help me please to find all letters in string without repeating using regex JS.
Examples:
let str = "abczacg";
str = str.match(/ pattern /); // return has to be: abczg
str = "aabbccdd" //return:abcd.
str = "hello world"//return: helo wrd
Is it possible?
Thank you!
Here is one approach. We can first reverse the input string. Then, do a global regex replacement on the following pattern:
(\w)(?=.*\1)
This will strip off any character for which we can find the same character later in the string. But, as we will be running this replacement on the reversed string, this has the actual effect of removing all duplicate letters other than their first occurrence. Finally, we reverse the remaining string again to arrive at the expected output.
var input = "abczacg";
var output = input.split("").reverse().join("").replace(/(\w)(?=.*\1)/g, "");
output = output.split("").reverse().join("");
console.log(output);
An alternative without using a regex using a Set:
[
"abczacg",
"aabbccdd",
"hello world"
].forEach(s => {
console.log([...new Set(s.split(''))].join(''))
})
Related
I'm trying to use a regex in JS to remove the last part of a string. This substring starts with &&&, is followed by something not &&&, and ends with .pdf.
So, for example, the final regex should take a string like:
parent&&&child&&&grandchild.pdf
and match
parent&&&child
I'm not that great with regex's, so my best effort has been something like:
.*?(?:&&&.*\.pdf)
Which matches the whole string. Can anyone help me out?
You may use this greedy regex either in replace or in match:
var s = 'parent&&&child&&&grandchild.pdf';
// using replace
var r = s.replace(/(.*)&&&.*\.pdf$/, '$1');
console.log(r);
//=> parent&&&child
// using match
var m = s.match(/(.*)&&&.*\.pdf$/)
if (m) {
console.log(m[1]);
//=> parent&&&child
}
By using greedy pattern .* before &&& we make sure to match **last instance of &&& in input.
You want to remove the last portion, so replace it
var str = "parent&&&child&&&grandchild.pdf"
var result = str.replace(/&&&[^&]+\.pdf$/, '')
console.log(result)
I'm trying to write a script to remove the end of a string after a user inserts a number of special characters.
An example would be: Remove the end of a string from the 3rd comma, including the 3rd comma, so:
// Hi, this, sentence, has, a, lot, of commas
would become:
// Hi, this, sentence
I haven't been able to accomplish this with indexOf() because I don't know where the 3rd comma will occur in the sentence and I don't want to use split because that would create a break at every comma.
You can use split/slice/join to get the desired part of the string:
const str = "Hi, this, sentence, has, a, lot, of commas";
const parts = str.split(",");
const firstThree = parts.slice(0,3);
const result = firstThree.join(",");
console.log(result, parts, firstThree);
In a one-liner, that would be:
const result = str.split(",").slice(0,3).join(",");
Another simple option would be a regex:
const str = "Hi, this, sentence, has, a, lot, of commas";
const match = str.match(/([^,]+,){3}/)[0]; // "Hi, this, sentence,"
console.log(match.slice(0, -1));
This one uses the string variant of slice.
This is how the regex works:
In a capturing group (),
Find me at least one (+) character that is not a comma ([^,]): [^,]+
Followed by a comma ,
Now give me 3 of those groups after each other {3}
You can use below regex to get the result
const str = 'Hi, this, sentence, has, a, lot, of commas';
const m = str.match(/^(?:[^,]*,){2}([^,]*)/)[0];
console.log(m);
I would like to split a word by numbers, but at the same time keep the numbers in node.js.
For example, take this following sentence:
var a = "shuan3jia4";
What I want is:
"shuan3 jia4"
However, if you use a regexp's split() function, the numbers that are used on the function are gone, for example:
s.split(/[0-9]/)
The result is:
[ 'shuan', 'jia', '' ]
So is there any way to keep the numbers that are used on the split?
You can use match to actually split it per your requirement:
var a = "shuan3jia4";
console.log(a.match(/[a-z]+[0-9]/ig));
use parenthesis around the match you wanna keep
see further details at Javascript and regex: split string and keep the separator
var s = "shuan3jia4";
var arr = s.split(/([0-9])/);
console.log(arr);
var s = "shuan3jia4";
var arr = s.split(/(?<=[0-9])/);
console.log(arr);
This will work as per your requirements. This answer was curated from #arhak and C# split string but keep split chars / separators
As #codybartfast said, (?<=PATTERN) is positive look-behind for PATTERN. It should match at any place where the preceding text fits PATTERN so there should be a match (and a split) after each occurrence of any of the characters.
Split, map, join, trim.
const a = 'shuan3jia4';
const splitUp = a.split('').map(function(char) {
if (parseInt(char)) return `${char} `;
return char;
});
const joined = splitUp.join('').trim();
console.log(joined);
Is it possible to use regex to find all words within a sentence that contains a substring?
Example:
var sentence = "hello my number is 344undefined848 undefinedundefined undefinedcalling whistleundefined";
I need to find all words in this sentence which contains 'undefined' and remove those words.
Output should be "hello my number is ";
FYI - currently I tokenize (javascript) and iterate through all the tokens to find and remove, then merge the final string. I need to use regex. Please help.
Thanks!
You can use:
str = str.replace(/ *\b\S*?undefined\S*\b/g, '');
RegEx Demo
It certainly is possible.
Something like start of word, zero or more letters, "undefined", zero or more letters, end of word should do it.
A word boundary is \b outside a character class, so:
\b\w*?undefined\w*?\b
using non-greedy repetition to avoid the letter matching tryig to match "undefined" and leading to lots of backtracking.
Edit switch [a-zA-Z] to \w because the example includes numbers in the "words".
\S*undefined\S*
Try this simple regex.Replace by empty string.See demo.
https://www.regex101.com/r/fG5pZ8/5
you can use str.replace function like this
str = str.replace(/undefined/g, '');
Since there are enough solutions with regular expressions, here is another one - using arrays and simple function that finds occurrence of a string in a string :)
Even though the code looks more "dirty", it actually works faster than regular expression, so it might make sense to consider it when dealing with LARGE strings
var sentence = "hello my number is 344undefined848 undefinedundefined undefinedcalling whistleundefined";
var array = sentence.split(' ');
var sanitizedArray = [];
for (var i = 0; i <= array.length; i++) {
if (undefined !== array[i] && array[i].indexOf('undefined') == -1) {
sanitizedArray.push(array[i]);
}
}
var sanitizedSentence = sanitizedArray.join(' ');
alert(sanitizedSentence);
Fiddle: http://jsfiddle.net/448bbumh/
I'm trying to build a regular expression that parses a string and skips things in brackets.
Something like
string = "A bc defg hi [hi] jkl mnop.";
The .match() should return "hi" but not [hi]. I've spent 5 hours running through RE's but I'm throwing in the towel.
Also this is for javascript or jquery if that matters.
Any help is appreciated. Also I'm working on getting my questions formatted correctly : )
EDIT:
Ok I just had a eureka moment and figured out that the original RegExp I was using actually did work. But when I was replaces the matches with the [matches] it simply replaced the first match in the string... over and over. I thought this was my regex refusing to skip the brackets but after much time of trying almost all of the solutions below, I realized that I was derping Hardcore.
When .replace was working its magic it was on the first match, so I quite simply added a space to the end of the result word as follows:
var result = string.match(regex);
var modifiedResult = '[' + result[0].toString() + ']';
string.replace(result[0].toString() + ' ', modifiedResult + ' ');
This got it to stop targeting the original word in the string and stop adding a new set of brackets to it with every match. Thank you all for your help. I am going to give answer credit to the post that prodded me in the right direction.
preprocess the target string by removing everything between brackets before trying to match your RE
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*\]/, "")
then apply your RE to tmpstring
correction: made the match for brackets eager per nhahtd comment below, and also, made the RE global
string = "A bc defg hi [hi] jkl mnop."
tmpstring = string.replace(/\[.*?\]/g, "")
You don't necessarily need regex for this. Simply use string manipulation:
var arr = string.split("[");
var final = arr[0] + arr[1].split("]")[1];
If there are multiple bracketed expressions, use a loop:
while (string.indexOf("[") != -1){
var arr = string.split("[");
string = arr[0] + arr.slice(1).join("[").split("]").slice(1).join("]");
}
Using only Regular Expressions, you can use:
hi(?!])
as an example.
Look here about negative lookahead: http://www.regular-expressions.info/lookaround.html
Unfortunately, javascript does not support negative lookbehind.
I used http://regexpal.com/ to test, abcd[hi]jkhilmnop as test data, hi(?!]) as the regex to find. It matched 'hi' without matching '[hi]'. Basically it matched the 'hi' so long as there was not a following ']' character.
This of course, can be expanded if needed. This has a benefit of not requiring any pre-processing for the string.
r"\[(.*)\]"
Just play arounds with this if you wanto to use regular expressions.
What do yo uwant to do with it? If you want to selectively replace parts like "hi" except when it's "[hi]", then I often use a system where I match what I want to avoid first and then what I want to watch; if it matches what I want to avoid then I return the match, otherwise I return the processed match.
Like this:
return string.replace(/(\[\w+\])|(\w+)/g, function(all, m1, m2) {return m1 || m2.toUpperCase()});
which, with the given string, returns:
"A BC DEFG HI [hi] JKL MNOP."
Thus: it replaces every word with uppercase (m1 is empty), except if the word is between square brackets (m1 is not empty).
This builds an array of all the strings contained in [ ]:
var regex = /\[([^\]]*)\]/;
var string = "A bc defg hi [hi] [jkl] mnop.";
var results=[], result;
while(result = regex.exec(string))
results.push(result[1]);
edit
To answer to the question, this regex returns the string less all is in [ ], and trim whitespaces:
"A bc defg [hi] mnop [jkl].".replace(/(\s{0,1})\[[^\]]*\](\s{0,1})/g,'$1')
Instead of skipping the match you can probably try something different - match everything but do not capture the string within square brackets (inclusive) with something like this:
var r = /(?:\[.*?[^\[\]]\])|(.)/g;
var result;
var str = [];
while((result = r.exec(s)) !== null){
if(result[1] !== undefined){ //true if [string] matched but not captured
str.push(result[1]);
}
}
console.log(str.join(''));
The last line will print parts of the string which do not match the [string] pattern. For example, when called with the input "A [bc] [defg] hi [hi] j[kl]u m[no]p." the code prints "A hi ju mp." with whitespaces intact.
You can try different things with this code e.g. replacing etc.