CODEWITHSUNDEEP
javascriptsubstringsubstr
Anjana Sharma
Anjana Sharma

Reputation: 4745

How to grab substring before a specified character in JavaScript?

I am trying to extract everything before the ',' comma. How do I do this in JavaScript or jQuery? I tried this and not working..

1345 albany street, Bellevue WA 42344

I just want to grab the street address.

var streetaddress= substr(addy, 0, index(addy, '.'));

Upvotes: 402

Views: 713128

Answers (13)

bgondy
bgondy

Reputation: 1328

According to the ESLint rule prefer-string-slice, here is my implementation using slice() method:

function getStreetAddress(str: string, separator = ','): string {
  const separatorIndex = str.indexOf(separator);

  return separatorIndex === -1 ? str : str.slice(0, separatorIndex);
}

It returns the whole string if the separator has not been found.

Worth to mention, as it seems you're trying to parse a kind of CSV file row, this code may work in your case, but not when the street address contains special characters or the separator itself (,). To do so, fields in CSV files may be wrapped with quotes, which is not handled in this simple implementation.

Upvotes: 1

wheresrhys
wheresrhys

Reputation: 23500

const streetAddress = addy.substring(0, addy.indexOf(","));

While it’s not the best place for definitive information on what each method does (MDN Web Docs are better for that) W3Schools.com is good for introducing you to syntax.

Upvotes: 598

monkey
monkey

Reputation: 1597

If you are worried about catching the case where no comma is present, you could just do this:

let end = addy.indexOf(",") >= 0 ? addy.indexOf(",") : addy.length;
let streetaddress = addy.substr(0, end);

Nobody said it had to go on one line.

Upvotes: -1

Ramen_Lover912
Ramen_Lover912

Reputation: 57

You could use regex as this will give you the string if it matches the requirements. The code would be something like:

const address = "1345 albany street, Bellevue WA 42344";
const regex = /[1-9][0-9]* [a-zA-Z]+ [a-zA-Z]+/;
const matchedResult = address.match(regex);

console.log(matchedResult[0]); // This will give you 1345 albany street.

So to break the code down. [1-9][0-9]* basically means the first number cannot be a zero and has to be a number between 1-9 and the next number can be any number from 0-9 and can occur zero or more times as sometimes the number is just one digit and then it matches a space. [a-zA-Z] basically matches all capital letters to small letters and has to occur one or more times and this is repeated.

Upvotes: 2

Grant Miller
Grant Miller

Reputation: 28999

You can also use shift().

var streetaddress = addy.split(',').shift();

According to MDN Web Docs:

The shift() method removes the first element from an array and returns that removed element. This method changes the length of the array.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/shift

Upvotes: 21

Miles Florence
Miles Florence

Reputation: 427

//split string into an array and grab the first item

var streetaddress = addy.split(',')[0];

Also, I'd recommend naming your variables with camel-case(streetAddress) for better readability.

Upvotes: 41

m a
m a

Reputation: 139

almost the same thing as David G's answer but without the anonymous function, if you don't feel like including one.

s = s.substr(0, s.indexOf(',') === -1 ? s.length : s.indexOf(','));

in this case we make use of the fact that the second argument of substr is a length, and that we know our substring is starting at 0.

the top answer is not a generic solution because of the undesirable behavior if the string doesn't contain the character you are looking for.

if you want correct behavior in a generic case, use this method or David G's method, not the top answer

regex and split methods will also work, but may be somewhat slower / overkill for this specific problem.

Upvotes: 13

Gaurav
Gaurav

Reputation: 28755

var newString = string.substr(0,string.indexOf(','));

Upvotes: 9

user3336882
user3336882

Reputation: 3313

var streetaddress = addy.split(',')[0];

Upvotes: 263

David G
David G

Reputation: 5784

If you want to return the original string untouched if it does not contain the search character then you can use an anonymous function (a closure):

var streetaddress=(function(s){var i=s.indexOf(',');
   return i==-1 ? s : s.substr(0,i);})(addy);

This can be made more generic:

var streetaddress=(function(s,c){var i=s.indexOf(c);
   return i==-1 ? s : s.substr(0,i);})(addy,',');

Upvotes: 2

flu
flu

Reputation: 14683

If you like it short simply use a RegExp:

var streetAddress = /[^,]*/.exec(addy)[0];

Upvotes: 22

Mikey G
Mikey G

Reputation: 3491

try this:

streetaddress.substring(0, streetaddress.indexOf(','));

Upvotes: 47

Mira Weller
Mira Weller

Reputation: 2432

var streetaddress = addy.substr(0, addy.indexOf('.'));

(You should read through a javascript tutorial, esp. the part about String functions)

Upvotes: 8

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