I want to find in string a specific string surrounded by white spaces. For example I want receive the value true from:
Regex.IsMatch("I like ZaleK", "zalek",RegexOptions.IgnoreCase)
and value false from:
Regex.IsMatch("I likeZaleK", "zalek",RegexOptions.IgnoreCase)
Here is my code:
Regex.IsMatch(w_all_file, #"\b" + TB_string.Text.Trim() + #"\b", RegexOptions.IgnoreCase) ;
It does not work when in the w_all_file is string I am looking for followed by "-"
For example: if w_all_file = "I like zalek_" - the string "zalek" is not found, but if
w_all_file = "I like zalek-" - the string "zalek" is found
Any ideas why?
Thanks,
Zalek
The \b character in regex doesn't consider an underscore as word boundry. You might want to change it to something like this:
Regex.IsMatch(w_all_file, #"[\b_]" + TB_string.Text.Trim() + #"[\b_]", RegexOptions.IgnoreCase) ;
That's what you need?
string input = "type your name";
string pattern = "your";
Regex.IsMatch(input, " " + pattern + " ");
\b matches at a word boundary, which are defined as between a character that is included in \w and one that is not. \w is the same as [a-zA-Z0-9_], so it matches underscores.
So basically, \b will match after the "k" in zalek- but not in zalek_.
It sounds like you want the match to also fail on zalek-, which you can do by using lookaround. Just replace the \b at the beginning with (?<![\w-]), and replace the \b at the end with (?![\w-]):
Regex.IsMatch(w_all_file, #"(?<![\w-])" + TB_string.Text.Trim() + #"(?![\w-])", RegexOptions.IgnoreCase) ;
Note that if you add additional characters to the character class [\w-], you need to make sure that the "-" is the very last character, or that you escape it with a backslash (if you don't it will be interpreted as a range of characters).
Related
I'm trying to isolate a " character when (simultaneously):
it's not in the beginning of the line
it's not followed by the character ";"
it's not preceded by the character ";"
E.g.:
Line: "Best Before - NO MATCH
Line: Best Before"; - NO MATCH
Line: ;"Best "Before - NO MATCH
Line: Best "Before - MATCH
My best solution is (?<![;])([^^])(")(?![;]) but it's not working correctly.
I also tried (?<![;])(")(?![;]), but it's only partial (missing the "not at the beginning" part)
I don't understand why I'm spelling the "AND not at the beginning" wrong.
Where am I missing it?
If you want to allow partial matches, you can extend the lookbehind with an alternation not asserting the start of the string to the left.
The semi colon [;] does not have to be between square brackets.
(?<!;|^)"(?!;)
Regex demo
if you want to match the " when there is no occurrence of '" to the left and right, and a infinite quantifier in a lookbehind assertion is allowed:
(?<!^.*;(?=").*|^)"(?!;|.*;")
Regex demo
In notepad++ you can use
^.*(?:;"|";).*$(*SKIP)(*F)|(?<!^)"
Regex demo
You can use the fact that not preceded by ; means that it's also not the first character on the line to simplify things
[^;]"(?:[^;]|$)
This gives you
Match a character that's not a ; (so there must be a character and thus the next character can't be the start of the line)
Match a "
Match a character that's not a ; or the end of the line
I know you are asking for a regex solution, but, almost always, strings can also be filtered using string methods in whatever language you are working in.
For the sake of completeness, to show that regex is not your only available tool here, here is a short javascript using the string methods:
myString.charAt()
myString.includes()
Working Example:
const checkLine = (line) => {
switch (true) {
// DOUBLE QUOTES AT THE BEGINNING
case(line.charAt(0) === '"') :
return console.log(line, '// NO MATCH');
// DOUBLE QUOTES IMMEDIATELY FOLLOWED BY SEMI-COLON
case(line.includes('";')) :
return console.log(line, '// NO MATCH');
// DOUBLE QUOTES IMMEDIATELY PRECEDED BY SEMI-COLON
case(line.includes(';"')) :
return console.log(line, '// NO MATCH');
default:
return console.log(line, '// MATCH');
}
}
checkLine('"Best Before');
checkLine('Best Before";');
checkLine(';"Best "Before');
checkLine('Best "Before');
Further Reading:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/charAt
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/includes
I am trying to find substrings that are not surrounded by other a-zA-Z0-9 symbols.
For example: I want to find substring hello, so it won't match hello1 or hellow but will match Hello and heLLo!##$%.
And I have such sample below.
std::string s = "1mySymbol1, /_mySymbol_ mysymbol";
const std::string sub = "mysymbol";
std::regex rgx("[^a-zA-Z0-9]*" + sub + "[^a-zA-Z0-9]*", std::regex::icase);
std::smatch match;
while (std::regex_search(s, match, rgx)) {
std::cout << match.size() << "match: " << match[0] << '\n';
s = match.suffix();
}
The result is:
1match: mySymbol
1match: , /_mySymbol_
1match: mysymbol
But I don't understand why first occurance 1mySymbol1 also matches my regex?
How to create a proper regex that will ignore such strings?
UDP
If I do like this
std::string s = "mySymbol, /_mySymbol_ mysymbol";
const std::string sub = "mysymbol";
std::regex rgx("[^a-zA-Z0-9]+" + sub + "[^a-zA-Z0-9]+", std::regex::icase);
then I find only substring in the middle
1match: , /_mySymbol_
And don't find substrings at the beggining and at the end.
The regex [^a-zA-Z0-9]* will match 0 or more characters, so it's perfectly valid for [^a-zA-Z0-9]*mysymbol[^a-zA-Z0-9]* to match mysymbol in 1mySymbol1 (allowing for case insensitivity). As you saw, this is fixed when you use [^a-zA-Z0-9]+ (matching 1 or more characters) instead.
With your update, you see that this doesn't match strings at the beginning or end. That's because [^a-zA-Z0-9]+ has to match 1 or more characters (which don't exist at the beginning or end of the string).
You have a few options:
Use beginning/end anchors: (?:[^a-zA-Z0-9]+|^)mysymbol(?:[^a-zA-Z0-9]+|$) (non-alphanumeric OR beginning of string, followed by mysymbol, followed by non-alphanumeric OR end of string).
Use negative lookahead and negative lookbehind: (?<![a-zA-Z0-9])mysymbol(?![a-zA-Z0-9]) (match mysymbol which doesn't have an alphanumeric character before or after it). Note that using this the match won't include the characters before/after mysymbol.
I recommend using https://regex101.com/ to play around with regular expressions. It lists all the different constructs you can use.
I am trying to replace text in string using regex. I accomplished it in c# using the same pattern but in swift its not working as per needed.
Here is my code:
var pattern = "\\d(\\()*[x]"
let oldString = "2x + 3 + x2 +2(x)"
let newString = oldString.stringByReplacingOccurrencesOfString(pattern, withString:"*" as String, options:NSStringCompareOptions.RegularExpressionSearch, range:nil)
print(newString)
What I want after replacement is :
"2*x + 3 +x2 + 2*(x)"
What I am getting is :
"* + 3 + x2 +*)"
Try this:
(?<=\d)(?=x)|(?<=\d)(?=\()
This pattern matches not any characters in the given string, but zero width positions in between characters.
For example, (?<=\d)(?=x) This matches a position in between a digit and 'x'
(?<= is look behind assertion (?= is look ahead.
(?<=\d)(?=\() This matches the position between a digit and '('
So the pattern before escaping:
(?<=\d)(?=x)|(?<=\d)(?=\()
Pattern, after escaping the parentheses and '\'
\(?<=\\d\)\(?=x\)|\(?<=\\d\)\(?=\\\(\)
Test string:
best.string_a = true;
best.string_b + bad.string_c;
best.string_d ();
best.string_e );
I want to catch string that after '.' and followed by anything except '('. My expression:
\.\#<=[_a-z]\+\(\s*[^(]\)\#=
I want :
string_a
string_b
string_c
string_e
But it doesn't work and result :
string_a
string_b
string_c
string_d
string_e
I am new to vim regex and i dont know why :(
Make this \.\#<=\<[_a-z]\+\>\(\s*(\)\#!
This matches:
\.\#<= Assure a dot is in front of the match followed by
\<[_a-z]\+\> A word containing only lowercase or '_' chars
\(\s*(\)\#! not followed by (any amount of spaces in front of a '(')
this would work for your needs too:
\.\zs[_a-z]\+\>\ze\s*[^( ]
I am looking for a perl regex which will validate a string containing only the letters ACGT. For example "AACGGGTTA" should be valid while "AAYYGGTTA" should be invalid, since the second string has "YY" which is not one of A,C,G,T letters. I have the following code, but it validates both the above strings
if($userinput =~/[A|C|G|T]/i)
{
$validEntry = 1;
print "Valid\n";
}
Thanks
Use a character class, and make sure you check the whole string by using the start of string token, \A, and end of string token, \z.
You should also use * or + to indicate how many characters you want to match -- * means "zero or more" and + means "one or more."
Thus, the regex below is saying "between the start and the end of the (case insensitive) string, there should be one or more of the following characters only: a, c, g, t"
if($userinput =~ /\A[acgt]+\z/i)
{
$validEntry = 1;
print "Valid\n";
}
Using the character-counting tr operator:
if( $userinput !~ tr/ACGT//c )
{
$validEntry = 1;
print "Valid\n";
}
tr/characterset// counts how many characters in the string are in characterset; with the /c flag, it counts how many are not in the characterset. Using !~ instead of =~ negates the result, so it will be true if there are no characters not in characterset or false if there are characters not in characterset.
Your character class [A|C|G|T] contains |. | does not stand for alternation in a character class, it only stands for itself. Therefore, the character class would include the | character, which is not what you want.
Your pattern is not anchored. The pattern /[ACGT]+/ would match any string that contains one or more of any of those characters. Instead, you need to anchor your pattern, so that only strings that contain just those characters from beginning to end are matched.
$ can match a newline. To avoid that, use \z to anchor at the end. \A anchors at the beginning (although it doesn't make a difference whether you use that or ^ in this case, using \A provides a nice symmetry.
So, you check should be written:
if ($userinput =~ /\A [ACGT]+ \z/ix)
{
$validEntry = 1;
print "Valid\n";
}