptions.RemoveEmptyEntries)
For Each item As String In result
Console.WriteLine(item)
Next
The regex gets a little tricky since you will need to go through the Groups and Captures of the nested portions to get the proper order.
EDIT: here's my regex solution. It accepts multiple occurrences of R's and C's.
Dim input As String = "=Sheet1!R14C1R22C71junk"
Dim pattern As String = "=(?<Sheet>Sheet\d+)!(?:R(?<R>\d+)C(?<C>\d+))+"
Dim m As Match = Regex.Match(input, pattern)
If m.Success Then
Console.WriteLine(m.Groups("Sheet").Value)
For i = 0 To m.Groups("R").Captures.Count - 1
Console.WriteLine(m.Groups("R").Captures(i).Value)
Console.WriteLine(m.Groups("C").Captures(i).Value)
Next
End If
Pattern explanation:
"=(?Sheet\d+)" : matches an = sign followed by "Sheet" and digits. Uses named group of "Sheet"
"!(?:R(?\d+)C(?\d+))+" : matches the exclamation mark followed by at least one occurrence of the *R*xx*C*xx portion of the text. Named groups of "R" and "C" are used.
"(?:...)+" : this portion from the above portion matches but does not capture the inner pattern (i.e., the R/C part). This is to avoid unnecessarily capturing them while we are actually capturing them with the named groups.
More general regexes for R1C1 style:
^=(?:(?<Sheet>[^!]+)!)?(?:R((?<RAbs>\d+)|(?<RRel>\[-?\d+\]))C((?<CAbs>\d+)|(?<CRel>\[-?\d+\]))){1,2}$
And A1 style:
^=(?:(?<Sheet>[^!]+)!)?(?:(?<Col1>\$?[a-z]+)(?<Row1>\$?\d+))(?:\:(?<Col2>\$?[a-z]+)(?<Row2>\$?\d+))?$
It doesn't match external references like =[Book1]Sheet1!A1 though.