How do I remove a character from a string and remove the previous character as well?
Example:
"ABCXDEXFGHXIJK"
I want to split the string by "X" and remove the previous character which returns
"ABDFGIJK" // CX, EX, HX are removed
I found this thread but it removes everything before rather than a specific amount of characters: How to remove part of a string before a ":" in javascript?
I can run a for loop but I was wondering if there was a better/simpler way to achieve this
const remove = function(str){
for(let i = 0; i < str.length; i++){
if(str[i] === "X") str = str.slice(0, i - 1) + str.slice(i + 1);
}
return str
}
console.log(remove("ABCXDEXFGHXIJK")) // ABDFGIJK
You can use String.prototype.replace and regex.
"ABCXDEXFGHXIJK".replace(/.X/g, '')
The g at the end is to replace every occurrence of .X. You can use replaceAll as well, but it has less support.
"ABCXDEXFGHXIJK".replaceAll(/.X/g, '')
If you want it to be case insensitive, use the i flag as well.
"ABCXDEXFGHXIJK".replace(/.x/gi, '')
The simplest way is to use a regular expression inside replace.
"ABCXDEXFGHXIJK".replace(/.X/g, "")
.X means "match the combination of X and any single character before it, g flag after the expression body repeats the process globally (instead of doing it once).
While not the most computationally efficient, you could use the following one-liner that may meet your definition of "a better/simpler way to achieve this":
const remove = str => str.split("X").map((ele, idx) => idx !== str.split("X").length - 1 ? ele.slice(0, ele.length - 1) : ele).join("");
console.log(remove("ABCXDEXFGHXIJK"));
Maybe you can use recursion.
function removeChar(str, char){
const index = str.indexOf(char);
if(index < 0) return str;
// removes 2 characters from string
return removeChar(str.split('').splice(index - 2, index).join());
}
Try this way (Descriptive comments are added in the below code snippet itself) :
// Input string
const str = "ABCXDEXFGHXIJK";
// split the input string based on 'X' and then remove the last item from each element by using String.slice() method.
const splittedStrArr = str.split('X').map(item => item = item.slice(0, -1));
// Output by joining the modified array elements.
console.log(splittedStr.join(''))
By using RegEx :
// Input string
const str = "ABCXDEXFGHXIJK";
// Replace the input string by matching the 'X' and one character before that with an empty string.
const modifiedStr = str.replace(/.X/g, "")
// Output
console.log(modifiedStr)
I want to change a char in string with many values
I have string like this :
date_format = "%m/%d/%Y";
And i want to replace ever % with the char which after, so the date variable should be like this:
date_format="mm/dd/YY";
Here is what I tried so far, but i can't get it to work, so I need some help here:
function replaceon(str, index, chr) {
if (index > str.length - 1) return str;
return str.substr(0, index) + chr + str.substr(index + 1);
}
function locations(substring, string) {
var a = [],
i = -1;
while ((i = string.indexOf(substring, i + 1)) >= 0) a.push(i);
return a;
}
function corrent_format(date_format) {
var my_locations = locations('%', date_format);
console.log(my_locations.length);
for (var i = 0; i < my_locations.length; i++) {
replaceon(date_format, my_locations[i], date_format[my_locations[i] + 1]);
}
return date_format;
}
console.log(corrent_format(date_format));
You can try this:
"%m/%d/%Y".replace(/%([^%])/g,"$1$1")
Hope this hepls.
You can use a regular expression for that:
var date_format="%m/%d/%Y";
var res = date_format.replace(/%(.)/g, "$1$1");
console.log(res);
function act(str) {
var res = "";
for (var i = 0; i < (str.length - 1); i++) {
if (str[i] === "%")
res += str[i + 1];
else
res += str[i];
}
res += str[i];
return res;
}
var date_format = "%m/%d/%Y";
console.log(act(date_format));
Your code is not working because the date_format variable is not being modified by the corrent_format function. The replaceon function returns a new string. If you assign the result to date_format, you should get the expected result:
for (var i = 0; i < my_locations.length; i++) {
date_format = replaceon(date_format, my_locations[i], date_format[my_locations[i]+1])
}
Alternatively, you could perform the replacement using String.replace and a regular expression:
date_format.replace(/%(.)/g, '$1$1');
For the regex-challenged among us, here's a translation of /%(.)/g, '$1$1':
/ means that the next part is going to be regex.
% find a %.
. any single character, so %. would match %m, %d, and/or %Y.
(.) putting it in parens means to capture the value to use later on.
/g get all the matches in the source string (instead of just the first one).
?1 references the value we captured before in (.).
?1?1 repeat the captured value twice.
So, replace every %. with whatever's in the ., times two.
Now, this regex expression is the most concise and quickest way to do the job at hand. But maybe you can't use regular expressions. Maybe you have a dyslexic boss who has outlawed their use. (Dyslexia and regex are uneasy companions at best.) Maybe you haven't put in the 47 hours screaming at regular expressions that aren't doing what you want, that you're required to put in before you're allowed to use them. Or maybe you just hate regular expressions.
If any of these apply, you can also do this:
var x = '%m/%d/%y';
x = x.replace('%', 'm');
x = x.replace('%', 'd');
x = x.replace('%', 'y');
alert(x);
This takes advantage of the fact that the replace function only replaces the first match found.
But seriously, don't use this. Use regex. It's always better to invest that 20 hours working out a regex expression that condenses the 20 lines of code you wrote in 15 minutes down to one. Unless you have to get it done sometime tonight, and whatever you're trying just doesn't work, and it's getting close to midnight, and you're getting tired...well, no. Use regex. Really. Resist the temptation to avoid finding a regex solution. You'll be glad you did when you wake up at your desk in the morning, having missed your deadline, and get to go home and spend more time with your family, courtesy of your generous severance package.
I have an string, but at certain points in the string it has dynamic values.
I want to find those points in the string and replace them with values from an array.
Let's say I have:
var array = ['string', 'value'];
var string = 'This is a {1}, and this is a {2}.';
Now I need to figure out a way to replace {1}, with the value of the first index in the array and replace {2} with the value of the second index of the array.
So the string would like this:
This is a string, and this is a value.
This seems so simple, but I am breaking my head on this. I can't find a way to do this easily.
You can use replace with a regular expression and a function:
string.replace(/{(\d+)}/g, function(match, n) {
return array[n-1];
});
You can also check n < 1 || n > array.length to provide a fallback value in case the number is out of bounds.
Alternatively, the new ES6 way is using tagged template strings
function tag(strings, ...values) {
var parts = [];
for(var i=0; i<strings.length; ++i)
parts.push(strings[i], array[values[i]-1]);
return parts.join('');
}
tag`This is a ${1}, and this is a ${2}.`
like this
string.replace(/\{(\d+)\}/g, function(m, n){
return array[n-1]
})
You can use Array.prototype.reduce:
var StringHelper = {
format: function(format, args) {
return args.reduce(function(result, currentReplace, currentReplaceIndex) {
result = result.replace("{" + (currentReplaceIndex + 1) + "}", args[currentReplaceIndex]);
return result;
}, format);
}
};
var sourceString = "My name is {1} and I'm {2} years old";
var replaceItems = ["Matías", "31"];
var parsedString = StringHelper.format(sourceString, replaceItems);
alert(parsedString);
It's a good alternative to regular expressions, and I'm not absolutely sure if I did a jsPerf test in the right way, but it shows that this approach outperforms the regular expression one.
I have a string, 12345.00, and I would like it to return 12345.0.
I have looked at trim, but it looks like it is only trimming whitespace and slice which I don't see how this would work. Any suggestions?
You can use the substring function:
let str = "12345.00";
str = str.substring(0, str.length - 1);
console.log(str);
This is the accepted answer, but as per the conversations below, the slice syntax is much clearer:
let str = "12345.00";
str = str.slice(0, -1);
console.log(str);
You can use slice! You just have to make sure you know how to use it. Positive #s are relative to the beginning, negative numbers are relative to the end.
js>"12345.00".slice(0,-1)
12345.0
You can use the substring method of JavaScript string objects:
s = s.substring(0, s.length - 4)
It unconditionally removes the last four characters from string s.
However, if you want to conditionally remove the last four characters, only if they are exactly _bar:
var re = /_bar$/;
s.replace(re, "");
The easiest method is to use the slice method of the string, which allows negative positions (corresponding to offsets from the end of the string):
const s = "your string";
const withoutLastFourChars = s.slice(0, -4);
If you needed something more general to remove everything after (and including) the last underscore, you could do the following (so long as s is guaranteed to contain at least one underscore):
const s = "your_string";
const withoutLastChunk = s.slice(0, s.lastIndexOf("_"));
console.log(withoutLastChunk);
For a number like your example, I would recommend doing this over substring:
console.log(parseFloat('12345.00').toFixed(1));
Do note that this will actually round the number, though, which I would imagine is desired but maybe not:
console.log(parseFloat('12345.46').toFixed(1));
Be aware that String.prototype.{ split, slice, substr, substring } operate on UTF-16 encoded strings
None of the previous answers are Unicode-aware.
Strings are encoded as UTF-16 in most modern JavaScript engines, but higher Unicode code points require surrogate pairs, so older, pre-existing string methods operate on UTF-16 code units, not Unicode code points.
See: Do NOT use .split('').
const string = "ẞ🦊";
console.log(string.slice(0, -1)); // "ẞ\ud83e"
console.log(string.substr(0, string.length - 1)); // "ẞ\ud83e"
console.log(string.substring(0, string.length - 1)); // "ẞ\ud83e"
console.log(string.replace(/.$/, "")); // "ẞ\ud83e"
console.log(string.match(/(.*).$/)[1]); // "ẞ\ud83e"
const utf16Chars = string.split("");
utf16Chars.pop();
console.log(utf16Chars.join("")); // "ẞ\ud83e"
In addition, RegExp methods, as suggested in older answers, don’t match line breaks at the end:
const string = "Hello, world!\n";
console.log(string.replace(/.$/, "").endsWith("\n")); // true
console.log(string.match(/(.*).$/) === null); // true
Use the string iterator to iterate characters
Unicode-aware code utilizes the string’s iterator; see Array.from and ... spread.
string[Symbol.iterator] can be used (e.g. instead of string) as well.
Also see How to split Unicode string to characters in JavaScript.
Examples:
const string = "ẞ🦊";
console.log(Array.from(string).slice(0, -1).join("")); // "ẞ"
console.log([
...string
].slice(0, -1).join("")); // "ẞ"
Use the s and u flags on a RegExp
The dotAll or s flag makes . match line break characters, the unicode or u flag enables certain Unicode-related features.
Note that, when using the u flag, you eliminate unnecessary identity escapes, as these are invalid in a u regex, e.g. \[ is fine, as it would start a character class without the backslash, but \: isn’t, as it’s a : with or without the backslash, so you need to remove the backslash.
Examples:
const unicodeString = "ẞ🦊",
lineBreakString = "Hello, world!\n";
console.log(lineBreakString.replace(/.$/s, "").endsWith("\n")); // false
console.log(lineBreakString.match(/(.*).$/s) === null); // false
console.log(unicodeString.replace(/.$/su, "")); // ẞ
console.log(unicodeString.match(/(.*).$/su)[1]); // ẞ
// Now `split` can be made Unicode-aware:
const unicodeCharacterArray = unicodeString.split(/(?:)/su),
lineBreakCharacterArray = lineBreakString.split(/(?:)/su);
unicodeCharacterArray.pop();
lineBreakCharacterArray.pop();
console.log(unicodeCharacterArray.join("")); // "ẞ"
console.log(lineBreakCharacterArray.join("").endsWith("\n")); // false
Note that some graphemes consist of more than one code point, e.g. 🏳️🌈 which consists of the sequence 🏳 (U+1F3F3), VS16 (U+FE0F), ZWJ (U+200D), 🌈 (U+1F308).
Here, even Array.from will split this into four “characters”.
Matching those is made easier with the RegExp set notation and properties of strings proposal.
Using JavaScript's slice function:
let string = 'foo_bar';
string = string.slice(0, -4); // Slice off last four characters here
console.log(string);
This could be used to remove '_bar' at end of a string, of any length.
A regular expression is what you are looking for:
let str = "foo_bar";
console.log(str.replace(/_bar$/, ""));
Try this:
const myString = "Hello World!";
console.log(myString.slice(0, -1));
Performance
Today 2020.05.13 I perform tests of chosen solutions on Chrome v81.0, Safari v13.1 and Firefox v76.0 on MacOs High Sierra v10.13.6.
Conclusions
the slice(0,-1)(D) is fast or fastest solution for short and long strings and it is recommended as fast cross-browser solution
solutions based on substring (C) and substr(E) are fast
solutions based on regular expressions (A,B) are slow/medium fast
solutions B, F and G are slow for long strings
solution F is slowest for short strings, G is slowest for long strings
Details
I perform two tests for solutions A, B, C, D, E(ext), F, G(my)
for 8-char short string (from OP question) - you can run it HERE
for 1M long string - you can run it HERE
Solutions are presented in below snippet
function A(str) {
return str.replace(/.$/, '');
}
function B(str) {
return str.match(/(.*).$/)[1];
}
function C(str) {
return str.substring(0, str.length - 1);
}
function D(str) {
return str.slice(0, -1);
}
function E(str) {
return str.substr(0, str.length - 1);
}
function F(str) {
let s= str.split("");
s.pop();
return s.join("");
}
function G(str) {
let s='';
for(let i=0; i<str.length-1; i++) s+=str[i];
return s;
}
// ---------
// TEST
// ---------
let log = (f)=>console.log(`${f.name}: ${f("12345.00")}`);
[A,B,C,D,E,F,G].map(f=>log(f));
This snippet only presents soutions
Here are example results for Chrome for short string
Use regex:
let aStr = "12345.00";
aStr = aStr.replace(/.$/, '');
console.log(aStr);
How about:
let myString = "12345.00";
console.log(myString.substring(0, myString.length - 1));
1. (.*), captures any character multiple times:
console.log("a string".match(/(.*).$/)[1]);
2. ., matches last character, in this case:
console.log("a string".match(/(.*).$/));
3. $, matches the end of the string:
console.log("a string".match(/(.*).{2}$/)[1]);
https://stackoverflow.com/questions/34817546/javascript-how-to-delete-last-two-characters-in-a-string
Just use trim if you don't want spaces
"11.01 °C".slice(0,-2).trim()
Here is an alternative that i don't think i've seen in the other answers, just for fun.
var strArr = "hello i'm a string".split("");
strArr.pop();
document.write(strArr.join(""));
Not as legible or simple as slice or substring but does allow you to play with the string using some nice array methods, so worth knowing.
debris = string.split("_") //explode string into array of strings indexed by "_"
debris.pop(); //pop last element off the array (which you didn't want)
result = debris.join("_"); //fuse the remainng items together like the sun
If you want to do generic rounding of floats, instead of just trimming the last character:
var float1 = 12345.00,
float2 = 12345.4567,
float3 = 12345.982;
var MoreMath = {
/**
* Rounds a value to the specified number of decimals
* #param float value The value to be rounded
* #param int nrDecimals The number of decimals to round value to
* #return float value rounded to nrDecimals decimals
*/
round: function (value, nrDecimals) {
var x = nrDecimals > 0 ? 10 * parseInt(nrDecimals, 10) : 1;
return Math.round(value * x) / x;
}
}
MoreMath.round(float1, 1) => 12345.0
MoreMath.round(float2, 1) => 12345.5
MoreMath.round(float3, 1) => 12346.0
EDIT: Seems like there exists a built in function for this, as Paolo points out. That solution is obviously much cleaner than mine. Use parseFloat followed by toFixed
if(str.substring(str.length - 4) == "_bar")
{
str = str.substring(0, str.length - 4);
}
Via slice(indexStart, indexEnd) method - note, this does NOT CHANGE the existing string, it creates a copy and changes the copy.
console.clear();
let str = "12345.00";
let a = str.slice(0, str.length -1)
console.log(a, "<= a");
console.log(str, "<= str is NOT changed");
Via Regular Expression method - note, this does NOT CHANGE the existing string, it creates a copy and changes the copy.
console.clear();
let regExp = /.$/g
let b = str.replace(regExp,"")
console.log(b, "<= b");
console.log(str, "<= str is NOT changed");
Via array.splice() method -> this only works on arrays, and it CHANGES, the existing array (so careful with this one), you'll need to convert a string to an array first, then back.
console.clear();
let str = "12345.00";
let strToArray = str.split("")
console.log(strToArray, "<= strToArray");
let spliceMethod = strToArray.splice(str.length-1, 1)
str = strToArray.join("")
console.log(str, "<= str is changed now");
In cases where you want to remove something that is close to the end of a string (in case of variable sized strings) you can combine slice() and substr().
I had a string with markup, dynamically built, with a list of anchor tags separated by comma. The string was something like:
var str = "<a>text 1,</a><a>text 2,</a><a>text 2.3,</a><a>text abc,</a>";
To remove the last comma I did the following:
str = str.slice(0, -5) + str.substr(-4);
You can, in fact, remove the last arr.length - 2 items of an array using arr.length = 2, which if the array length was 5, would remove the last 3 items.
Sadly, this does not work for strings, but we can use split() to split the string, and then join() to join the string after we've made any modifications.
var str = 'string'
String.prototype.removeLast = function(n) {
var string = this.split('')
string.length = string.length - n
return string.join('')
}
console.log(str.removeLast(3))
Try to use toFixed
const str = "12345.00";
return (+str).toFixed(1);
Try this:
<script>
var x="foo_foo_foo_bar";
for (var i=0; i<=x.length; i++) {
if (x[i]=="_" && x[i+1]=="b") {
break;
}
else {
document.write(x[i]);
}
}
</script>
You can also try the live working example on http://jsfiddle.net/informativejavascript/F7WTn/87/.
#Jason S:
You can use slice! You just have to
make sure you know how to use it.
Positive #s are relative to the
beginning, negative numbers are
relative to the end.
js>"12345.00".slice(0,-1)
12345.0
Sorry for my graphomany but post was tagged 'jquery' earlier. So, you can't use slice() inside jQuery because slice() is jQuery method for operations with DOM elements, not substrings ...
In other words answer #Jon Erickson suggest really perfect solution.
However, your method will works out of jQuery function, inside simple Javascript.
Need to say due to last discussion in comments, that jQuery is very much more often renewable extension of JS than his own parent most known ECMAScript.
Here also exist two methods:
as our:
string.substring(from,to) as plus if 'to' index nulled returns the rest of string. so:
string.substring(from) positive or negative ...
and some other - substr() - which provide range of substring and 'length' can be positive only:
string.substr(start,length)
Also some maintainers suggest that last method string.substr(start,length) do not works or work with error for MSIE.
Use substring to get everything to the left of _bar. But first you have to get the instr of _bar in the string:
str.substring(3, 7);
3 is that start and 7 is the length.
Say, I have a string
"hello is it me you're looking for"
I want to cut part of this string out and return the new string, something like
s = string.cut(0,3);
s would now be equal to:
"lo is it me you're looking for"
EDIT: It may not be from 0 to 3. It could be from 5 to 7.
s = string.cut(5,7);
would return
"hellos it me you're looking for"
You're almost there. What you want is:
http://www.w3schools.com/jsref/jsref_substr.asp
So, in your example:
Var string = "hello is it me you're looking for";
s = string.substr(3);
As only providing a start (the first arg) takes from that index to the end of the string.
Update, how about something like:
function cut(str, cutStart, cutEnd){
return str.substr(0,cutStart) + str.substr(cutEnd+1);
}
Use
substring
function
Returns a subset of a string between
one index and another, or through the
end of the string.
substring(indexA, [indexB]);
indexA
An integer between 0 and one less than the length of the string.
indexB
(optional) An integer between 0 and the length of the string.
substring extracts characters from indexA up to but not including indexB. In particular:
* If indexA equals indexB, substring returns an empty string.
* If indexB is omitted, substring extracts characters to the end
of the string.
* If either argument is less than 0 or is NaN, it is treated as if
it were 0.
* If either argument is greater than stringName.length, it is treated as
if it were stringName.length.
If indexA is larger than indexB, then the effect of substring is as if the two arguments were swapped; for example, str.substring(1, 0) == str.substring(0, 1).
Some other more modern alternatives are:
Split and join
function cutFromString(oldStr, fullStr) {
return fullStr.split(oldStr).join('');
}
cutFromString('there ', 'Hello there world!'); // "Hello world!"
Adapted from MDN example
String.replace(), which uses regex. This means it can be more flexible with case sensitivity.
function cutFromString(oldStrRegex, fullStr) {
return fullStr.replace(oldStrRegex, '');
}
cutFromString(/there /i , 'Hello THERE world!'); // "Hello world!"
s = string.cut(5,7);
I'd prefer to do it as a separate function, but if you really want to be able to call it directly on a String from the prototype:
String.prototype.cut= function(i0, i1) {
return this.substring(0, i0)+this.substring(i1);
}
string.substring() is what you want.
Just as a reference for anyone looking for similar function, I have a String.prototype.bisect implementation that splits a string 3-ways using a regex/string delimiter and returns the before,delimiter-match and after parts of the string....
/*
Splits a string 3-ways along delimiter.
Delimiter can be a regex or a string.
Returns an array with [before,delimiter,after]
*/
String.prototype.bisect = function( delimiter){
var i,m,l=1;
if(typeof delimiter == 'string') i = this.indexOf(delimiter);
if(delimiter.exec){
m = this.match(delimiter);
i = m.index;
l = m[0].length
}
if(!i) i = this.length/2;
var res=[],temp;
if(temp = this.substring(0,i)) res.push(temp);
if(temp = this.substr(i,l)) res.push(temp);
if(temp = this.substring(i+l)) res.push(temp);
if(res.length == 3) return res;
return null;
};
/* though one could achieve similar and more optimal results for above with: */
"my string to split and get the before after splitting on and once".split(/and(.+)/,2)
// outputs => ["my string to split ", " get the before after splitting on and once"]
As stated here: https://developer.mozilla.org/en/Core_JavaScript_1.5_Reference/Objects/String/split
If separator is a regular expression that contains capturing parentheses, then each time separator is matched the results (including any undefined results) of the capturing parentheses are spliced into the output array. However, not all browsers support this capability.
You need to do something like the following:
var s = "I am a string";
var sSubstring = s.substring(2); // sSubstring now equals "am a string".
You have two options about how to go about it:
http://www.quirksmode.org/js/strings.html#substring
http://www.quirksmode.org/js/strings.html#substr
Try the following:
var str="hello is it me you're looking for";
document.write(str.substring(3)+"<br />");
You can check this link
this works well
function stringCutter(str,cutCount,caretPos){
let firstPart = str.substring(0,caretPos-cutCount);
let secondPart = str.substring(caretPos,str.length);
return firstPart + secondPart;
}