I want to replace the underscore that precedes a date in a string e.g.
thequick_brown_20210813_fox
To:
thequick_brown_red_20210813_fox
I.e. replace that underscore with _red_
This captures the date part: (20\d{2})(\d{2})(\d{2})
And to replace I assume I can just use str.replace
But not sure how I can capture the underscore that precedes it.
You can try the following RegEx with Positive Lookahead:
/_(?=\d{8})/
Where:
_ matches the character _
(?=\d{8}) - Positive Lookahead
\d - matches a digit (equivalent to [0-9])
{8} - matches the previous token exactly 8 times
var str = 'thequick_brown_20210813_fox';
var patt = /_(?=\d{8})/;
str = str.replace(patt, '_red_');
console.log(str);
This works
And as bonus I have a super robust regex for dates
const str = 'thequick_brown_20210813_fox';
const re = /_(?=\d{4}(0[1-9]|1[0-2])(0[1-9]|[12][0-9]|3[01]))/;
const newStr = str.replace(re, '_red_');
console.log(newStr);
The regex is from here
This is what I came up with using the RegEx you provided.
let str = "thequick_brown_20210813_fox";
let newStr = "";
let regex = /(20\d{2})(\d{2})(\d{2})/;
let splits = str.split(regex);
newStr += splits[0] + "red_";
for(let i=1; i<splits.length; i++){
newStr += splits[i];
}
console.log(newStr);
I would use as a matching regex:
_(?=20[0-9]{6})
_ Matches underscore if ...
(?=20[0-9]{6}) This is a lookahead assertion that the following characters must be '20' followed by 6 digits.
Note that the above only matches the underscore when followed by '20' followed by an additional 6 digits. So we simply replace the underscore with '_red_':
let s = 'thequick_brown_20210813_fox';
s = s.replace(/_(?=20[0-9]{6})/g, '_red_');
console.log(s);
How to Replace -84 in a string: my-name-is-dude-84 with '' Regex?
I means the last '-' + number
I tried :
string = 'my-name-is-dude-84';
let regex = /[^\-*][1-9]/;
let specialChar = string.replace(regex, '');
then I received is my-name-is-dude-
I expect my string will be: my-name-is-dude
You're close, but this is what you need to do (I guess)
string = 'my-name-is-dude-84';
let regex = /-\d+$/;
let specialChar = string.replace(regex, '');
document.write(specialChar);
Your [^\-*] tries to match all characters but \, - and *. Also [1-9] only matches one digit (between 1 and 9). Use \d (all digits), and add a + to make it match one or more. Also, adding an end of string anchor $ to it makes it only match the hyphen+number at the end of the string.
You can use this regex (.*?)-\d+$
regex demo
JavaScript demo
string = 'my-name-is-99-dude-84';
let regex = /(.*?)-\d+$/;
let specialChar = string.replace(regex, "$1");
document.write(specialChar);
I know the regex that separates two words as following:
input:
'WonderWorld'
output:
'Wonder World'
"WonderWorld".replace(/([A-Z])/g, ' $1');
Now I am looking to remove number in year format from string, what changes should be done in the above code to get:
input
'WonderWorld 2016'
output
'Wonder World'
You can match the location before an uppercase letter (but excluding the beginning of a line) with \B(?=[A-Z]) and match the trailing spaces if any with 4 digits right before the end (\s*\b\d{4}\b). In a callback, check if the match is not empty, and replace accordingly. If a match is empty, we matched the location before an uppercase letter (=> replace with a space) and if not, we matched the year at the end (=> replace with empty string). The four digit chunks are only matched as whole words due to the \b word boundaries around the \d{4}.
var re = /\B(?=[A-Z])|\s*\d{4}\b/g;
var str = 'WonderWorld 2016';
var result = str.replace(re, function(match) {
return match ? "" : " ";
});
document.body.innerHTML = "<pre>'" + result + "'</pre>";
A similar approach, just a different pattern for matching glued words (might turn out more reliable):
var re = /([a-z])(?=[A-Z])|\s*\b\d{4}\b/g;
var str = 'WonderWorld 2016';
var result = str.replace(re, function(match, group1) {
return group1 ? group1 + " " : "";
});
document.body.innerHTML = "<pre>'" + result + "'</pre>";
Here, ([a-z])(?=[A-Z]) matches and captures into Group 1 a lowercase letter that is followed with an uppercase one, and inside the callback, we check if Group 1 matched (with group1 ?). If it matched, we return the group1 + a space. If not, we matched the year at the end, and remove it.
Try this:
"WonderWorld 2016".replace(/([A-Z])|\b[0-9]{4}\b/g, ' $1')
How about this, a single regex to do what you want:
"WonderWorld 2016".replace(/([A-Z][a-z]+)([A-Z].*)\s.*/g, '$1 $2');
"Wonder World"
get everything apart from digits and spaces.
re-code of #Wiktor Stribiżew's solution:
str can be any "WonderWorld 2016" | "OneTwo 1000 ThreeFour" | "Ruby 1999 IamOnline"
str.replace(/([a-z])(?=[A-Z])|\s*\d{4}\b/g, function(m, g) {
return g ? g + " " : "";
});
import re
remove_year_regex = re.compile(r"[0-9]{4}")
Test regex expression here
I have the following URL structure:
https://api.bestschool.com/student/1102003120009/tests/json
I want to cut the student ID from the URL. So far I've came up with this:
/(student\/.*[^\/]*)/
which returns
student/1102003120009/tests/json
I only want the ID.
Your regex (student\/.*[^\/]*) matches and captures into Group 1 a literal sequence student/, then matches any characters other than a newline, 0 or more occurrences (.*) - that can match the whole line at once! - and then 0 or more characters other than /. It does not work because of .*. Also, a capturing group should be moved to the [^\/]* pattern.
You can use the following regex and grab Group 1 value:
student\/([^\/]*)
See regex demo
The regex matches student/ literally, and then matches and captures into Group 1 zero or more symbols other than /.
Alternatively, if you want to avoid using capturing, and assuming that the ID is always numeric and is followed by /tests/, you can use the following regex:
\d+(?=\/tests\/)
The \d+ matches 1 or more digits, and (?=\/tests\/) checks if right after the digits there is a /tests/ character sequence.
var re = /student\/([^\/]*)/;
var str = 'https://api.bestschool.com/student/1102003120009/tests/json';
var m = str.match(re);
if (m !== null) {
document.getElementById("r").innerHTML = "First method : " + m[1] + "<br/>";
}
var m2 = str.match(/\d+(?=\/tests\/)/);
if (m2 !== null) {
document.getElementById("r").innerHTML += "Second method: " + m2;
}
<div id="r"/>
Using JavaScript, how can I remove the last comma, but only if the comma is the last character or if there is only white space after the comma? This is my code.
I got a working fiddle. But it has a bug.
var str = 'This, is a test.';
alert( removeLastComma(str) ); // should remain unchanged
var str = 'This, is a test,';
alert( removeLastComma(str) ); // should remove the last comma
var str = 'This is a test, ';
alert( removeLastComma(str) ); // should remove the last comma
function removeLastComma(strng){
var n=strng.lastIndexOf(",");
var a=strng.substring(0,n)
return a;
}
This will remove the last comma and any whitespace after it:
str = str.replace(/,\s*$/, "");
It uses a regular expression:
The / mark the beginning and end of the regular expression
The , matches the comma
The \s means whitespace characters (space, tab, etc) and the * means 0 or more
The $ at the end signifies the end of the string
you can remove last comma from a string by using slice() method, find the below example:
var strVal = $.trim($('.txtValue').val());
var lastChar = strVal.slice(-1);
if (lastChar == ',') {
strVal = strVal.slice(0, -1);
}
Here is an Example
function myFunction() {
var strVal = $.trim($('.txtValue').text());
var lastChar = strVal.slice(-1);
if (lastChar == ',') { // check last character is string
strVal = strVal.slice(0, -1); // trim last character
$("#demo").text(strVal);
}
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<p class="txtValue">Striing with Commma,</p>
<button onclick="myFunction()">Try it</button>
<p id="demo"></p>
function removeLastComma(str) {
return str.replace(/,(\s+)?$/, '');
}
In case its useful or a better way:
str = str.replace(/(\s*,?\s*)*$/, "");
It will replace all following combination end of the string:
1. ,<no space>
2. ,<spaces>
3. , , , , ,
4. <spaces>
5. <spaces>,
6. <spaces>,<spaces>
The greatly upvoted answer removes not only the final comma, but also any spaces that follow. But removing those following spaces was not what was part of the original problem. So:
let str = 'abc,def,ghi, ';
let str2 = str.replace(/,(?=\s*$)/, '');
alert("'" + str2 + "'");
'abc,def,ghi '
https://jsfiddle.net/dc8moa3k/
long shot here
var sentence="I got,. commas, here,";
var pattern=/,/g;
var currentIndex;
while (pattern.test(sentence)==true) {
currentIndex=pattern.lastIndex;
}
if(currentIndex==sentence.trim().length)
alert(sentence.substring(0,currentIndex-1));
else
alert(sentence);
Remove last comma. Working example
function truncateText() {
var str= document.getElementById('input').value;
str = str.replace(/,\s*$/, "");
console.log(str);
}
<input id="input" value="address line one,"/>
<button onclick="truncateText()">Truncate</button>
First, one should check if the last character is a comma.
If it exists, remove it.
if (str.indexOf(',', this.length - ','.length) !== -1) {
str = str.substring(0, str.length - 1);
}
NOTE str.indexOf(',', this.length - ','.length) can be simplified to str.indexOf(',', this.length - 1)
you can remove last comma:
var sentence = "I got,. commas, here,";
sentence = sentence.replace(/(.+),$/, '$1');
console.log(sentence);
To remove the last comma from a string, you need
text.replace(/,(?=[^,]*$)/, '')
text.replace(/,(?![^,]*,)/, '')
See the regex demo. Details:
,(?=[^,]*$) - a comma that is immediately followed with any zero or more chars other than a comma till end of string.
,(?![^,]*,) - a comma that is not immediately followed with any zero or more chars other than a comma and then another comma.
See the JavaScript demo:
const text = '1,This is a test, and this is another, ...';
console.log(text.replace(/,(?=[^,]*$)/, ''));
console.log(text.replace(/,(?![^,]*,)/, ''));
Remove whitespace and comma at the end use this,
var str = "Hello TecAdmin, ";
str = str.trim().replace(/,(?![^,]*,)/, '')
// Output
"Hello TecAdmin"
The problem is that you remove the last comma in the string, not the comma if it's the last thing in the string. So you should put an if to check if the last char is ',' and change it if it is.
EDIT: Is it really that confusing?
'This, is a random string'
Your code finds the last comma from the string and stores only 'This, ' because, the last comma is after 'This' not at the end of the string.
With or without Regex.
I suggest two processes and also consider removing space as well. Today I got this problem and I fixed this by writing the below code.
I hope this code will help others.
//With the help of Regex
var str = " I am in Pakistan, I am in India, I am in Japan, ";
var newstr = str.replace(/[, ]+$/, "").trim();
console.log(newstr);
//Without Regex
function removeSpaceAndLastComa(str) {
var newstr = str.trim();
var tabId = newstr.split(",");
strAry = [];
tabId.forEach(function(i, e) {
if (i != "") {
strAry.push(i);
}
})
console.log(strAry.join(","));
}
removeSpaceAndLastComa(str);
If you are targeting es6, then you can simply do this
str = Array.from( str ).splice(0, str.length - 1).join('');
This Array.from(str) converts the string to an array (so we can slice it)
This splice( 0 , str.length - 1 ) returns an array with the items from the array sequentially except the last item in the array
This join('') joins the entries in the array to form a string
Then if you want to make sure that a comma actually ends the string before performing the operation, you can do something like this
str = str.endsWith(',') ? Array.from(str).splice(0,str.length - 1).join('') : str;