am tyring to write a regex pattern that detects if a word starts with var and ends with a semicolon. I have the regex code in place but when run it matches from where the last semicolon is instead of matching each word ending with a semicolon. here is my regex pattern.
dim pattern as string = "^[var].+;$"; help please.
your regex now matching whole line which starts with 'v', 'a' or 'r' and ends with semicolon.
if you want match whole lines start with var and ends with semicolon this is the way:
"^var.+;$"
if only variable definitions inside line then:
"var.+;"
this second way will match following:
var a;
var b, c;
a = 5; var b, c = a; //comment
a = 5; //comment var ;
bold indicates match
Seems like you want something like this,
(?<!\S)var\S*;(?!\S)
(?<!\S) Asserts that the match var won't be preceded by a non-space character. \S* matches zero or more non-space characters. ;(?!\S) Matches the semicolon only if it's not followed by a non-space character.
DEMO
OR
(?<!\S)var.*?;(?!\S)
DEMO
Related
Let's start with the following input.
Input = 'blue, blueblue, b l u e'
I want to match everything that is not the string 'blue'. Note that blueblue should not match, but single characters should (even if present in match string).
From this, If I replace the matches with an empty string, it should return:
Result = 'blueblueblue'
I have tried with [^\bblue\b]+
but this matches the last four single characters 'b', 'l','u','e'
Another solution:
(?<=blue)(?:(?!blue).)+(?=blue|$)|^(?:(?!blue).)+(?=blue|$)
Regex demo
If you regex engine support the \K flag, then we can try:
/blue\K|.*?(?=blue|$)/gm
Demo
This pattern says to match:
blue match "blue"
\K but then forget that match
| OR
.*? match anything else until reaching
(?=blue|$) the next "blue" or the end of the string
Edit:
On JavaScript, we can try the following replacement:
var input = "blue, blueblue, b l u e";
var output = input.replace(/blue|.*?(?=blue|$)/g, (x) => x != "blue" ? "" : "blue");
console.log(output);
I'm trying to isolate a " character when (simultaneously):
it's not in the beginning of the line
it's not followed by the character ";"
it's not preceded by the character ";"
E.g.:
Line: "Best Before - NO MATCH
Line: Best Before"; - NO MATCH
Line: ;"Best "Before - NO MATCH
Line: Best "Before - MATCH
My best solution is (?<![;])([^^])(")(?![;]) but it's not working correctly.
I also tried (?<![;])(")(?![;]), but it's only partial (missing the "not at the beginning" part)
I don't understand why I'm spelling the "AND not at the beginning" wrong.
Where am I missing it?
If you want to allow partial matches, you can extend the lookbehind with an alternation not asserting the start of the string to the left.
The semi colon [;] does not have to be between square brackets.
(?<!;|^)"(?!;)
Regex demo
if you want to match the " when there is no occurrence of '" to the left and right, and a infinite quantifier in a lookbehind assertion is allowed:
(?<!^.*;(?=").*|^)"(?!;|.*;")
Regex demo
In notepad++ you can use
^.*(?:;"|";).*$(*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
[^;]"(?:[^;]|$)
This gives you
Match a character that's not a ; (so there must be a character and thus the next character can't be the start of the line)
Match a "
Match a character that's not a ; or the end of the line
I know you are asking for a regex solution, but, almost always, strings can also be filtered using string methods in whatever language you are working in.
For the sake of completeness, to show that regex is not your only available tool here, here is a short javascript using the string methods:
myString.charAt()
myString.includes()
Working Example:
const checkLine = (line) => {
switch (true) {
// DOUBLE QUOTES AT THE BEGINNING
case(line.charAt(0) === '"') :
return console.log(line, '// NO MATCH');
// DOUBLE QUOTES IMMEDIATELY FOLLOWED BY SEMI-COLON
case(line.includes('";')) :
return console.log(line, '// NO MATCH');
// DOUBLE QUOTES IMMEDIATELY PRECEDED BY SEMI-COLON
case(line.includes(';"')) :
return console.log(line, '// NO MATCH');
default:
return console.log(line, '// MATCH');
}
}
checkLine('"Best Before');
checkLine('Best Before";');
checkLine(';"Best "Before');
checkLine('Best "Before');
Further Reading:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/charAt
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/includes
I defined a regular expression to check if the string only contains alphabetic characters and with length 5:
use regex::Regex;
fn main() {
let re = Regex::new("[a-zA-Z]{5}").unwrap();
println!("{}", re.is_match("this-shouldn't-return-true#"));
}
The text I use contains many illegal characters and is longer than 5 characters, so why does this return true?
You have to put it inside ^...$ 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 have a regex that replaces all characters with the letter "z" that live outside curly braces. however I only want to replace characters a-z and A-Z. How can I modify the regex to do so?
in javascript:
let str = "hours, plural, =1 {hour} other {hours}";
str.replace(/[^{}](?=([^{}]*\{[^{}]*\})*[^{}]*$)/g, 'z');
this results in:
zzzzzzzzzzzzzzzzzz{hour}zzzzzzz{hours}
but should be:
zzzzz, zzzzzz, =1 {hour} zzzzz {hours}
thanks!
You can use this lookahead regex for search:
/[a-zA-Z](?![^{]*})/g
RegEx Demo
Description:
[a-zA-Z]: Match a letter a-z or A-Z
(?![^{]*}): Negative lookahead to assert that we don't have a closing } ahead of us without opening { in between.
Code:
const regex = /[a-zA-Z](?![^{]*})/g;
const str = `hours, plural, =1 {hour} other {hours}`;
const result = str.replace(regex, 'z');
console.log(result);
//=> zzzzz, zzzzzz, =1 {hour} zzzzz {hours}
I want to check if the string is not empty (having whitespaces only also counts as empty). How to compose the regular expression in actionscript?
The pattern should be something like /^\s*$/ (for a single line string); ^ and $ represent the start and end of the line and \s* means match zero or more whitespace characters. For example:
var s:String = /* ... */;
var allWhitespaceOrEmpty:RegExp = /^\s*$/;
if (allWhitespaceOrEmpty.test(s))
{
// is empty or all whitespace
}
else
{
// is non-empty with at least 1 non-whitespace char
}
Perhaps a simpler way as commenter Alexander Farber points out is to check for any character except a whitespace character, which is matched by \S in regex:
var nonWhitespaceChar:RegExp = /\S/;
if (nonWhitespaceChar.test(s))
{
// is non-empty with at least 1 non-whitespace char
}