I need to replace all instances of a substring with a modified version of the substring, can I do something like:
const regex = /[0-9]{4}[A-Z]{3}/g; // format: 0000ABC
myString = myString.replaceAll(regex, regex + " I'm modified");
Abstract example
If myString is
5000ABC, 250XYZ, GEN3000
and I want to modify certain 4 digit - 3 letter patterns, my expected output is
5000ABC I'm modified, 250XYZ, 1000DEF I'm modified, GEN3000
Not sure I understand the output format you want, but guessing from your example it looks like you want to append some string after each match. Here we go:
const input = 'Some text with 1234ABC and 5678XYZ';
const regex = /([0-9]{4}[A-Z]{3})/g; // format: 0000ABC
var result = input.replace(regex, "$1 I'm modified");
console.log('input: ' + input);
console.log('result: ' + result);
Output:
input: Some text with 1234ABC and 5678XYZ
result: Some text with 1234ABC I'm modified and 5678XYZ I'm modified
Explanation:
capture your pattern with parenthesis for later use
use a string .replace() where you can reference the captured pattern with $1, and prefix/append any text you want
The workaround I have at the moment is
const regex = /[0-9]{4}[A-Z]{3}/g;
var matches = myString.match(regex);
if (matches && matches.length > 0) {
for (var match of matches) {
var modified = match + "I'm modified";
myString = myString.replaceAll(match, modified)
}
}
I am trying to handle input groups similar to:
'...A.B.' and want to output '.....AB'.
Another example:
'.C..Z..B.' ==> '......CZB'
I have been working with the following:
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$1")
returns:
"....."
and
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$2")
returns:
"AB"
but
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$1$2")
returns
"...A.B."
Is there a way to return
"....AB"
with a single regexp?
I have only been able to accomplish this with:
'...A.B.'.replace(/(\.*)([A-Z]*)/g, "$1") + '...A.B.'.replace(/(\.*)([A-Z]*)/g, "$2")
==> ".....AB"
If the goal is to move all of the . to the beginning and all of the A-Z to the end, then I believe the answer to
with a single regexp?
is "no."
Separately, I don't think there's a simpler, more efficient way than two replace calls — but not the two you've shown. Instead:
var str = "...A..B...C.";
var result = str.replace(/[A-Z]/g, "") + str.replace(/\./g, "");
console.log(result);
(I don't know what you want to do with non-., non-A-Z characters, so I've ignored them.)
If you really want to do it with a single call to replace (e.g., a single pass through the string matters), you can, but I'm fairly sure you'd have to use the function callback and state variables:
var str = "...A..B...C.";
var dots = "";
var nondots = "";
var result = str.replace(/\.|[A-Z]|$/g, function(m) {
if (!m) {
// Matched the end of input; return the
// strings we've been building up
return dots + nondots;
}
// Matched a dot or letter, add to relevant
// string and return nothing
if (m === ".") {
dots += m;
} else {
nondots += m;
}
return "";
});
console.log(result);
That is, of course, incredibly ugly. :-)
I have a search box and i need to grab the value of this search and match all DIV which data-* value is starting with the search value.
Cases:
Search value: 201
Should match: data-year="2011", data-year="2012", data-year="2013"
Should fail: data-year="2009", data-year="2001"
This is what i come up with so far:
\\b(?=\\w*[" + token + "])\\w+\\b
token is a dynamic value from the search box. Therefore i need to use RegExp
This is working but it match all the value which contain 2 or 0 or 1 (for my understanding). so 2009 is valid match as well. :/
I also try to add the caret at the beginning in order to match the characthers just at the beginning of the world but clearly i'm missing something here:
^\\b(?=\\w*[" + token + "])\\w+\\b
The whole code is:
var token = '200'; // should fail
var tokenTwo = '201'; // shoudl work
var dataAtt = $('#div').data('year').toString();
var regexExpression ="^\\b(?=\\w*\\d*[" + token + "])\\w+\\d+\\b";
var regEXPRES = "^.*" + token + ".*$";
var regex = new RegExp(regexExpression, "i");
if( dataAtt.match(regex) ){
console.log(dataAtt);
alert('yey!!!');
} else {
alert('nope!! )')
}
and here is the JsFiddle http://jsfiddle.net/tk5m8coo/
p.s. I shouldn't have any cases where token is precede or follow by other characters, but if anyone as idea how to check also this, would be great. Just in case of any typo like s2015.
Problem is that you're putting your search value inside a character class when you enclose your regex by wrapping around [...].
You also don't need a lookahead. You can just use:
var regex = new RegExp("\\b\\w*" + value + "\\w*\\b");
To make sure search value is matched within a full word. (\w includes digits also so no need to use \d).
Full code:
var value = '20'; //token
var html = $("#div").html(); //data.()
var dataAtt = $('#div').data('year').toString();
var regex = new RegExp("\\b\\w*" + value + "\\w*\\b");
if( dataAtt.match(regex) ){
console.log(dataAtt + " matched");
} else {
console.log('nope')
}
Updated JS Fiddle
Or you can use indexOf():
var value = '20';
var html = $("#div").html();
var dataAtt = $('#div').data('year').toString();
if( dataAtt.indexOf(value) >= 0 ){
console.log('yey!!!');
} else {
console.log('nein!! )')
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id ="div" data-year="2015">
what bozo where 2015
</div>
Your regex is using a character class which will match any of the characters inside the square brackets. Remove the square brackets:
^\\b(?=\\w*" + token + ".*)\\w+\\b
I want to remove the few characters in a string using index.
for example:
My input is: "5,4,3,2,1"
I want to remove the 1th to 2nd index position characters(here , to 4).
The Output should be 5,3,2,1.
Is there any predefined function in jquery or javascript to done this?
You can try to use substring function like this:
var mystring = "5,4,3,2,1";
alert( mystring.substr(0, 1) + mystring.substr(3));
JSFIDDLE DEMO
I would juse use javascripts split function for that.
So if you have
var string = "5,4,3,2,1";
than you just need to do
var splitted = string.split(",");
whereas the character in the brackets is the one you want to split on. After you did that, you can just make a new string, and build it with the array elements.
So you do something like
var string2 = splitted[0] + "," + splitted[2] + "," + splitted[3] + "," splitted[4];
Given a string
'1.2.3.4.5'
I would like to get this output
'1.2345'
(In case there are no dots in the string, the string should be returned unchanged.)
I wrote this
function process( input ) {
var index = input.indexOf( '.' );
if ( index > -1 ) {
input = input.substr( 0, index + 1 ) +
input.slice( index ).replace( /\./g, '' );
}
return input;
}
Live demo: http://jsfiddle.net/EDTNK/1/
It works but I was hoping for a slightly more elegant solution...
There is a pretty short solution (assuming input is your string):
var output = input.split('.');
output = output.shift() + '.' + output.join('');
If input is "1.2.3.4", then output will be equal to "1.234".
See this jsfiddle for a proof. Of course you can enclose it in a function, if you find it necessary.
EDIT:
Taking into account your additional requirement (to not modify the output if there is no dot found), the solution could look like this:
var output = input.split('.');
output = output.shift() + (output.length ? '.' + output.join('') : '');
which will leave eg. "1234" (no dot found) unchanged. See this jsfiddle for updated code.
It would be a lot easier with reg exp if browsers supported look behinds.
One way with a regular expression:
function process( str ) {
return str.replace( /^([^.]*\.)(.*)$/, function ( a, b, c ) {
return b + c.replace( /\./g, '' );
});
}
You can try something like this:
str = str.replace(/\./,"#").replace(/\./g,"").replace(/#/,".");
But you have to be sure that the character # is not used in the string; or replace it accordingly.
Or this, without the above limitation:
str = str.replace(/^(.*?\.)(.*)$/, function($0, $1, $2) {
return $1 + $2.replace(/\./g,"");
});
You could also do something like this, i also don't know if this is "simpler", but it uses just indexOf, replace and substr.
var str = "7.8.9.2.3";
var strBak = str;
var firstDot = str.indexOf(".");
str = str.replace(/\./g,"");
str = str.substr(0,firstDot)+"."+str.substr(1,str.length-1);
document.write(str);
Shai.
Here is another approach:
function process(input) {
var n = 0;
return input.replace(/\./g, function() { return n++ > 0 ? '' : '.'; });
}
But one could say that this is based on side effects and therefore not really elegant.
This isn't necessarily more elegant, but it's another way to skin the cat:
var process = function (input) {
var output = input;
if (typeof input === 'string' && input !== '') {
input = input.split('.');
if (input.length > 1) {
output = [input.shift(), input.join('')].join('.');
}
}
return output;
};
Not sure what is supposed to happen if "." is the first character, I'd check for -1 in indexOf, also if you use substr once might as well use it twice.
if ( index != -1 ) {
input = input.substr( 0, index + 1 ) + input.substr(index + 1).replace( /\./g, '' );
}
var i = s.indexOf(".");
var result = s.substr(0, i+1) + s.substr(i+1).replace(/\./g, "");
Somewhat tricky. Works using the fact that indexOf returns -1 if the item is not found.
Trying to keep this as short and readable as possible, you can do the following:
JavaScript
var match = string.match(/^[^.]*\.|[^.]+/g);
string = match ? match.join('') : string;
Requires a second line of code, because if match() returns null, we'll get an exception trying to call join() on null. (Improvements welcome.)
Objective-J / Cappuccino (superset of JavaScript)
string = [string.match(/^[^.]*\.|[^.]+/g) componentsJoinedByString:''] || string;
Can do it in a single line, because its selectors (such as componentsJoinedByString:) simply return null when sent to a null value, rather than throwing an exception.
As for the regular expression, I'm matching all substrings consisting of either (a) the start of the string + any potential number of non-dot characters + a dot, or (b) any existing number of non-dot characters. When we join all matches back together, we have essentially removed any dot except the first.
var input = '14.1.2';
reversed = input.split("").reverse().join("");
reversed = reversed.replace(\.(?=.*\.), '' );
input = reversed.split("").reverse().join("");
Based on #Tadek's answer above. This function takes other locales into consideration.
For example, some locales will use a comma for the decimal separator and a period for the thousand separator (e.g. -451.161,432e-12).
First we convert anything other than 1) numbers; 2) negative sign; 3) exponent sign into a period ("-451.161.432e-12").
Next we split by period (["-451", "161", "432e-12"]) and pop out the right-most value ("432e-12"), then join with the rest ("-451161.432e-12")
(Note that I'm tossing out the thousand separators, but those could easily be added in the join step (.join(','))
var ensureDecimalSeparatorIsPeriod = function (value) {
var numericString = value.toString();
var splitByDecimal = numericString.replace(/[^\d.e-]/g, '.').split('.');
if (splitByDecimal.length < 2) {
return numericString;
}
var rightOfDecimalPlace = splitByDecimal.pop();
return splitByDecimal.join('') + '.' + rightOfDecimalPlace;
};
let str = "12.1223....1322311..";
let finStr = str.replace(/(\d*.)(.*)/, '$1') + str.replace(/(\d*.)(.*)/, '$2').replace(/\./g,'');
console.log(finStr)
const [integer, ...decimals] = '233.423.3.32.23.244.14...23'.split('.');
const result = [integer, decimals.join('')].join('.')
Same solution offered but using the spread operator.
It's a matter of opinion but I think it improves readability.