I need a regex that will only find matches where the entire string matches my query.
For instance if I do a search for movies with the name "Red October" I only want to match on that exact title (case insensitive) but not match titles like "The Hunt For Red October". Not quite sure I know how to do this. Anyone know?
Thanks!
Try the following regular expression:
^Red October$
By default, regular expressions are case sensitive. The ^ marks the start of the matching text and $ the end.
Generally, and with default settings, ^ and $ anchors are a good way of ensuring that a regex matches an entire string.
A few caveats, though:
If you have alternation in your regex, be sure to enclose your regex in a non-capturing group before surrounding it with ^ and $:
^foo|bar$
is of course different from
^(?:foo|bar)$
Also, ^ and $ can take on a different meaning (start/end of line instead of start/end of string) if certain options are set. In text editors that support regular expressions, this is usually the default behaviour. In some languages, especially Ruby, this behaviour cannot even be switched off.
Therefore there is another set of anchors that are guaranteed to only match at the start/end of the entire string:
\A matches at the start of the string.
\Z matches at the end of the string or before a final line break.
\z matches at the very end of the string.
But not all languages support these anchors, most notably JavaScript.
I know that this may be a little late to answer this, but maybe it will come handy for someone else.
Simplest way:
var someString = "...";
var someRegex = "...";
var match = Regex.Match(someString , someRegex );
if(match.Success && match.Value.Length == someString.Length){
//pass
} else {
//fail
}
Use the ^ and $ modifiers to denote where the regex pattern sits relative to the start and end of the string:
Regex.Match("Red October", "^Red October$"); // pass
Regex.Match("The Hunt for Red October", "^Red October$"); // fail
You need to enclose your regex in ^ (start of string) and $ (end of string):
^Red October$
If the string may contain regex metasymbols (. { } ( ) $ etc), I propose to use
^\QYourString\E$
\Q starts quoting all the characters until \E.
Otherwise the regex can be unappropriate or even invalid.
If the language uses regex as string parameter (as I see in the example), double slash should be used:
^\\QYourString\\E$
Hope this tip helps somebody.
Sorry, but that's a little unclear.
From what i read, you want to do simple string compare. You don't need regex for that.
string myTest = "Red October";
bool isMatch = (myTest.ToLower() == "Red October".ToLower());
Console.WriteLine(isMatch);
isMatch = (myTest.ToLower() == "The Hunt for Red October".ToLower());
You can do it like this Exemple if i only want to catch one time the letter minus a in a string and it can be check with myRegex.IsMatch()
^[^e][e]{1}[^e]$
Related
Good morning,
I want to find out in javascript with a query if a short string is included in a model number. The code now looks like this:
(ui.item.Partnumber).includes("C-")
But now I want to find out
if there is a number before the C
and a letter after the -.
Unfortunately I can't find anything on the web. Would that be possible with ${..} ? But also for this I found too little that it could help me.
Can someone help me please?
You can use a regular expression to test for those patterns.
The regex says:
Find a digit (\d) followed by C- followed by a letter from [A-Z].
If that's the entirety of the string you can bookend the regex with ^ (start of the string) and $ (end of the string). If you're looking to find uppercase or lowercase letters you can use [A-Za-z] instead.
const re = /\dC-[A-Z]/;
console.log(re.test('1C-3'));
console.log(re.test('1C-F'));
console.log(re.test('1C3'));
console.log(re.test('1F-3'));
console.log(re.test('6C-V'));
console.log(re.test('3C-T'));
Regex will be the best approach (see other answers)
But if you want to get number before and after do this. Use split and check the first and second array elements.
var result = ("123C-324").split("C-");
console.log(result);
I have an input string like this:
ABCDEFG[HIJKLMN]OPQRSTUVWXYZ
How can I replace each character in the string between the [] with an X (resulting in the same number of Xs as there were characters)?
For example, with the input above, I would like an output of:
ABCDEFG[XXXXXXX]OPQRSTUVWXYZ
I am using JavaScript's RegEx for this and would prefer if answers could be an implementation that does this using JavaScript's RegEx Replace function.
I am new to RegEx so please explain what you do and (if possible) link articles to where I can get further help.
Using replace() and passing the match to a function as parameter, and then Array(m.length).join("X") to generate the X's needed:
var str = "ABCDEFG[HIJKLMN]OPQRSTUVWXYZ"
str = str.replace(/\[[A-Z]*\]/g,(m)=>"["+Array(m.length-1).join("X")+"]")
console.log(str);
We could use also .* instead of [A-Z] in the regex to match any character.
About regular expressions there are thousands of resources, specifically in JavaScript, you could see Regular Expressions MDN but the best way to learn, in my opinion, is practicing, I find regex101 useful.
const str="ABCDEFG[HIJKLMN]OPQRSTUVWXYZ";
const run=str=>str.replace(/\[.*]/,(a,b,c)=>c=a.replace(/[^\[\]]/g,x=>x="X"));
console.log(run(str));
The first pattern /\[.*]/ is to select letters inside bracket [] and the second pattern /[^\[\]]/ is to replace the letters to "X"
We can observe that every individual letter you wish to match is followed by a series of zero or more non-'[' characters, until a ']' is found. This is quite simple to express in JavaScript-friendly regex:
/[A-Z](?=[^\[]*\])/g
regex101 example
(?= ) is a "positive lookahead assertion"; it peeks ahead of the current matching point, without consuming characters, to verify its contents are matched. In this case, "[^[]*]" matches exactly what I described above.
Now you can substitute each [A-Z] matched with a single 'X'.
You can use the following solution to replace a string between two square brackets:
const rxp = /\[.*?\]/g;
"ABCDEFG[HIJKLMN]OPQRSTUVWXYZ".replace(rxp, (x) => {
return x.replace(rxp, "X".repeat(x.length)-2);
});
I want to match all words which are starting with dollar sign but not slash and dollar sign.
I already try few regex.
(?:(?!\\)\$\w+)
\\(\\?\$\w+)\b
String
$10<i class="">$i01d</i>\$id
Expected result
*$10*
*$i01d*
but not this
*$id*
After find all expected matching word i want to replace this my object.
One option is to eliminate escape sequences first, and then match the cleaned-up string:
s = String.raw`$10<i class="">$i01d</i>\$id`
found = s.replace(/\\./g, '').match(/\$\w+/g)
console.log(found)
The big problem here is that you need a negative lookbehind, however, JavaScript does not support it. It's possible to emulate it crudely, but I will offer an alternative which, while not great, will work:
var input = '$10<i class="">$i01d</i>\\$id';
var regex = /\b\w+\b\$(?!\\)/g;
//sample implementation of a string reversal function. There are better implementations out there
function reverseString(string) {
return string.split("").reverse().join("");
}
var reverseInput = reverseString(input);
var matches = reverseInput
.match(regex)
.map(reverseString);
console.log(matches);
It is not elegant but it will do the job. Here is how it works:
JavaScript does support a lookahead expression ((?>)) and a negative lookahead ((?!)). Since this is the reverse of of a negative lookbehind, you can reverse the string and reverse the regex, which will match exactly what you want. Since all the matches are going to be in reverse, you need to also reverse them back to the original.
It is not elegant, as I said, since it does a lot of string manipulations but it does produce exactly what you want.
See this in action on Regex101
Regex explanation Normally, the "match x long as it's not preceded by y" will be expressed as (?<!y)x, so in your case, the regex will be
/(?<!\\)\$\b\w+\b/g
demonstration (not JavaScript)
where
(?<!\\) //do not match a preceding "\"
\$ //match literal "$"
\b //word boundary
\w+ //one or more word characters
\b //second word boundary, hence making the match a word
When the input is reversed, so do all the tokens in order to match. Furthermore, the negative lookbehind gets inverted into a negative lookahead of the form x(?!y) so the new regular expression is
/\b\w+\b\$(?!\\)/g;
This is more difficult than it appears at first blush. How like Regular Expressions!
If you have look-behind available, you can try:
/(?<!\\)\$\w+/g
This is NOT available in JS. Alternatively, you could specify a boundary that you know exists and use a capture group like:
/\s(\$\w+)/g
Unfortunately, you cannot rely on word boundaries via /b because there's no such boundary before '\'.
Also, this is a cool site for testing your regex expressions. And this explains the word boundary anchor.
If you're using a language that supports negative lookback assertions you can use something like this.
(?<!\\)\$\w+
I think this is the cleanest approach, but unfortunately it's not supported by all languages.
This is a hackier implementation that may work as well.
(?:(^\$\w+)|[^\\](\$\w+))
This matches either
A literal $ at the beginning of a line followed by multiple word characters. Or...
A literal $ this is preceded by any character except a backslash.
Here is a working example.
I am trying to find a pattern in a string that has a value that starts with ${ and ends with }. There will be a word between the curly brackets, but I won't know what word it is.
This is what I have \$\\{[a-zA-Z]\\}
${a} works, but ${aa} doesn't. It seems it's only looking for a single character.
I am unsure what I am doing wrong, or how to fix it and would appreciate any help anyone can provide.
I think this could help you
var str = "The quick brown ${fox} jumps over the lazy ${dog}";
var re = /\$\{([a-z]+)\}/gi;
var match;
while (match = re.exec(str)) {
console.log(match[1]);
}
Click Run code snippet and check your developer console for output
"fox"
"dog"
Explanation
+ means match 1 or more of the previous term — in this example, match 1 or more of [a-z]
the (...) parentheses will "capture" the match so you can actually do something with it — in my example, I'm just using console.log to output it
the i modifier (at the end of the regexp) means perform a case-insensitive match
the g modifier means match all instances of this regexp in the target string
The while loop will continue running for each match that re.exec finds. Once re.exec cannot match another instance, it will return null and the loop will exit.
Additional information
Try console.log(match) using the code above. Each match comes with other useful information such as the string index where the match occurred
Gotchas
This will not work for nested ${} sets
For example, this regexp will not work on "The quick brown ${fox jumps ${over}} the lazy ${dog}."
You're close!
All you need is to use a + to tell the expression that there will be one or more of whatever was just before it (in this case [a-zA-Z]) like this:
\${[a-zA-Z]+}
A good website for regex reference and testing is http://rubular.com/
It looks like you need to add a +, which tells the regex to look for one or more of a character.
Try: \${[a-zA-Z]+}
You need to use * (zero or more) or + (one or more). So this [a-zA-Z] would be [a-zA-Z]+, meaning 1 or more letters. The entire regex would look like:
\$\{[a-zA-Z]+\}
I have a string and I want to validate that string so that it must not contain certain characters like '/' '\' '&' ';' etc... How can I validate all that at once?
You can solve this with regular expressions!
mystring = "hello"
yourstring = "bad & string"
validRegEx = /^[^\\\/&]*$/
alert(mystring.match(validRegEx))
alert(yourstring.match(validRegEx))
matching against the regex returns the string if it is ok, or null if its invalid!
Explanation:
JavaScript RegEx Literals are delimited like strings, but with slashes (/'s) instead of quotes ("'s).
The first and last characters of the validRegEx cause it to match against the whole string, instead of just part, the carat anchors it to the beginning, and the dollar sign to the end.
The part between the brackets ([ and ]) are a character class, which matches any character so long as it's in the class. The first character inside that, a carat, means that the class is negated, to match the characters not mentioned in the character class. If it had been omited, the class would match the characters it specifies.
The next two sequences, \\ and \/ are backslash escaped because the backslash by itself would be an escape sequence for something else, and the forward slash would confuse the parser into thinking that it had reached the end of the regex, (exactly similar to escaping quotes in strings).
The ampersand (&) has no special meaning and is unescaped.
The remaining character, the kleene star, (*) means that whatever preceeded it should be matched zero or more times, so that the character class will eat as many characters that are not forward or backward slashes or ampersands, including none if it cant find any. If you wanted to make sure the matched string was non-empty, you can replace it with a plus (+).
I would use regular expressions.
See this guide from Mozillla.org. This article does also give a good introduction to regular expressions in JavaScript.
Here is a good article on Javascript validation. Remember you will need to validate on the server side too. Javascript validation can easily be circumvented, so it should never be used for security reasons such as preventing SQL Injection or XSS attacks.
You could learn regular expressions, or (probably simpler if you only check for one character at a time) you could have a list of characters and then some kind of sanitize function to remove each one from the string.
var myString = "An /invalid &string;";
var charList = ['/', '\\', '&', ';']; // etc...
function sanitize(input, list) {
for (char in list) {
input = input.replace(char, '');
}
return input
}
So then:
sanitize(myString, charList) // returns "An invalid string"
You can use the test method, with regular expressions:
function validString(input){
return !(/[\\/&;]/.test(input));
}
validString('test;') //false
You can use regex. For example if your string matches:
[\\/&;]+
then it is not valid. Look at:
http://www.regular-expressions.info/javascriptexample.html
You could probably use a regular expression.
As the others have answered you can solve this with regexp but remember to also check the value server-side. There is no guarantee that the user has JavaScript activated. Never trust user input!