cross-browser

Find the exact height and width of the viewport in a cross-browser way

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You might try this: function getViewport() { var viewPortWidth; var viewPortHeight; // the more standards compliant browsers (mozilla/netscape/opera/IE7) use window.innerWidth and window.innerHeight if (typeof window.innerWidth != ‘undefined’) { viewPortWidth = window.innerWidth, viewPortHeight = window.innerHeight } // IE6 in standards compliant mode (i.e. with a valid doctype as the first line in the document) else if … Read more

header/footer/nav tags – what happens to these in IE7, IE8 and browsers than don’t support HTML5?

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Place this is the <head> section of your page, before any CSS files are loaded. <!–[if lte IE 8]> <script src=”https://cdnjs.cloudflare.com/ajax/libs/html5shiv/3.7.3/html5shiv.min.js”></script> <![endif]–> html5shi(m|v) creates doc elements for all the html5 elements so the styles from your CSS can kick in. Default behaviour for IE is to ignore unknown elements. For more info see resig’s blog … Read more

Cross browser method to fit a child div to its parent’s width

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The solution is to simply not declare width: 100%. The default is width: auto, which for block-level elements (such as div), will take the “full space” available anyway (different to how width: 100% does it). See: http://jsfiddle.net/U7PhY/2/ Just in case it’s not already clear from my answer: just don’t set a width on the child … Read more

Making things unselectable in IE

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In IE, you need the unselectable attribute in HTML: <div id=”foo” unselectable=”on”>…</div> … or set it via JavaScript: document.getElementById(“foo”).setAttribute(“unselectable”, “on”); The thing to be aware of is that the unselectableness is not inherited by children of an unselectable element. This means you either have to put an attribute in the start tag of every element … Read more

Testing for multiple screens with javascript

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I didn’t find the answer I needed anywhere on StackOverflow, so responding late to the several related threads here. The solution is, first, check for secondary display with window.screen.isExtended. Then await window.getScreenDetails(), which returns an object that includes an array of screens. Very thorough explanation here: https://web.dev/multi-screen-window-placement/

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