CODEWITHSUNDEEP
d3.js
dwilbank
dwilbank

Reputation: 2520

what is this extra 'parse' call doing? (D3 time formatting)

In this block,

https://bl.ocks.org/mbostock/1667367

I see this

var parseDate = d3.time.format("%b %Y").parse;

which is used later like this

function type(d) {
  d.date = parseDate(d.date);
  d.price = +d.price;
  return d;
}

Why is .parse stuck on the end when every other example I've seen doesn't need it?

When I try to experiment with that syntax in other REPLs, I get a type error:

parseDate.parse is not a function

Upvotes: 0

Views: 87

Answers (1)

Eric Guan
Eric Guan

Reputation: 15982

It parses a string to a date.

https://github.com/mbostock/d3/wiki/Time-Formatting#parse

In the context of var parseDate = d3.time.format("%b %Y").parse;

parseDate.parse is not a function makes sense because parseDate is the function, not parseDate.parse. You call it like parseDate('1/1/2016'), not parseDate.parse('1/1/2016')

You don't need to stick parse at the end. You can do

var parseDate = d3.time.format("%b %Y")

and then

parseDate.parse('1/1/2015).

Example from D3 docs.

var format = d3.time.format("%Y-%m-%d");
format.parse("2011-01-01"); // returns a Date
format(new Date(2011, 0, 1)); // returns a string

This allows you to format and parse a string. The first example only allows you to parse.

Upvotes: 2

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