To get all colors on a page with jQuery, you can use the “*” selector to get all DOM elements, call “getComputedStyle” on each element, and then de-duplicate the results:<\/p>\n
\nvar elements = $('*');\n\nfunction getColor(i, v) { \n return window.getComputedStyle(v).color \n}\n\ncolors = $.unique(elements.map(m));\n\n\nUnfortunately this gives you just CSS, and not images, but still pretty useful:\n\n\n[\"rgb(66, 137, 186)\", \n \"rgb(131, 131, 131)\", \n \"rgb(119, 119, 119)\", \n \"rgb(0, 0, 0)\", \n \"rgb(66, 137, 186)\", \n \"rgb(119, 119, 119)\", \n \"rgb(101, 101, 101)\", \n \"rgb(255, 255, 255)\", \n \"rgb(119, 119, 119)\"]\n<\/pre>\nAs a potentially better approach, you can group all the DOM elements together (which would let you change color later)<\/p>\n
\nfunction col(v) { \n return window.getComputedStyle(v).color;\n}\n\n_.groupBy(elements, col)\n<\/pre>\n","protected":false},"excerpt":{"rendered":"To get all colors on a page with jQuery, you can use the “*” selector to get all DOM elements, call “getComputedStyle” on each element, and then de-duplicate the results: var elements = $(‘*’); function getColor(i, v) { return window.getComputedStyle(v).color } colors = $.unique(elements.map(m)); Unfortunately this gives you just CSS, and not images, but still … <\/p>\n