I'm trying to get my Dart web app to: (1) determine if a particular string matches a given regex, and (2) if it does, extract a group/segment out of the string.
Specifically, I want to make sure that a given string is of the following form:
http://myapp.example.com/#<string-of-1-or-more-chars>[?param1=1¶m2=2]
Where <string-of-1-or-more-chars> is just that: any string of 1+ chars, and where the query string ([?param1=1¶m2=2]) is optional.
So:
Decide if the string matches the regex; and if so
Extract the <string-of-1-or-more-chars> group/segment out of the string
Here's my best attempt:
String testURL = "http://myapp.example.com/#fizz?a=1";
String regex = "^http://myapp.example.com/#.+(\?)+\$";
RegExp regexp= new RegExp(regex);
Iterable<Match> matches = regexp.allMatches(regex);
String viewName = null;
if(matches.length == 0) {
// testURL didn't match regex; throw error.
} else {
// It matched, now extract "fizz" from testURL...
viewName = ??? // (ex: matches.group(2)), etc.
}
In the above code, I know I'm using the RegExp API incorrectly (I'm not even using testURL anywhere), and on top of that, I have no clue how to use the RegExp API to extract (in this case) the "fizz" segment/group out of the URL.
The RegExp class comes with a convenience method for a single match:
RegExp regExp = new RegExp(r"^http://myapp.example.com/#([^?]+)");
var match = regExp.firstMatch("http://myapp.example.com/#fizz?a=1");
print(match[1]);
Note: I used anubhava's regular expression (yours was not escaping the ? correctly).
Note2: even though it's not necessary here, it is usually a good idea to use raw-strings for regular expressions since you don't need to escape $ and \ in them. Sometimes using triple-quote raw-strings are convenient too: new RegExp(r"""some'weird"regexp\$""").
Try this regex:
String regex = "^http://myapp.example.com/#([^?]+)";
And then grab: matches.group(1)
String regex = "^http://myapp.example.com/#([^?]+)";
Then:
var match = matches.elementAt(0);
print("${match.group(1)}"); // output : fizz
Related
Im a fresher to RegEx.
I want to get all Syllables out of my String using this RegEx:
/[^aeiouy]*[aeiouy]+(?:[^aeiouy]*\$|[^aeiouy](?=[^aeiouy]))?/gi
And I implemented it in Dart like this:
void main() {
String test = 'hairspray';
final RegExp syllableRegex = RegExp("/[^aeiouy]*[aeiouy]+(?:[^aeiouy]*\$|[^aeiouy](?=[^aeiouy]))?/gi");
print(test.split(syllableRegex));
}
The Problem:
Im getting the the word in the List not being splitted.
What do I need to change to get the Words divided as List.
I tested the RegEx on regex101 and it shows up to Matches.
But when Im using it in Dart with firstMatch I get null
You need to
Use a mere string pattern without regex delimiters in Dart as a regex pattern
Flags are not used, i is implemented as a caseSensitive option to RegExp and g is implemented as a RegExp#allMatches method
You need to match and extract, not split with your pattern.
You can use
String test = 'hairspray';
final RegExp syllableRegex = RegExp(r"[^aeiouy]*[aeiouy]+(?:[^aeiouy]*$|[^aeiouy](?=[^aeiouy]))?",
caseSensitive: true);
for (Match match in syllableRegex.allMatches(test)) {
print(match.group(0));
}
Output:
hair
spray
I am trying to get string using RegEx; here is the string:
window.runParams = {};
window.runParams = {blablabla};
How to get the second string {blablabla}? I am using REGEX:
(?<=window.runParams = ").*(?=;)
But that gets the first string {}.
If you want to get string with braces eg: {blablabla}
window.runParams = ({\w+})
If you want to get only the string inside braces eg: blablabla
window.runParams = {(\w+)}
Value of group 1 is your result
The following pattern captures only curly brackets with word character content:
(?<=window.runParams = ){\w+}(?=;)
and will only capture:
{blablabla}
when run against the text:
window.runParams = {};
window.runParams = {blablabla};
See results here:
https://regex101.com/r/mTwA64/1
try modifying your regex so it only accepts matches with non-empty curly brackets \{.+\} such as
(?<=window\.runParams = )(\{.+\})(?=;)
...there's probably ways to simplify the regex further, depending on you problem...my guess is you don't need the lookahead/lookbehind, e.g. in the example given \{.+\} will do just fine (returns {blablabla}) ....but it really depends on the format and content of your file...also remember braces, dots etc have a special meaning in regexes so you probably would want to escape them
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.
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.
I'd like to capture the value of the Initial Catalog in this string:
"blah blah Initial Catalog = MyCat'"
I'd like the result to be: MyCat
There could or could not be spaces before and after the equal sign and there could or could not be spaces before the single quote.
Tried this and various others but no go:
/Initial Catalog\s?=\s?.*\s?\'/
Using .Net.
You need to put parentheses around the part of the string that you would like to match:
/Initial Catalog\s*=\s*(.*?)\s*'/
Also you would like to exclude as many spaces as possible before the ', so you need \s* rather than \s?. The .*? means that the extracted part of the string doesn't take those spaces, since it is now lazy.
This is a nice regex
= *(.*?) *'
Use the idea and add \s and more literal text as needed.
In C# group 1 will contain the match
string resultString = null;
try {
Regex regexObj = new Regex("= *(.*?) *'");
resultString = regexObj.Match(subjectString).Groups[1].Value;
} catch (ArgumentException ex) {
// Syntax error in the regular expression
}
Regex rgx = new Regex(#"=\s*([A-z]+)\s*'");
String result = rgx.Match(text).Groups[1].Value;