var cartstring = "27,00 - R"
How can I remove spaces and "-" and "R" using only regex (not allowed to use slice etc.)? I need to make strings cartstring1 and cartstring2 which should both be equal to "27,00", first by removing spaces and "-" and "R", and second by allowing only numbers and ",".
cartstring1 = cartstring.replace(/\s/g, "");
cartstring2 = cartstring.replace(/\D/g, "");
Please help me modify these regular expressions to have a working code. I tried to read about regex but still cannot quite get it. Thank you very much in advance.
you can just capture just what you are interested in number and comma:
let re = /[\d,]+/g
let result = "27,00 - R".match(re)
console.log(result)
You can group the characters you want to remove:
var cartstring = "27,00 - R"
let res = cartstring.replace(/(\s|-|R)/g, "")
console.log(res)
Or alternatively, split the string by a space and get the first item:
var cartstring = "27,00 - R"
let res = cartstring.split(" ")[0]
console.log(res)
You are using 2 replacements, one replacing all whitespace chars \s and the other replacing all non digits \D, but note that \D also matches \s so you could omit the first call.
Using \D will also remove the comma that you want to keep, so you can match all chars except digits or a comma using [^\d,]+ in a single replacement instead:
var cartstring = "27,00 - R";
console.log(cartstring.replace(/[^\d,]+/g, ''));
I have problem with simple rexex. I have example strings like:
Something1\sth2\n649 sth\n670 sth x
Sth1\n\something2\n42 036 sth\n42 896 sth y
I want to extract these numbers from strings. So From first example I need two groups: 649 and 670. From second example: 42 036 and 42 896. Then I will remove space.
Currently I have something like this:
\d+ ?\d+
But it is not a good solution.
You can use
\n\d+(?: \d+)?
\n - Match new line
\d+ - Match digit from 0 to 9 one or more time
(?: \d+)? - Match space followed by digit one or more time. ( ? makes it optional )
let strs = ["Something1\sth2\n649 sth\n670 sth x","Sth1\n\something2\n42 036 sth\n42 896 sth y"]
let extractNumbers = str => {
return str.match(/\n\d+(?: \d+)?/g).map(m => m.replace(/\s+/g,''))
}
strs.forEach(str=> console.log(extractNumbers(str)))
If you need to remove the spaces. Then the easiest way for you to do this would be to remove the spaces and then scrape the numbers using 2 different regex.
str.replace(/\s+/, '').match(/\\n(\d+)/g)
First you remove spaces using the \s token with a + quantifier using replace.
Then you capture the numbers using \\n(\d+).
The first part of the regex helps us make sure we are not capturing numbers that are not following a new line, using \ to escape the \ from \n.
The second part (\d+) is the actual match group.
var str1 = "Something1\sth2\n649 sth\n670 sth x";
var str2 = "Sth1\n\something2\n42 036 sth\n42 896 sth y";
var reg = /(?<=\n)(\d+)(?: (\d+))?/g;
var d;
while(d = reg.exec(str1)){
console.log(d[2] ? d[1]+d[2] : d[1]);
}
console.log("****************************");
while(d = reg.exec(str2)){
console.log(d[2] ? d[1]+d[2] : d[1]);
}
I have string
Started: 11.11.2014 11:19:28.376<br/>Ended: 1.1.4<br/>1:9:8.378<br/>Request took: 0:0:0.2
I need to add zeros in case I encounter 1:1:8 it should be 01:01:08 same goes for date. I tried using
/((:|\.|\s)[0-9](:|\.))/g
but it did not give all possible overlapping matches. How to fix it?
var str = "Started: 11.11.2014 11:19:28.376<br/>Ended: 11.11.2014<br/>11:19:28.378<br/>Request took: 0:0:0.2";
var re = /((:|\.|\s)[0-9](:|\.))/g
while ((match = re.exec(str)) != null) {
//alert("match found at " + match.index);
str = [str.slice(0,match.index), '0', str.slice(match.index+1,str.length)];
}
alert(str);
This will probably do what you want:
str.replace(/\b\d\b/g, "0$&")
It searches for lone digits \d, and pad 0 in front.
The first word boundary \b checks that there is no [a-zA-Z0-9_] in front, and the second checks there is no [a-zA-Z0-9_] behind the digit.
$& in the replacement string refers to the whole match.
If you want to pad 0 as long as the character before and after are not digits:
str.replace(/(^|\D)(\d)(?!\d)/g, "$10$2")
I need to split a keyword string and turn it into a comma delimited string. However, I need to get rid of extra spaces and any commas that the user has already input.
var keywordString = "ford tempo, with,,, sunroof";
Output to this string:
ford,tempo,with,sunroof,
I need the trailing comma and no spaces in the final output.
Not sure if I should go Regex or a string splitting function.
Anyone do something like this already?
I need to use javascript (or JQ).
EDIT (working solution):
var keywordString = ", ,, ford, tempo, with,,, sunroof,, ,";
//remove all commas; remove preceeding and trailing spaces; replace spaces with comma
str1 = keywordString.replace(/,/g , '').replace(/^\s\s*/, '').replace(/\s\s*$/, '').replace(/[\s,]+/g, ',');
//add a comma at the end
str1 = str1 + ',';
console.log(str1);
You will need a regular expression in both cases. You could split and join the string:
str = str.split(/[\s,]+/).join();
This splits on and consumes any consecutive white spaces and commas. Similarly, you could just match and replace these characters:
str = str.replace(/[\s,]+/g, ',');
For the trailing comma, just append one
str = .... + ',';
If you have preceding and trailing white spaces, you should remove those first.
Reference: .split, .replace, Regular Expressions
In ES6:
var temp = str.split(",").map((item)=>item.trim());
In addition to Felix Kling's answer
If you have preceding and trailing white spaces, you should remove
those first.
It's possible to add an "extension method" to a JavaScript String by hooking into it's prototype. I've been using the following to trim preceding and trailing white-spaces, and thus far it's worked a treat:
// trims the leading and proceeding white-space
String.prototype.trim = function()
{
return this.replace(/^\s\s*/, '').replace(/\s\s*$/, '');
};
I would keep it simple, and just match anything not allowed instead to join on:
str.split(/[^a-zA-Z-]+/g).filter(v=>v);
This matches all the gaps, no matter what non-allowed characters are in between. To get rid of the empty entry at the beginning and end, a simple filter for non-null values will do. See detailed explanation on regex101.
var str = ", ,, ford, tempo, with,,, sunroof,, ,";
var result = str.split(/[^a-zA-Z-]+/g).filter(v=>v).join(',');
console.info(result);
let query = "split me by space and remove trailing spaces and store in an array ";
let words = query.trim().split(" ");
console.log(words)
Output :
[
'split', 'me', 'by', 'space','and','remove', 'trailing', 'spaces', 'and', 'store', 'in', 'an', 'array'
]
If you just want to split, trim and join keeping the whitespaces, you can do this with lodash:
// The string to fix
var stringToFix = "The Wizard of Oz,Casablanca,The Green Mile";
// split, trim and join back without removing all the whitespaces between
var fixedString = _.map(stringToFix.split(','), _.trim).join(' == ');
// output: "The Wizard of Oz == Casablanca == The Green Mile"
console.log(fixedString);
<script src="https://cdn.jsdelivr.net/lodash/4.16.3/lodash.min.js"></script>
I'm new to using regexp, can someone give me the regexp that will strip out everything but an integer from a string in javascript?
I would like to take the string "http://www.foo.com/something/1234/somethingelse" and get it down to 1234 as an integer.
Thanks
var str = "something 123 foo 432";
// Replace all non-digits:
str = str.replace(/\D/g, '');
alert(str); // alerts "123432"
In response to your edited question, extracting a string of digits from a string can be simple, depending on whether you want to target a specific area of the string or if you simply want to extract the first-occurring string of digits. Try this:
var url = "http://www.foo.com/something/1234/somethingelse";
var digitMatch = url.match(/\d+/); // matches one or more digits
alert(digitMatch[0]); // alerts "1234"
// or:
var url = "http://x/y/1234/z/456/v/890";
var digitMatch = url.match(/\d+/g); // matches one or more digits [global search]
digitMatch; // => ['1234', '456', '890']
This is just for integers:
[0-9]+
The + means match 1 or more, and the [0-9] means match any character from the range 0 to 9.
uri = "http://www.foo.com/something/1234/somethingelse";
alert(uri.replace(/.+?\/(\d+)\/.+/, "$1"))
Just define a character-class that requires the values to be numbers.
/[^0-9]/g // matches anything that is NOT 0-9 (only numbers will remain)