Checking whether website showDirectoryPicker JS function with Puppeteer

Question

Hello I want to get check whether the website has showDirectoryPicker function with the puppeteer.

Currently my code looks like this:

    'use strict';

const puppeteer = require('puppeteer');

(async function main() {
  try {
    const browser = await puppeteer.launch({ headless:false,executablePath: '/Applications/Google\ Chrome.app/Contents/MacOS/Google\ Chrome', });
    const [page] = await browser.pages();

    await page.goto('https://example.com');


    console.log(await page.evaluate(() => typeof showDirectoryPicker === 'function'));
 

    await browser.close();
  } catch (err) {
    console.error(err);
  }
})();

Currently this statement

    console.log(await page.evaluate(() => typeof showDirectoryPicker === 'function'));

returns True for the every website since it is a valid JS function. However, I want to get True if the analyzed website has the showDirectoryPicker function.

score 0 · Answer 1 · answered Feb 23 '21 at 14:44

If I understand your question correctly, you are trying to evaluate if the page calls the showDirectoryPicker() method, not if the browser supports it. One way to approach this would be to override the method with your own implementation that then reports back to Puppeteer if it gets called by the page. See my StackOverflow answer on overriding a function with a variant that logs whenever it gets called. You can then catch this log output with Puppeteer:

page.on('console', (message) => {
  /*
    Check that the message is what your overridden
    custom variant logs.
  */
});

Checking whether website showDirectoryPicker JS function with Puppeteer

1 Answers1