I'm trying to implement the escape character functionality in a macro generator I'm writing in Dart. For example, I would like the program to grab all the occurrences of '¶m' in my string and replace it with 'John', unless the '&' character is preceded with the escape character '\'. Example: "My name is ¶m and my parameter is called \¶m." -> "My name is John and my parameter is called ¶m". What would be the regular expression to catch all the substrings that contain the '&', then my parameter's name, and without the preceding '\'?
It's possible to match that, even avoiding escapes of backslashes, as:
var re = RegExp(r"(?<!(?:^|[^\\])(?:\\{2})*\\)&\w+");
This uses negative lookbehind to find a & followed by word-characters, and not preceded by an odd number of backslashes.
More likely, you want to also recognize double-backslashes and convert them to single-backslashes. That's actually easier if you try to find all matches, because then you know all preceding double-backslashes are part of an earlier match:
var re = RegExp(r"\\\\|(?<!\\)&\w+");
This, when used as re.allMatches will find all occurrences of \\ and &word where the latter is not preceded by an odd number of backslashes.
var _re = RegExp(r"\\\\|(?<!\\)&(\w+)");
String template(String input, Map<String, String> values) {
return input.replaceAllMapped(_re, (m) {
var match = m[0]!;
if (match == r"\\") return r"\";
var replacement = values[m[1]!];
if (replacement != null) return replacement;
// do nothing for undefined words.
return match;
});
}
(You might also want to allow something like &{foo} if parameters can occur next to other characters, like &{amount)USD).
To keep the character before ¶m when it matches a non-backslash character you need to use so called capturing groups. These are are subexpressions of a regular expression inside parentheses. To use capturing groups in Dard you need to use the method replaceAllMapped. We also have the case when the template starts with ¶m and in this case we match at the beginning of the string instead.
Try this:
void main() {
final template = 'My name is ¶m and my parameter is called \\¶m.';
final populatedTemplate = template.replaceAllMapped(RegExp(r'(^|[^\\])¶m\b'), (match) {
return '${match.group(1)}John';
});
final result = populatedTemplate.replaceAll(RegExp(r'\\¶m\b'), 'John');
print(result);
}
Related
i am new to scala and hate regex :D
cuurently i am debuggig a piece of code
def validateReslutions(reslutions: String): Unit = {
val regex = "(\\d+-\\d+[d,w,m,h,y],?)*"
if (!reslutions.matches(regex)) {
throw new Error("no match")
} else {
print("matched")
}
}
validateReslutions(reslutions = "(20-1w,100-1w)")
}
the problem is it produces no match for this input , so how to correct the regex to match this input
Your (20-1w,100-1w) string contains a pair of parentheses at the start and end, and the rest matches with your (\d+-\d+[d,w,m,h,y],?)* regex. Since String#matches requires a full string match, you get an exception.
Include the parentheses patterns to the regex to avoid the exception:
def validateReslutions(reslutions: String): Unit = {
val regex = """\((\d+-\d+[dwmhy],?)*\)"""
if (!reslutions.matches(regex)) {
throw new Error("no match")
} else {
print("matched")
}
}
validateReslutions(reslutions = "(20-1w,100-1w)")
// => matched
See the Scala demo.
Note the triple quotes used to define the string literal, inside which you can use single backslashes to define literal backslash chars.
Also, mind the absence of commas in the character class, they match literal commas in the text, they do not mean "or" inside character classes.
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 make a lot of changes on a original csv string. there is a lot of comma delimiter. I have to replace by a ";" either only the commas inside the expression || ....|| or only the commas outside this expression. i need to do this change in order to have different delimiter in the expression ||....|| compare to the rest of the string.
Example:
(.*)(?:\|\|)(?:.*)(,)(?:.*)\|\|
After I use
var regex = /myregex/g;
var str = str.replace(regex, ',')
thanks
You can use
const string = "aba,bjlj,alj,ljlj||name1,name2,name3||jflkj,glfgjlf,jflg,fjlfd||name1,name2||fd,sdfsfd,dfs||name1,name2,name3,name4,name5||";
console.log( string.replace(/\|{2}[\w\W]*?\|{2}/g, (x) => x.replace(/,/g, ';')) );
The regex is
/\|{2}.*?\|{2}/gs // matches any text between two double pipes
/\|{2}[\w\W]*?\|{2}/g // matches any text between two double pipes
/\|{2}.*?\|{2}/g // matches any text but line breaks between two double pipes
Note the . does not match line breaks without the s modifier flag.
The regex matches double pipe, then any zero or more chars, as few as possible up to the next double pipe.
Then, x, the whole match value, is passed as an argument to the anonymous callback function used as a replacement argument, and all commas are replaced with ; only inside the matches.
The "contrary" solution is to match and capture the strings between double pipes and only match commas in all other contexts so that you could keep the captures and replace those commas:
const string = "aba,bjlj,alj,ljlj||name1,name2,name3||jflkj,glfgjlf,jflg,fjlfd||name1,name2||fd,sdfsfd,dfs||name1,name2,name3,name4,name5||";
console.log( string.replace(/(\|{2}[\w\W]*?\|{2})|,/g, (x,y) => y || ';') );
Big Thanks.
I also find
var newStr = str.replace(/\|{2}.*?\|{2}/g, function(match) {
return match.replace(/,/g,";");
});
Do you think is it possible to do the contrary and change all the comma outside the occurence ||...|| ?
I tried looking for an answer to this question but just couldn't finding anything and I hope that there's an easy solution for this. I have and using the following code in C#,
String pattern = ("(hello|hello world)");
Regex regex = new Regex(pattern, RegexOptions.IgnoreCase);
var matches = regex.Matches("hello world");
Question is, is there a way for the matches method to return the longest pattern first? In this case, I want to get "hello world" as my match as opposed to just "hello". This is just an example but my pattern list consist of decent amount of words in it.
If you already know the lengths of the words beforehand, then put the longest first. For example:
String pattern = ("(hello world|hello)");
The longest will be matched first. If you don't know the lengths beforehand, this isn't possible.
An alternative approach would be to store all the matches in an array/hash/list and pick the longest one manually, using the language's built-in functions.
Regular expressions (will try) to match patterns from left to right. If you want to make sure you get the longest possible match first, you'll need to change the order of your patterns. The leftmost pattern is tried first. If a match is found against that pattern, the regular expression engine will attempt to match the rest of the pattern against the rest of the string; the next pattern will be tried only if no match can be found.
String pattern = ("(hello world|hello wor|hello)");
Make two different regex matches. The first will match your longer option, and if that does not work, the second will match your shorter option.
string input = "hello world";
string patternFull = "hello world";
Regex regexFull = new Regex(patternFull, RegexOptions.IgnoreCase);
var matches = regexFull.Matches(input);
if (matches.Count == 0)
{
string patternShort = "hello";
Regex regexShort = new Regex(patternShort, RegexOptions.IgnoreCase);
matches = regexShort.Matches(input);
}
At the end, matches will be be the output of "full" or "short", but "full" will be checked first and will short-circuit if it is true.
You can wrap the logic in a function if you plan on calling it many times. This is something I came up with (but there are plenty of other ways you can do this).
public bool HasRegexMatchInOrder(string input, params string[] patterns)
{
foreach (var pattern in patterns)
{
Regex regex = new Regex(pattern, RegexOptions.IgnoreCase);
if (regex.IsMatch(input))
{
return true;
}
}
return false;
}
string input = "hello world";
bool hasAMatch = HasRegexMatchInOrder(input, "hello world", "hello", ...);
What is the regular expression (in JavaScript if it matters) to only match if the text is an exact match? That is, there should be no extra characters at other end of the string.
For example, if I'm trying to match for abc, then 1abc1, 1abc, and abc1 would not match.
Use the start and end delimiters: ^abc$
It depends. You could
string.match(/^abc$/)
But that would not match the following string: 'the first 3 letters of the alphabet are abc. not abc123'
I think you would want to use \b (word boundaries):
var str = 'the first 3 letters of the alphabet are abc. not abc123';
var pat = /\b(abc)\b/g;
console.log(str.match(pat));
Live example: http://jsfiddle.net/uu5VJ/
If the former solution works for you, I would advise against using it.
That means you may have something like the following:
var strs = ['abc', 'abc1', 'abc2']
for (var i = 0; i < strs.length; i++) {
if (strs[i] == 'abc') {
//do something
}
else {
//do something else
}
}
While you could use
if (str[i].match(/^abc$/g)) {
//do something
}
It would be considerably more resource-intensive. For me, a general rule of thumb is for a simple string comparison use a conditional expression, for a more dynamic pattern use a regular expression.
More on JavaScript regexes: https://developer.mozilla.org/en/JavaScript/Guide/Regular_Expressions
"^" For the begining of the line "$" for the end of it. Eg.:
var re = /^abc$/;
Would match "abc" but not "1abc" or "abc1". You can learn more at https://developer.mozilla.org/en-US/docs/Web/JavaScript/Guide/Regular_Expressions