Here is an example that demonstrates a flawed method:
const HTML = require('html-element');
const doc = `<body>
</body>`;
const page = HTML.document.createElement(doc)
page.appendChild('<div>1</div>')
page.appendChild('<div>2</div>')
page.appendChild('<div>3</div>')
console.log(page)
However, this approach is incorrect because
page.document.querySelectorAll("div")
does not work. Is there a proper way to accomplish this task?
To access all the divs in a page, utilize the childNodes property of the page object.
// var document = require('html-element').document;
const HTML = require('html-element');
const doc = `<body></body>`;
const page = HTML.document.createElement(doc)
page.appendChild('<div>1</div>')
page.appendChild('<div>2</div>')
page.appendChild('<div>3</div>')
// childNodes ✅
const divs = page.childNodes;
for (const div of divs) {
console.log(`div =`, div)
}
screenshot
https://i.stack.imgur.com/u1XLE.png
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