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I'm trying to figure out how to remove every second character (starting from the first one) from a string in Javascript.
For example, the string "This is a test!" should become "hsi etTi sats!"
I also want to save every deleted character into another array.
I have tried using replace method and splice method, but wasn't able to get them to work properly. Mostly because replace only replaces the first character.
function encrypt(text, n) {
if (text === "NULL") return n;
if (n <= 0) return text;
var encArr = [];
var newString = text.split("");
var j = 0;
for (var i = 0; i < text.length; i += 2) {
encArr[j++] = text[i];
newString.splice(i, 1); // this line doesn't work properly
}
}
You could reduce the characters of the string and group them to separate arrays using the % operator. Use destructuring to get the 2D array returned to separate variables
let str = "This is a test!";
const [even, odd] = [...str].reduce((r,char,i) => (r[i%2].push(char), r), [[],[]])
console.log(odd.join(''))
console.log(even.join(''))
Using a for loop:
let str = "This is a test!",
odd = [],
even = [];
for (var i = 0; i < str.length; i++) {
i % 2 === 0
? even.push(str[i])
: odd.push(str[i])
}
console.log(odd.join(''))
console.log(even.join(''))
It would probably be easier to use a regular expression and .replace: capture two characters in separate capturing groups, add the first character to a string, and replace with the second character. Then, you'll have first half of the output you need in one string, and the second in another: just concatenate them together and return:
function encrypt(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
console.log(encrypt('This is a test!'));
Pretty simple with .reduce() to create the two arrays you seem to want.
function encrypt(text) {
return text.split("")
.reduce(({odd, even}, c, i) =>
i % 2 ? {odd: [...odd, c], even} : {odd, even: [...even, c]}
, {odd: [], even: []})
}
console.log(encrypt("This is a test!"));
They can be converted to strings by using .join("") if you desire.
I think you were on the right track. What you missed is replace is using either a string or RegExp.
The replace() method returns a new string with some or all matches of a pattern replaced by a replacement. The pattern can be a string or a RegExp, and the replacement can be a string or a function to be called for each match. If pattern is a string, only the first occurrence will be replaced.
Source: String.prototype.replace()
If you are replacing a value (and not a regular expression), only the first instance of the value will be replaced. To replace all occurrences of a specified value, use the global (g) modifier
Source: JavaScript String replace() Method
So my suggestion would be to continue still with replace and pass the right RegExp to the function, I guess you can figure out from this example - this removes every second occurrence for char 't':
let count = 0;
let testString = 'test test test test';
console.log('original', testString);
// global modifier in RegExp
let result = testString.replace(/t/g, function (match) {
count++;
return (count % 2 === 0) ? '' : match;
});
console.log('removed', result);
like this?
var text = "This is a test!"
var result = ""
var rest = ""
for(var i = 0; i < text.length; i++){
if( (i%2) != 0 ){
result += text[i]
} else{
rest += text[i]
}
}
console.log(result+rest)
Maybe with split, filter and join:
const remaining = myString.split('').filter((char, i) => i % 2 !== 0).join('');
const deleted = myString.split('').filter((char, i) => i % 2 === 0).join('');
You could take an array and splice and push each second item to the end of the array.
function encrypt(string) {
var array = [...string],
i = 0,
l = array.length >> 1;
while (i <= l) array.push(array.splice(i++, 1)[0]);
return array.join('');
}
console.log(encrypt("This is a test!"));
function encrypt(text) {
text = text.split("");
var removed = []
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed.push(letter)
return false;
}
return true
}).join("")
return {
full: encrypted + removed.join(""),
encrypted: encrypted,
removed: removed
}
}
console.log(encrypt("This is a test!"))
Splice does not work, because if you remove an element from an array in for loop indexes most probably will be wrong when removing another element.
I don't know how much you care about performance, but using regex is not very efficient.
Simple test for quite a long string shows that using filter function is on average about 3 times faster, which can make quite a difference when performed on very long strings or on many, many shorts ones.
function test(func, n){
var text = "";
for(var i = 0; i < n; ++i){
text += "a";
}
var start = new Date().getTime();
func(text);
var end = new Date().getTime();
var time = (end-start) / 1000.0;
console.log(func.name, " took ", time, " seconds")
return time;
}
function encryptREGEX(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
function encrypt(text) {
text = text.split("");
var removed = "";
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed += letter;
return false;
}
return true
}).join("")
return encrypted + removed
}
var timeREGEX = test(encryptREGEX, 10000000);
var timeFilter = test(encrypt, 10000000);
console.log("Using filter is faster ", timeREGEX/timeFilter, " times")
Using actually an array for storing removed letters and then joining them is much more efficient, than using a string and concatenating letters to it.
I changed an array to string in filter solution to make it the same like in regex solution, so they are more comparable.
Other SO 'Replace string with alphabet positions' questions didn't utilize map, which is what I'm trying to learn how to use to solve this.
Problem:
Given a string, replace every letter with its position in the alphabet.
If anything in the text isn't a letter, ignore it and don't return it.
"a" = 1, "b" = 2, etc.
What I've tried is:
-looping over a new array instance and setting the index value to String.fromCharCode()
- taking input string making it lowercase
-splitting to array
-return array.map().join(' ')
function alphabetPosition(text) {
let alphabet = new Array(26);
for (let i = 0; i<26; ++i) {
let char = String.fromCharCode(97 + i);
alphabet[i] = char;
}
text = text.toLowerCase();
let arr = text.split('');
return arr.map(element => { return element = alphabet.indexOf(element+1) }).join(' ');
}
expected it to return a string of alphabet positions, but got nothing at all. What is wrong with my implementation of Array.map()?
In your map element would be a letter, "a" for example. Then you add (concat) 1 to it, which results in "a1" which is not in your alphabet. Also element = is unneccessary, returning the position is enough.
You've complicated the solution, the simplest approach would be to just find the charcode and return that.
function alphabetPosition(text) {
let str = '';
for (var i = 0; i < text.length; i++) {
str += (text[i] + (text.charCodeAt(i) - 96));
}
return str;
}
I totally understand that is a coding challenge, interview question or likewise so if you really need to use map() you should only return the result of the callback passed to map as follows :
return arr.map(x => alphabet.indexOf(x) + 1).join(' ')
However reduce() seems more appropriate in your case :
return arr.reduce((ac, cv) => ac + (alphabet.indexOf(cv) + 1) + ' ', '')
Your map() last line of the function was returning the value of
an assignment.
return arr.map(element => { return element = alphabet.indexOf(element+1) }).join(' ');
Just alphabet.indexOf(element) would have sufficed.
This will give you the result you want:
alphabetPosition = text => {
let alphabet = new Array(26);
for (let i = 0; i < 26; ++i) {
let char = String.fromCharCode(97 + i);
alphabet[i] = char;
}
return text.toLowerCase().split('').map(element =>
alphabet.indexOf(element)
).join(' ');
}
console.log(alphabetPosition("This is a string"));
Hope this helps,
Let's say I have this 3 emojis in a string: 😀🎃👪
There are not any spaces or any other character except emojis in the string.
How can I remove the last emoji in javascript?
The answer below doesn't use any special package and safely removes the last emoji
function safeEmojiBackspace(str)
{
let initialRealCount = fancyCount(str);
while(str.length > 0 && fancyCount(str) !== initialRealCount - 1)
{
str = str.substring(0,str.length - 1);
}
return str;
}
function fancyCount(str){
const joiner = "\u{200D}";
const split = str.split(joiner);
let count = 0;
for(const s of split){
//removing the variation selectors
const num = Array.from(s.split(/[\ufe00-\ufe0f]/).join("")).length;
count += num;
}
//assuming the joiners are used appropriately
return count / split.length;
}
Sample usage
let str = "something😀🎃👪";
str = safeEmojiBackspace(str);//"something😀🎃"
You can do this. It will always remove the last emoji.
function removeEmoji() {
var emoStringArray = document.getElementById('emoji').innerHTML;
var lastIndex = emoStringArray.lastIndexOf(" ");
var stripedEmoStringArray = emoStringArray.substring(0, lastIndex);
document.getElementById('emoji').innerHTML = stripedEmoStringArray;
}
<p id="emoji">
😀 🎃 👪
</p>
<button onclick="removeEmoji()">Remove</button>
I hope this is what you want.
var emoString = "😀 🎃 👪";
emoString = emoString.slice(0, -2);
However, this would work only if you have 3 emojis in total. Hence to achieve a generalised solution, you can use the underscore functions split() and javascript function join() :
var emoString = "😀 🎃 👪";
emoString = _.rest(emoString.split(' ')).join(' ')
Hope this will solve your issue.
Ok, here is how I solved it:
function deleteEmoji(emojiStr) {
let emojisArray = emojiStr.match(/([\uD800-\uDBFF][\uDC00-\uDFFF])/g);
emojisArray = emojisArray.splice(0, emojisArray.length - 1);
return emojisArray.join("");
}
let emojitext = "😀🎃👪";
console.log(deleteEmoji(emojitext));
I was actually surprised that unicode in this day an age is still not fully supported in browsers. I assume a lot of this is down to windows and it's version of UTF-16.
The OP I believe has found his own solution to the original problem, but I thought there has to be a more generic solution to surrogate pair unicode characters.
Anyway, so my solution is convert the text into a UTF-32 array, these can then be manipulated must easier, using slice etc.
After you have done what you want to the array, just convert back.
Below is an example.
Some of the code I got from -> Is it possible to convert a string containing "high" unicode chars to an array consisting of dec values derived from utf-32 ("real") codes?
and http://speakingjs.com/es5/ch24.html
function decodeUnicode(str) {
const r = [];
let i = 0;
while(i < str.length) {
let chr = str.charCodeAt(i++);
if(chr >= 0xD800 && chr <= 0xDBFF) {
var low = str.charCodeAt(i++);
r.push(0x10000 +
((chr - 0xD800) << 10) | (low - 0xDC00));
} else {
r.push(chr);
}
}
return r;
}
function toUTF16(codePoint) {
const TEN_BITS = parseInt('1111111111', 2);
if (codePoint <= 0xFFFF) { return codePoint; }
codePoint -= 0x10000;
const leadingSurrogate = 0xD800 | (codePoint >> 10);
const trailingSurrogate = 0xDC00 | (codePoint & TEN_BITS);
return String.fromCharCode(leadingSurrogate) +
String.fromCharCode(trailingSurrogate);
}
function encodeUnicode(data) {
return data.reduce((a, v) => {
a += toUTF16(v);
return a;
},"");
}
var unicode = decodeUnicode("😀🎃👪");
for (let l = 0; l < unicode.length; l ++)
console.log(encodeUnicode(
unicode.slice(0, l ? -l : unicode.length)));
console.log("pick some random ones");
let str = "";
for (let l = 0; l < 20; l ++) {
let rnd = Math.trunc(Math.random()*unicode.length);
str += encodeUnicode(unicode.slice(rnd,rnd+1));
}
console.log(str);
I have a string similar to "<p></p>". Now, I want to split this string, so I have 2 tags. If I do
var arr = "<p></p>".split("><") , I get an array that looks like
["<p", "/p>"]
Is there a simple way to keep the separator in this split? NOT a REGEX (Not a dupe) I want :
["<p>","</p>"]
Since javascript regex doesn't support look behind assertion it's not possible with String#split method. Use String#match method to get the complete string.
var arr = "<p></p>".match(/[\s\S]+?>(?=<|$)/g)
console.log(arr)
Without regex and using split you can do something like this.
var arr = "<p></p>".split('><').map(function(v, i, arr1) {
if (i != 0)
v = '<' + v;
if (i < arr1.length - 1)
v += '>';
return v;
})
// using ternary
var arr1 = "<p></p>".split('><').map(function(v, i, arr1) {
return (i != 0 ? '<' : '') + v + (i < arr1.length - 1 ? '>' : '');
})
console.log(arr);
console.log(arr1);
To do this without a regular expression, you'll need some kind of parser. Inspect every character, build up chunks and store them in an array. You may then want to process the bits, looking for tokens or doing other processing. E.g.
/* Break string into chunks of <...>, </...> and anything in between.
** #param {string} s - string to parse
** #returns {Array} chunks of string
*/
function getChunks(s) {
var parsed = [];
var limit = s.length - 1;
s.split('').reduce(function(buffer, char, i) {
var startTag = char == '<';
var endTag = char == '/';
var closeTag = char == '>';
if (startTag) {
if (buffer.length) {
parsed.push(buffer);
}
buffer = char;
} else if (endTag) {
buffer += char;
} else if (closeTag) {
parsed.push(buffer + char)
buffer = '';
} else {
buffer += char;
}
if (i == limit && buffer.length) {
parsed.push(buffer);
}
return buffer;
}, '');
return parsed;
}
['<p></p>',
'<div>More complex</div>',
'<span>broken tag</sp'
].forEach(function(s){
console.log(s + ' => [' + getChunks(s) + ']')
});
Note that this is very simple and just looks for <...> and </...> where ... can be anything.
What's the JavaScript equivalent to this C# Method:
var x = "|f|oo||";
var y = x.Trim('|'); // "f|oo"
C# trims the selected character only at the beginning and end of the string!
One line is enough:
var x = '|f|oo||';
var y = x.replace(/^\|+|\|+$/g, '');
document.write(x + '<br />' + y);
^ beginning of the string
\|+ pipe, one or more times
| or
\|+ pipe, one or more times
$ end of the string
A general solution:
function trim (s, c) {
if (c === "]") c = "\\]";
if (c === "^") c = "\\^";
if (c === "\\") c = "\\\\";
return s.replace(new RegExp(
"^[" + c + "]+|[" + c + "]+$", "g"
), "");
}
chars = ".|]\\^";
for (c of chars) {
s = c + "foo" + c + c + "oo" + c + c + c;
console.log(s, "->", trim(s, c));
}
Parameter c is expected to be a character (a string of length 1).
As mentionned in the comments, it might be useful to support multiple characters, as it's quite common to trim multiple whitespace-like characters for example. To do this, MightyPork suggests to replace the ifs with the following line of code:
c = c.replace(/[-/\\^$*+?.()|[\]{}]/g, '\\$&');
This part [-/\\^$*+?.()|[\]{}] is a set of special characters in regular expression syntax, and $& is a placeholder which stands for the matching character, meaning that the replace function escapes special characters. Try in your browser console:
> "{[hello]}".replace(/[-/\\^$*+?.()|[\]{}]/g, '\\$&')
"\{\[hello\]\}"
Update: Was curious around the performance of different solutions and so I've updated a basic benchmark here:
https://www.measurethat.net/Benchmarks/Show/12738/0/trimming-leadingtrailing-characters
Some interesting and unexpected results running under Chrome.
https://www.measurethat.net/Benchmarks/ShowResult/182877
+-----------------------------------+-----------------------+
| Test name | Executions per second |
+-----------------------------------+-----------------------+
| Index Version (Jason Larke) | 949979.7 Ops/sec |
| Substring Version (Pho3niX83) | 197548.9 Ops/sec |
| Regex Version (leaf) | 107357.2 Ops/sec |
| Boolean Filter Version (mbaer3000)| 94162.3 Ops/sec |
| Spread Version (Robin F.) | 4242.8 Ops/sec |
+-----------------------------------+-----------------------+
Please note; tests were carried out on only a single test string (with both leading and trailing characters that needed trimming). In addition, this benchmark only gives an indication of raw speed; other factors like memory usage are also important to consider.
If you're dealing with longer strings I believe this should outperform most of the other options by reducing the number of allocated strings to either zero or one:
function trim(str, ch) {
var start = 0,
end = str.length;
while(start < end && str[start] === ch)
++start;
while(end > start && str[end - 1] === ch)
--end;
return (start > 0 || end < str.length) ? str.substring(start, end) : str;
}
// Usage:
trim('|hello|world|', '|'); // => 'hello|world'
Or if you want to trim from a set of multiple characters:
function trimAny(str, chars) {
var start = 0,
end = str.length;
while(start < end && chars.indexOf(str[start]) >= 0)
++start;
while(end > start && chars.indexOf(str[end - 1]) >= 0)
--end;
return (start > 0 || end < str.length) ? str.substring(start, end) : str;
}
// Usage:
trimAny('|hello|world ', [ '|', ' ' ]); // => 'hello|world'
// because '.indexOf' is used, you could also pass a string for the 2nd parameter:
trimAny('|hello| world ', '| '); // => 'hello|world'
EDIT: For fun, trim words (rather than individual characters)
// Helper function to detect if a string contains another string
// at a specific position.
// Equivalent to using `str.indexOf(substr, pos) === pos` but *should* be more efficient on longer strings as it can exit early (needs benchmarks to back this up).
function hasSubstringAt(str, substr, pos) {
var idx = 0, len = substr.length;
for (var max = str.length; idx < len; ++idx) {
if ((pos + idx) >= max || str[pos + idx] != substr[idx])
break;
}
return idx === len;
}
function trimWord(str, word) {
var start = 0,
end = str.length,
len = word.length;
while (start < end && hasSubstringAt(str, word, start))
start += word.length;
while (end > start && hasSubstringAt(str, word, end - len))
end -= word.length
return (start > 0 || end < str.length) ? str.substring(start, end) : str;
}
// Usage:
trimWord('blahrealmessageblah', 'blah');
If I understood well, you want to remove a specific character only if it is at the beginning or at the end of the string (ex: ||fo||oo|||| should become foo||oo). You can create an ad hoc function as follows:
function trimChar(string, charToRemove) {
while(string.charAt(0)==charToRemove) {
string = string.substring(1);
}
while(string.charAt(string.length-1)==charToRemove) {
string = string.substring(0,string.length-1);
}
return string;
}
I tested this function with the code below:
var str = "|f|oo||";
$( "#original" ).html( "Original String: '" + str + "'" );
$( "#trimmed" ).html( "Trimmed: '" + trimChar(str, "|") + "'" );
You can use a regular expression such as:
var x = "|f|oo||";
var y = x.replace(/^\|+|\|+$/g, "");
alert(y); // f|oo
UPDATE:
Should you wish to generalize this into a function, you can do the following:
var escapeRegExp = function(strToEscape) {
// Escape special characters for use in a regular expression
return strToEscape.replace(/[\-\[\]\/\{\}\(\)\*\+\?\.\\\^\$\|]/g, "\\$&");
};
var trimChar = function(origString, charToTrim) {
charToTrim = escapeRegExp(charToTrim);
var regEx = new RegExp("^[" + charToTrim + "]+|[" + charToTrim + "]+$", "g");
return origString.replace(regEx, "");
};
var x = "|f|oo||";
var y = trimChar(x, "|");
alert(y); // f|oo
A regex-less version which is easy on the eye:
const trim = (str, chars) => str.split(chars).filter(Boolean).join(chars);
For use cases where we're certain that there's no repetition of the chars off the edges.
to keep this question up to date:
here is an approach i'd choose over the regex function using the ES6 spread operator.
function trimByChar(string, character) {
const first = [...string].findIndex(char => char !== character);
const last = [...string].reverse().findIndex(char => char !== character);
return string.substring(first, string.length - last);
}
Improved version after #fabian 's comment (can handle strings containing the same character only)
function trimByChar1(string, character) {
const arr = Array.from(string);
const first = arr.findIndex(char => char !== character);
const last = arr.reverse().findIndex(char => char !== character);
return (first === -1 && last === -1) ? '' : string.substring(first, string.length - last);
}
This can trim several characters at a time:
function trimChars (str, c) {
var re = new RegExp("^[" + c + "]+|[" + c + "]+$", "g");
return str.replace(re,"");
}
var x = "|f|oo||";
x = trimChars(x, '|'); // f|oo
var y = "..++|f|oo||++..";
y = trimChars(y, '|.+'); // f|oo
var z = "\\f|oo\\"; // \f|oo\
// For backslash, remember to double-escape:
z = trimChars(z, "\\\\"); // f|oo
For use in your own script and if you don't mind changing the prototype, this can be a convenient "hack":
String.prototype.trimChars = function (c) {
var re = new RegExp("^[" + c + "]+|[" + c + "]+$", "g");
return this.replace(re,"");
}
var x = "|f|oo||";
x = x.trimChars('|'); // f|oo
Since I use the trimChars function extensively in one of my scripts, I prefer this solution. But there are potential issues with modifying an object's prototype.
If you define these functions in your program, your strings will have an upgraded version of trim that can trim all given characters:
String.prototype.trimLeft = function(charlist) {
if (charlist === undefined)
charlist = "\s";
return this.replace(new RegExp("^[" + charlist + "]+"), "");
};
String.prototype.trim = function(charlist) {
return this.trimLeft(charlist).trimRight(charlist);
};
String.prototype.trimRight = function(charlist) {
if (charlist === undefined)
charlist = "\s";
return this.replace(new RegExp("[" + charlist + "]+$"), "");
};
var withChars = "/-center-/"
var withoutChars = withChars.trim("/-")
document.write(withoutChars)
Source
https://www.sitepoint.com/trimming-strings-in-javascript/
const trim = (str, char) => {
let i = 0;
let j = str.length-1;
while (str[i] === char) i++;
while (str[j] === char) j--;
return str.slice(i,j+1);
}
console.log(trim('|f|oo|', '|')); // f|oo
Non-regex solution.
Two pointers: i (beginning) & j (end).
Only move pointers if they match char and stop when they don't.
Return remaining string.
I would suggest looking at lodash and how they implemented the trim function.
See Lodash Trim for the documentation and the source to see the exact code that does the trimming.
I know this does not provide an exact answer your question, but I think it's good to set a reference to a library on such a question since others might find it useful.
This one trims all leading and trailing delimeters
const trim = (str, delimiter) => {
const pattern = `[^\\${delimiter}]`;
const start = str.search(pattern);
const stop = str.length - str.split('').reverse().join('').search(pattern);
return str.substring(start, stop);
}
const test = '||2|aaaa12bb3ccc|||||';
console.log(trim(test, '|')); // 2|aaaa12bb3ccc
I like the solution from #Pho3niX83...
Let's extend it with "word" instead of "char"...
function trimWord(_string, _word) {
var splitted = _string.split(_word);
while (splitted.length && splitted[0] === "") {
splitted.shift();
}
while (splitted.length && splitted[splitted.length - 1] === "") {
splitted.pop();
}
return splitted.join(_word);
};
The best way to resolve this task is (similar with PHP trim function):
function trim( str, charlist ) {
if ( typeof charlist == 'undefined' ) {
charlist = '\\s';
}
var pattern = '^[' + charlist + ']*(.*?)[' + charlist + ']*$';
return str.replace( new RegExp( pattern ) , '$1' )
}
document.getElementById( 'run' ).onclick = function() {
document.getElementById( 'result' ).value =
trim( document.getElementById( 'input' ).value,
document.getElementById( 'charlist' ).value);
}
<div>
<label for="input">Text to trim:</label><br>
<input id="input" type="text" placeholder="Text to trim" value="dfstextfsd"><br>
<label for="charlist">Charlist:</label><br>
<input id="charlist" type="text" placeholder="Charlist" value="dfs"><br>
<label for="result">Result:</label><br>
<input id="result" type="text" placeholder="Result" disabled><br>
<button type="button" id="run">Trim it!</button>
</div>
P.S.: why i posted my answer, when most people already done it before? Because i found "the best" mistake in all of there answers: all used the '+' meta instead of '*', 'cause trim must remove chars IF THEY ARE IN START AND/OR END, but it return original string in else case.
Another version to use regular expression.
No or(|) used and no global(g) used.
function escapeRegexp(s) {
return s.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');
}
function trimSpecific(value, find) {
const find2 = escapeRegexp(find);
return value.replace(new RegExp(`^[${find2}]*(.*?)[${find2}]*$`), '$1')
}
console.log(trimSpecific('"a"b"', '"') === 'a"b');
console.log(trimSpecific('""ab"""', '"') === 'ab');
console.log(trimSpecific('"', '"') === '');
console.log(trimSpecific('"a', '"') === 'a');
console.log(trimSpecific('a"', '"') === 'a');
console.log(trimSpecific('[a]', '[]') === 'a');
console.log(trimSpecific('{[a]}', '[{}]') === 'a');
expanding on #leaf 's answer, here's one that can take multiple characters:
var trim = function (s, t) {
var tr, sr
tr = t.split('').map(e => `\\\\${e}`).join('')
sr = s.replace(new RegExp(`^[${tr}]+|[${tr}]+$`, 'g'), '')
return sr
}
function trim(text, val) {
return text.replace(new RegExp('^'+val+'+|'+val+'+$','g'), '');
}
"|Howdy".replace(new RegExp("^\\|"),"");
(note the double escaping. \\ needed, to have an actually single slash in the string, that then leads to escaping of | in the regExp).
Only few characters need regExp-Escaping., among them the pipe operator.
const special = ':;"<>?/!`~##$%^&*()+=-_ '.split("");
const trim = (input) => {
const inTrim = (str) => {
const spStr = str.split("");
let deleteTill = 0;
let startChar = spStr[deleteTill];
while (special.some((s) => s === startChar)) {
deleteTill++;
if (deleteTill <= spStr.length) {
startChar = spStr[deleteTill];
} else {
deleteTill--;
break;
}
}
spStr.splice(0, deleteTill);
return spStr.join("");
};
input = inTrim(input);
input = inTrim(input.split("").reverse().join("")).split("").reverse().join("");
return input;
};
alert(trim('##This is what I use$%'));
String.prototype.TrimStart = function (n) {
if (this.charAt(0) == n)
return this.substr(1);
};
String.prototype.TrimEnd = function (n) {
if (this.slice(-1) == n)
return this.slice(0, -1);
};
To my knowledge, jQuery doesnt have a built in function the method your are asking about.
With javascript however, you can just use replace to change the content of your string:
x.replace(/|/i, ""));
This will replace all occurences of | with nothing.
try:
console.log(x.replace(/\|/g,''));
Try this method:
var a = "anan güzel mi?";
if (a.endsWith("?")) a = a.slice(0, -1);
document.body.innerHTML = a;