Complementary function to util.format() in Node.js

Question

I know how to format a string using %f, %d etc. using util.format(). Can someone tell me which is the complementary function that enables SCANNING from a string (not from console input).

For example:

Running...

const util = require('util');
var weatherStr = util.format(`The temperature at %d o' clock was %f deg. C and the humidity was %f.`, 5, 23.9, 0.5);
console.log(weatherStr);

...generates...

The temperature at 5 o' clock was 23.9 deg. C and the humidity was 0.5.

I was expecting a util function which would work such that running the following code...

const util = require('util');
var weatherStr = 'The temperature at 5 o' clock was 23.9 deg. C and the humidity was 0.5.';
console.log(util.????(tempStr, `humidity was %f.`));

...generates...

0.5

Which is the util function that does this? I don't think "parseFloat" will work because it will extract 23.9.

I'm new to JS and Node but I expected a "scan" function. I know there is a scanf npm library but it seems to work with console input rather than existing strings. I have been doing searches for "%f" among JS and Node functions and surprisingly util.format seems to be the only one with a mention of it.

trincot · Accepted Answer · 2017-08-26 20:58:58Z

I don't know of any scan library like that, but you can use regular expressions. Here are some patterns you could use:

Integer: [+-]?\d+
Decimal: [+-]?\d+(?:\.\d+)?

If you put these in a capture group, you can access the corresponding matches from the array that String#match returns:

var weatherStr = "The temperature at 5 o'clock was 23.9 deg. C and the humidity was 0.5.";
console.log(+weatherStr.match(/humidity was ([+-]?\d+(?:\.\d+)?)./)[1]);

You could create a utility function that can deal with %d and %f:

function scanf(input, find) {
    var pattern = {
        "d": "(\\d+)",
        "f": "(\\d+(?:\\.\\d+)?)"
    };
    find = find
        // Escape characters for use in RegExp constructor:
        .replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&')
        // Replace %-patterns
        .replace(/((?:\\)*)%([a-z])/g, function (m, a, b) {
            return a.length % 4 == 0 && b in pattern ? a + pattern[b] : m;
        });
    var match = input.match(new RegExp(find));
    return match && match.slice(1).map(Number);
}

var weatherStr = "The temperature at 5 o'clock was 23.9 deg. C and the humidity was 0.5.";
console.log(scanf(weatherStr, "humidity was %f"));
console.log(scanf(weatherStr, "at %d o'clock was %f"));

Akash Sharma · Accepted Answer · 2017-08-27 10:45:26Z

0

Thanks trincot!

Actually it turns out the scanf npm library (https://www.npmjs.com/package/scanf) DOES solve my problem. I just hadn't read it all the way through. I had to install "sscanf" (note the double-s) as well. The sscanf method (listed at the bottom of the package page) works just as I expected.

I'm surprised this package is not more popular, but it is what I need. Thanks again!

answered Aug 27, 2017 at 10:45

Akash Sharma

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