I want to write a regular expression, in JavaScript, for finding the string starting and ending with :.
For example "hello :smile: :sleeping:" from this string I need to find the strings which are starting and ending with the : characters. I tried the expression below, but it didn't work:
^:.*\:$
My guess is that you not only want to find the string, but also replace it. For that you should look at using a capture in the regexp combined with a replacement function.
const emojiPattern = /:(\w+):/g
function replaceEmojiTags(text) {
return text.replace(emojiPattern, function (tag, emotion) {
// The emotion will be the captured word between your tags,
// so either "sleep" or "sleeping" in your example
//
// In this function you would take that emotion and return
// whatever you want based on the input parameter and the
// whole tag would be replaced
//
// As an example, let's say you had a bunch of GIF images
// for the different emotions:
return '<img src="/img/emoji/' + emotion + '.gif" />';
});
}
With that code you could then run your function on any input string and replace the tags to get the HTML for the actual images in them. As in your example:
replaceEmojiTags('hello :smile: :sleeping:')
// 'hello <img src="/img/emoji/smile.gif" /> <img src="/img/emoji/sleeping.gif" />'
EDIT: To support hyphens within the emotion, as in "big-smile", the pattern needs to be changed since it is only looking for word characters. For this there is probably also a restriction such that the hyphen must join two words so that it shouldn't accept "-big-smile" or "big-smile-". For that you need to change the pattern to:
const emojiPattern = /:(\w+(-\w+)*):/g
That pattern is looking for any word that is then followed by zero or more instances of a hyphen followed by a word. It would match any of the following: "smile", "big-smile", "big-smile-bigger".
The ^ and $ are anchors (start and end respectively). These cause your regex to explicitly match an entire string which starts with : has anything between it and ends with :.
If you want to match characters within a string you can remove the anchors.
Your * indicates zero or more so you'll be matching :: as well. It'll be better to change this to + which means one or more. In fact if you're just looking for text you may want to use a range [a-z0-9] with a case insensitive modifier.
If we put it all together we'll have regex like this /:([a-z0-9]+):/gmi
match a string beginning with : with any alphanumeric character one or more times ending in : with the modifiers g globally, m multi-line and i case insensitive for things like :FacePalm:.
Using it in JavaScript we can end up with:
var mytext = 'Hello :smile: and jolly :wave:';
var matches = mytext.match(/:([a-z0-9]+):/gmi);
// matches = [':smile:', ':wave:'];
You'll have an array with each match found.
Related
I need to parse the tokens from a GS1 UDI format string:
"(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888"
I would like to split that string with a regex on the "(nnn)" and have the delimiter included with the split values, like this:
[ "(20)987111", "(240)A", "(10)ABC123", "(17)2022-04-01", "(21)888888888888888" ]
Below is a JSFiddle with examples, but in case you want to see it right here:
// This includes the delimiter match in the results, but I want the delimiter included WITH the value
// after it, e.g.: ["(20)987111", ...]
str = "(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888";
console.log(str.split(/(\(\d{2,}\))/).filter(Boolean))
// Result: ["(20)", "987111", "(240)", "A", "(10)", "ABC123", "(17)", "2022-04-01", "(21)", "888888888888888"]
// If I include a pattern that should (I think) match the content following the delimiter I will
// only get a single result that is the full string:
str = "(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888";
console.log(str.split(/(\(\d{2,}\)\W+)/).filter(Boolean))
// Result: ["(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888"]
// I think this is because I'm effectively mathching the entire string, hence a single result.
// So now I'll try to match only up to the start of the next "(":
str = "(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888";
console.log(str.split(/(\(\d{2,}\)(^\())/).filter(Boolean))
// Result: ["(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888"]
I've found and read this question, however the examples there are matching literals and I'm using character classes and getting different results.
I'm failing to create a regex pattern that will provide what I'm after. Here's a JSFiddle of some of the things I've tried: https://jsfiddle.net/6bogpqLy/
I can't guarantee the order of the "application identifiers" in the input string and as such, match with named captures isn't an attractive option.
You can split on positions where parenthesised element follows, by using a zero-length lookahead assertion:
const text = "(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888"
const parts = text.split(/(?=\(\d+\))/)
console.log(parts)
Instead of split use match to create the array. Then find 1) digits in parenthesis, followed by a group that might contain a digit, a letter, or a hyphen, and then 2) group that whole query.
(PS. I often find a site like Regex101 really helps when it comes to testing out expressions outside of a development environment.)
const re = /(\(\d+\)[\d\-A-Z]+)/g;
const str = '(20)987111(240)A(10)ABC123(17)2022-04-01(21)888888888888888';
console.log(str.match(re));
Is it possible for a regex character set ([abc]) to have characters with special meanings (like $ for end of the line)? For example, is it possible to make a characters set that matches either a / or the start of the line ^?
I tried /[^\/]/g but it just checks for a literal ^ or a /.
Note: I'm using JavaScript.
No you can't create custom special meaning, but you can be clever about the way you use the regex.
When doing complicated regex I test if any of these special characters exist in the string I'm testing -
const str = '¡¢my string <>some content <>between</> fragment</> and other data...'
const markers = ['\u00A1','\u00A2','\u00A4']; // ['¡','¢','¤'] you can add others
const safeMarker = markers.find(marker => str.indexOf(marker) === -1) // ¤ - this is not in the string so I can use it as a marker
if (safeMarker) {
const replaced = str.replace(/<\/>/g, safeMarker); // output = 'my string <>some content <>between¤ fragment¤ and other data...'
// do something with this regex and so on...
// then replace the string back
}
The beauty of this is that you can convert any combination of characters into a single marker, which means you can use it in your Negation expression like this:
/<>[^\u00A4]*\u00A4/g
Which would have been equivalent to (ie get the content between the tags)
<>[^</>]*</>
I have a string like the following:
SOME TEXT (BI1) SOME MORE TEXT (BI17) SOME FINAL TEXT (BI1234)
Question
I am trying to make a regex to get just the information between the curly brackets, for example the end string would look like:
BI1 BI17 BI1234
I have found this example on stackoverflow which will get the first value BI1, but will ignore the rest after.
Get text between two rounded brackets
this is the REGEX I created from the above link: /\(([^)]+)\)/g but it includes the brackets, I want to remove these.
I am using this website to attempt to solve this query which has a testing window to see if the regex entered works:
http://www.regexr.com
Additional Information
there can be any amount of numbers also, which is why I have given 3 different examples.
this is a continous string, not on seperate lines
thanks for any help on this matter.
While this isn't possible using just regexes, you can do it with string#split and the following regex:
\).*?\(|^.*?\(|\).*?$
Yielding code that looks a bit like this:
function getBracketed(str) {
return str.split(/\).*?\(|^.*?\(|\).*?$/).filter(Boolean);
}
(You need to filter out the empty strings that'll appear at the beginning and end if you do it this way - hence the extra operation).
Regex demo on Regex101
Code demo on Repl.it
If you need to keep all inside parentheses and remove everything else, you might use
var str = "SOME TEXT (BI1) SOME MORE TEXT (BI17) SOME FINAL TEXT (BI1234)";
var result = str.replace(/.*?\(([^()]*)\)/g, " $1").trim();
console.log(result);
If you need to get only the BI+digits pattern inside parentheses, use
/.*?\((BI\d+)\)/g
Details:
.*? - match any 0+ chars other than linebreak symbols
\( - match a (
(BI\d+) - Group 1 capturing BI + 1 or more digits (\d+) (or [^()]* - zero or more chars other than ( and ))
\) - a closing ).
To get all the values as array (say, for later joining), use
var str = "SOME TEXT (BI1) SOME MORE TEXT (BI17) SOME FINAL TEXT (BI1234)";
var re = /\((BI\d+)\)/g;
var res =str.match(re).map(function(s) {return s.substring(1, s.length-1);})
console.log(res);
console.log(res.join(" "));
I've got a question concerning regex.
I was wondering how one could replace an encapsulated text, something like {key:23} to something like <span class="highlightable">23</span, so that the entity will still remain encapsulated, but with something else.
I will do this in JS, but the regex is what is important, I have been searching for a while, probably searching for the wrong terms, I should probably learn more about regex, generally.
In any case, is there someone who knows how to perform this operation with simplicity?
Thanks!
It's important that you find {key:23} in your text first, and then replace it with your wanted syntax, this way you avoid replacing {key:'sometext'} with that syntax which is unwanted.
var str = "some random text {key:23} some random text {key:name}";
var n = str.replace(/\{key:[\d]+\}/gi, function myFunction(x){return x.replace(/\{key:/,'<span>').replace(/\}/, '</span>');});
this way only {key:AnyNumber} gets replaced, and {key:AnyThingOtherThanNumbers} don't get touched.
It seems you are new to regex. You need to learn more about character classes and capturing groups and backreferences.
The regex is somewhat basic in your case if you do not need any nested encapsulated text support.
Let's start:
The beginning is {key: - it will match the substring literally. Note that { can be a special character (denoting start of a limiting quantifier), thus, it is a good idea to escape it: {key:.
([^}]+) - This is a bit more interesting: the round brackets around are a capturing group that let us later back-reference the matched text. The [^}]+ means 1 or more characters (due to +) other than } (as [^}] is a negated character class where ^ means not)
} matches a } literally.
In the replacement string, we'll get the captured text using a backreference $1.
So, the entire regex will look like:
{key:([^}]+)}
See demo on regex101.com
Code snippet:
var re = /{key:([^}]+)}/g;
var str = '{key:23}';
var subst = '<span class="highlightable">$1</span>';
document.getElementById("res").innerHTML = str.replace(re, subst);
.highlightable
{
color: red;
}
<div id="res"/>
If you want to use a different behavior based on the value of key, then you'll need to adjust the regex to either match digits only (with \d+) or letters only (say, with [a-zA-Z] for English), or other shorthand classes, ranges (= character classes), or their combinations.
If your string is in var a, then:
var test = a.replace( /\{key:(\d+)\}/g, "<span class='highlightable'>$1</span>");
How it is better to check (test) if text contains only characters from set (for example if text contains only punctuation marks)
var regex = /[\.,-\/#!$%\^&\*;:{}=\-_`~()]/g
res = text.replace(regex, '')
if (res) return false
so I made it with replace is it possible to do it with regex.test?
Yes it is. There are two possibilities. One is, that you use anchors to assert that the full string is made up of these:
var regex = /^[\.,-\/#!$%\^&\*;:{}=\-_`~()]+$/;
if(regex.test(text))
Alternatively you can use a negated character class and see whether it matches and then again negate the result
var regex = /[^\.,-\/#!$%\^&\*;:{}=\-_`~()]/;
if(!regex.test(text))
Note that ,-\/ is a range that includes ,-./. This is redundant and may become a source of errors if the character class is ever changed. You might want to simplify your character class to:
[.,\/#!$%^&*;:{}=_`~()-]
(Or the negated version of that, depending on which approach you choose.)