Enhance your JQuery skills by implementing variables within the .css() function

Attempting to randomly assign values to an image's left and right properties using JQuery. Here is the code snippet:

var sides = ["left", "right"]
var currentSide = sides[Math.floor(Math.random() * sides.length)];
$("#"+currentImage).css({currentSide : "80px"});

An array called 'sides' holds either 'left' or 'right', with one being randomly selected to set the CSS property of the element to either right: "80px"; or left: 80px;

However, this implementation is not functioning as expected. Manually setting it e.g. 

$("#"+currentImage).css({"left" : "80px"});
works fine, indicating that passing variables to the .css() function may be causing the issue.

Any suggestions on how to resolve this?

javascriptjquerycss

Answer №1

If you intend to generate an object with a dynamic key, then bracket notation is required. In this scenario, the object being created has a key of currentSide and not left/right.

To achieve this, you can utilize:

$("#"+currentImage).css(currentSide, "80px");

Answer №2

To prevent encountering similar issues with various applications that lack an alternative syntax like $().css(), I will provide a more general answer. This method allows you to apply multiple styles using just one .css() function call.

let directions = ["up", "down"]
let currentDirection = directions[Math.floor(Math.random() * directions.length)];
let styles = {};
styles[currentDirection] = "50px";
$("#"+currentImageID).css(styles);

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