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An elegant way of working with iterables

WARNING: this is a beta version. Feel free to play around with it but don't use it in production.

ESLinq aims to be a complete, robust, thoroughly tested port of Linq to Objects to ECMAScript 2015 (ES6).

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A first example

Problem: We have an array of users. A user can have multiple email addresses, and not all users are verified. We need a list of the email addresses of the verified users, and check if all of them end in ".com".

const users = [
        "id": 12,
        "name": "John Smith",
        "emails": ["[email protected]", "[email protected]", "[email protected]"],
        "verified": true
        "id": 56,
        "name": "Jennifer Brown",
        "emails": ["[email protected]", "[email protected]"],
        "verified": false
        "id": 98,
        "name": "Kate Newton",
        "emails": ["[email protected]", "kate.newton.3[email protected]"],
        "verified": true

First, we need to restrict the list to only contain verified users:

const verifiedUsers = from(users).where(user => user.verified);

Then, we need to get a concatenated list of the email addresses of verified users:

const verifiedUserEmails =
        .where(user => user.verified)
        .selectMany(user => user.emails);

It is important to note that verifiedUserEmails is a lazily evaluated iterable: filtering and transformation is only done once we start iterating it, e. g. with a for..of loop:

for (let email of verifiedUserEmails) {
    // The first time this loop executes is the first time the original iterable is read from.
    // Filtering (`where`) and transformation (`selectMany`) run element-by-element during iteration.

Now let's check if all emails end with ".com":

const allEmailsEndWithCom =
        .where(user => user.verified)
        .selectMany(user => user.emails)
        .all(email => email.endsWith(".com"));

The result is a bool value indicating whether all emails end with the substring ".com". The all operator is an eager one: it iterates through all elements to compute its value.