使用带或不带Jquery的javascript,我需要根据开始和结束颜色创建渐变颜色。这可以通过编程方式进行吗?
最终颜色只会是起始颜色的深色,而且是无序列表,我无法控制li项目的数量。我正在寻找一种解决方案,允许我选择开始和结束颜色,将十六进制值转换为RGB,以便可以在代码中进行操作。起始RGB值增加一个基于项目数计算的步长值。
因此,如果列表中有8个项目,那么它需要以8个步骤递增单独的红绿蓝值以获得最终颜色。有没有更好的方法来做到这一点,如果是这样,我在哪里可以找到一些示例代码?
答案 0 :(得分:78)
我创建了一个JS库RainbowVis-JS来解决这个一般性问题。您只需使用setNumberRange
设置项目数,然后使用setSpectrum
设置开始和结束颜色。然后,您将获得带有colourAt
的十六进制颜色代码。
var numberOfItems = 8;
var rainbow = new Rainbow();
rainbow.setNumberRange(1, numberOfItems);
rainbow.setSpectrum('red', 'black');
var s = '';
for (var i = 1; i <= numberOfItems; i++) {
var hexColour = rainbow.colourAt(i);
s += '#' + hexColour + ', ';
}
document.write(s);
// gives:
// #ff0000, #db0000, #b60000, #920000, #6d0000, #490000, #240000, #000000,
欢迎您查看图书馆的源代码。 :)
答案 1 :(得分:29)
正确生成颜色数组的功能!
function hex (c) {
var s = "0123456789abcdef";
var i = parseInt (c);
if (i == 0 || isNaN (c))
return "00";
i = Math.round (Math.min (Math.max (0, i), 255));
return s.charAt ((i - i % 16) / 16) + s.charAt (i % 16);
}
/* Convert an RGB triplet to a hex string */
function convertToHex (rgb) {
return hex(rgb[0]) + hex(rgb[1]) + hex(rgb[2]);
}
/* Remove '#' in color hex string */
function trim (s) { return (s.charAt(0) == '#') ? s.substring(1, 7) : s }
/* Convert a hex string to an RGB triplet */
function convertToRGB (hex) {
var color = [];
color[0] = parseInt ((trim(hex)).substring (0, 2), 16);
color[1] = parseInt ((trim(hex)).substring (2, 4), 16);
color[2] = parseInt ((trim(hex)).substring (4, 6), 16);
return color;
}
function generateColor(colorStart,colorEnd,colorCount){
// The beginning of your gradient
var start = convertToRGB (colorStart);
// The end of your gradient
var end = convertToRGB (colorEnd);
// The number of colors to compute
var len = colorCount;
//Alpha blending amount
var alpha = 0.0;
var saida = [];
for (i = 0; i < len; i++) {
var c = [];
alpha += (1.0/len);
c[0] = start[0] * alpha + (1 - alpha) * end[0];
c[1] = start[1] * alpha + (1 - alpha) * end[1];
c[2] = start[2] * alpha + (1 - alpha) * end[2];
saida.push(convertToHex (c));
}
return saida;
}
// Exemplo de como usar
var tmp = generateColor('#000000','#ff0ff0',10);
for (cor in tmp) {
$('#result_show').append("<div style='padding:8px;color:#FFF;background-color:#"+tmp[cor]+"'>COLOR "+cor+"° - #"+tmp[cor]+"</div>")
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="result_show"></div>
答案 2 :(得分:22)
是的,绝对。
我在Java中这样做,在JavaScript中也应该相当简单。
首先,您需要将颜色分解为RGB组件。
然后计算组件的开始和结束之间的差异。
最后,计算百分比差异并乘以每个组件的起始颜色,然后将其添加到起始颜色。
假设你可以得到RGB值,这应该这样做:
var diffRed = endColor.red - startColor.red;
var diffGreen = endColor.green - startColor.green;
var diffBlue = endColor.blue - startColor.blue;
diffRed = (diffRed * percentFade) + startColor.red;
diffGreen = (diffGreen * percentFade) + startColor.green;
diffBlue = (diffBlue * percentFade) + startColor.blue;
“percentFade”是一个浮动小数,表示淡入“endColor”的距离。 1将是完全淡入淡出(从而产生最终颜色)。 0将不是淡入淡出(起始颜色)。
答案 3 :(得分:11)
我根据@desau答案使用此功能:
getGradientColor = function(start_color, end_color, percent) {
// strip the leading # if it's there
start_color = start_color.replace(/^\s*#|\s*$/g, '');
end_color = end_color.replace(/^\s*#|\s*$/g, '');
// convert 3 char codes --> 6, e.g. `E0F` --> `EE00FF`
if(start_color.length == 3){
start_color = start_color.replace(/(.)/g, '$1$1');
}
if(end_color.length == 3){
end_color = end_color.replace(/(.)/g, '$1$1');
}
// get colors
var start_red = parseInt(start_color.substr(0, 2), 16),
start_green = parseInt(start_color.substr(2, 2), 16),
start_blue = parseInt(start_color.substr(4, 2), 16);
var end_red = parseInt(end_color.substr(0, 2), 16),
end_green = parseInt(end_color.substr(2, 2), 16),
end_blue = parseInt(end_color.substr(4, 2), 16);
// calculate new color
var diff_red = end_red - start_red;
var diff_green = end_green - start_green;
var diff_blue = end_blue - start_blue;
diff_red = ( (diff_red * percent) + start_red ).toString(16).split('.')[0];
diff_green = ( (diff_green * percent) + start_green ).toString(16).split('.')[0];
diff_blue = ( (diff_blue * percent) + start_blue ).toString(16).split('.')[0];
// ensure 2 digits by color
if( diff_red.length == 1 ) diff_red = '0' + diff_red
if( diff_green.length == 1 ) diff_green = '0' + diff_green
if( diff_blue.length == 1 ) diff_blue = '0' + diff_blue
return '#' + diff_red + diff_green + diff_blue;
};
示例:
getGradientColor('#FF0000', '#00FF00', 0.4);
=> "#996600"
答案 4 :(得分:4)
desau's answer很棒。这是javascript:
function hexToRgb(hex) {
var result = /^#?([a-f\d]{2})([a-f\d]{2})([a-f\d]{2})$/i.exec(hex);
return result ? {
r: parseInt(result[1], 16),
g: parseInt(result[2], 16),
b: parseInt(result[3], 16)
} : null;
}
function map(value, fromSource, toSource, fromTarget, toTarget) {
return (value - fromSource) / (toSource - fromSource) * (toTarget - fromTarget) + fromTarget;
}
function getColour(startColour, endColour, min, max, value) {
var startRGB = hexToRgb(startColour);
var endRGB = hexToRgb(endColour);
var percentFade = map(value, min, max, 0, 1);
var diffRed = endRGB.r - startRGB.r;
var diffGreen = endRGB.g - startRGB.g;
var diffBlue = endRGB.b - startRGB.b;
diffRed = (diffRed * percentFade) + startRGB.r;
diffGreen = (diffGreen * percentFade) + startRGB.g;
diffBlue = (diffBlue * percentFade) + startRGB.b;
var result = "rgb(" + Math.round(diffRed) + ", " + Math.round(diffGreen) + ", " + Math.round(diffBlue) + ")";
return result;
}
function changeBackgroundColour() {
var count = 0;
window.setInterval(function() {
count = (count + 1) % 200;
var newColour = getColour("#00FF00", "#FF0000", 0, 200, count);
document.body.style.backgroundColor = newColour;
}, 20);
}
changeBackgroundColour();
答案 5 :(得分:2)
xolor library具有渐变功能。这将创建一个从开始颜色到结束颜色的渐变中有8种颜色的数组:
var gradientColors = []
var startColor = "rgb(100,200,50)", endColor = "green"
var start = xolor(startColor)
for(var n=0; n<8; n++) {
gradientColors.push(start.gradient(endColor, n/8))
}
在github上查看更多信息:https://github.com/fresheneesz/xolor
答案 6 :(得分:1)
不是那么强大,但在大多数情况下工作,除了jQuery之外,您不必包含任何其他库以用于以下代码:
HTML:
<div id="colors"></div>
JavaScript的:
function rainbow(value, s, l, max, min, start, end) {
value = ((value - min) * (start - end) / max)+end;
return 'hsl(' + value + ','+s+'%,'+l+'%)';
}
function createRainbowDiv(start,end){
var gradient = $("<div>").css({display:"flex", "flex-direction":"row",height:"100%"});
for (var i = start; ((i <= end) && (i >= start)) || ((i >= end) && (i <= start));
i += (end-start) / Math.abs(end-start)){
gradient.append($("<div>").css({float:"left","background-color":rainbow(i, 100,50, Math.max(start,end), Math.min(start,end), start,end),flex:1}));
}
return gradient;
}
$("#colors").append(createRainbowDiv(0,150));
$("#colors").css("width","100%").css("height","10px");
这应该是一个包含彩虹的div。见http://jsfiddle.net/rootandy/54rV7/
答案 7 :(得分:1)
我需要为一组未知的动态元素创建足够大的颜色选项数组,但是我需要每个元素通过开始颜色和结束颜色递增。这种情况遵循&#34;%褪色&#34;方法,除了我有一个困难的时间遵循这个逻辑。这就是我使用两个rgb颜色值的输入并计算页面上元素数量的方法。
这是展示概念的link to a codepen。
以下是问题的代码段。
<style>
#test {
width:200px;
height:100px;
border:solid 1px #000;
}
.test {
width:49%;
height:100px;
border:solid 1px #000;
display: inline-block;
}
</style>
</head>
<body>
<div id="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<div class="test"></div>
<script>
var GColor = function(r,g,b) {
r = (typeof r === 'undefined')?0:r;
g = (typeof g === 'undefined')?0:g;
b = (typeof b === 'undefined')?0:b;
return {r:r, g:g, b:b};
};
// increases each channel by the difference of the two
// divided by 255 (the number of colors stored in the range array)
// but only stores a whole number
// This should respect any rgb combinations
// for start and end colors
var createColorRange = function(c1) {
var colorList = [], tmpColor, rr = 0, gg = 0, bb = 0;
for (var i=0; i<255; i++) {
tmpColor = new GColor();
if (rExp >= 0) {
tmpColor.r = Math.floor(c1.r - rr);
rr += rAdditive;
} else {
tmpColor.r = Math.floor(c1.r + rr);
rr += rAdditive;
}
if (gExp >= 0) {
tmpColor.g = Math.floor(c1.g - gg);
gg += gAdditive;
} else {
tmpColor.g = Math.floor(c1.g + gg);
gg += gAdditive;
}
if (bExp >= 0) {
tmpColor.b = Math.floor(c1.b - bb);
bb += bAdditive;
} else {
tmpColor.b = Math.floor(c1.b + bb);
bb += bAdditive;
}
console.log(tmpColor);
colorList.push(tmpColor);
}
return colorList;
};
/* ==================
Testing Code Below
================== */
var firstColor = new GColor(255, 24, 0);
var secondColor = new GColor(255, 182, 0);
// Determine the difference
var rExp = firstColor.r - secondColor.r;
// Divide that difference by length of the array
// you would like to create (255 in this case)
var rAdditive = Math.abs(rExp)/255;
var gExp = firstColor.g - secondColor.g;
var gAdditive = Math.abs(gExp)/255;
var bExp = firstColor.b - secondColor.b;
var bAdditive = Math.abs(bExp)/255;
var range = createColorRange(firstColor, secondColor);
console.log(range);
var pointer = 0;
// This gently cycles through
// all the colors on a single element
function rotateColors() {
var currentColor = range[pointer];
document.getElementById("test").style.backgroundColor = "rgb("+currentColor.r+","+currentColor.g+","+currentColor.b+")";
pointer++;
if (pointer < range.length) window.setTimeout(rotateColors, 5);
}
rotateColors();
// say I have 5 elements
// so I need 5 colors
// I already have my first and last colors
// but I need to locate the colors between
// my start color and my end color
// inside of this range
// so I divide the range's length by the
// number of colors I need
// and I store the index values of the middle values
// those index numbers will then act as my keys to retrieve those values
// and apply them to my element
var myColors = {};
var objects = document.querySelectorAll('.test');
myColors.num = objects.length;
var determineColors = function(numOfColors, colorArray) {
var colors = numOfColors;
var cRange = colorArray;
var distance = Math.floor(cRange.length/colors);
var object = document.querySelectorAll('.test');
var j = 0;
for (var i = 0; i < 255; i += distance) {
if ( (i === (distance*colors)) ) {
object[j].style.backgroundColor = "rgb(" + range[255].r + ", " + range[255].g + ", " + range[255].b + ")";
j = 0;
// console.log(range[i]);
} else {
// Apply to color to the element
object[j].style.backgroundColor = "rgb(" + range[i].r + ", " + range[i].g + ", " + range[i].b + ")";
// Have each element bleed into the next with a gradient
// object[j].style.background = "linear-gradient( 90deg, rgb(" + range[i].r + ", " + range[i].g + ", " + range[i].b + "), rgb(" + range[i+distance].r + ", " + range[i+distance].g + ", " + range[i+distance].b + "))";
j++;
}
}
};
setTimeout( determineColors(myColors.num, range), 2000);
</script>
</body>
答案 8 :(得分:0)
您可以检索元素列表。我不熟悉jQuery,但prototypejs有Element.childElements(),它将返回一个数组。一旦知道了数组的长度,就可以确定每个步骤更改像素组件的程度。下面的一些代码我没有以我提出的形式进行测试,但它应该会给你一个想法。
function hex (c) {
var s = "0123456789abcdef";
var i = parseInt (c);
if (i == 0 || isNaN (c))
return "00";
i = Math.round (Math.min (Math.max (0, i), 255));
return s.charAt ((i - i % 16) / 16) + s.charAt (i % 16);
}
/* Convert an RGB triplet to a hex string */
function convertToHex (rgb) {
return hex(rgb[0]) + hex(rgb[1]) + hex(rgb[2]);
}
/* Remove '#' in color hex string */
function trim (s) { return (s.charAt(0) == '#') ? s.substring(1, 7) : s }
/* Convert a hex string to an RGB triplet */
function convertToRGB (hex) {
var color[];
color[0] = parseInt ((trim(hex)).substring (0, 2), 16);
color[1] = parseInt ((trim(hex)).substring (2, 4), 16);
color[2] = parseInt ((trim(hex)).substring (4, 6), 16);
}
/* The start of your code. */
var start = convertToRGB ('#000000'); /* The beginning of your gradient */
var end = convertToRGB ('#ffffff'); /* The end of your gradient */
var arr = $('.gradientList').childElements();
var len = arr.length(); /* The number of colors to compute */
var alpha = 0.5; /* Alpha blending amount */
for (i = 0; i < len; i++) {
var c = [];
c[0] = start[0] * alpha + (1 - alpha) * end[0];
c[1] = start[1] * alpha + (1 - alpha) * end[1];
c[2] = start[2] * alpha + (1 - alpha) * end[2];
/* Set the background color of this element */
arr[i].setStyle ({ 'background-color': convertToHex (c) });
}
答案 9 :(得分:0)
这是一个现成的函数,用于将元素背景设置为渐变
Element.prototype.setGradient = function( from, to, vertical ){
this.style.background = 'linear-gradient(to '+(vertical ? 'top' : 'left')+', '+from+', '+to+' 100%)';
}
和用法:
document.querySelector('.mydiv').setGradient('red','green');
已通过chrome测试,我将尝试为其他浏览器进行更新
最基本的水平是:
Element.prototype.setGradient = function( fromColor, toColor ){
var canvas = document.createElement('canvas');
var ctx = canvas.getContext('2d');
var b = this.getBoundingClientRect();
var grd = ctx.createLinearGradient(0, 0, b.width, 0);
canvas.width = b.width;
canvas.height = b.height;
grd.addColorStop(0, fromColor);
grd.addColorStop(1, toColor);
ctx.fillStyle = grd;
ctx.fillRect(0, 0, b.width, b.height);
this.style.backgroundImage = 'url('+canvas.toDataURL()+')';
}
和用法:
document.querySelector('.mydiv').setGradient('red','green');
小提琴: https://jsfiddle.net/jch39bey/
-
设置垂直的简单标志
Element.prototype.setGradient = function( fromColor, toColor, vertical ){
var canvas = document.createElement('canvas');
var ctx = canvas.getContext('2d');
var b = this.getBoundingClientRect();
var grd = ctx.createLinearGradient(0, 0, vertical ? 0 : b.width, vertical ? b.height : 0);
canvas.width = b.width;
canvas.height = b.height;
grd.addColorStop(0, fromColor);
grd.addColorStop(1, toColor);
ctx.fillStyle = grd;
ctx.fillRect(0, 0, b.width, b.height);
this.style.backgroundImage = 'url('+canvas.toDataURL()+')';
}
和用法:
document.querySelector('.mydiv').setGradient('red','green',true);
答案 10 :(得分:0)
基于@drinor的答案-TypeScript支持
const getGradientColor = (startColor: string, endColor: string, percent: number) => {
// strip the leading # if it's there
startColor = startColor.replace(/^\s*#|\s*$/g, '');
endColor = endColor.replace(/^\s*#|\s*$/g, '');
// convert 3 char codes --> 6, e.g. `E0F` --> `EE00FF`
if (startColor.length === 3) {
startColor = startColor.replace(/(.)/g, '$1$1');
}
if (endColor.length === 3) {
endColor = endColor.replace(/(.)/g, '$1$1');
}
// get colors
const startRed = parseInt(startColor.substr(0, 2), 16),
startGreen = parseInt(startColor.substr(2, 2), 16),
startBlue = parseInt(startColor.substr(4, 2), 16);
const endRed = parseInt(endColor.substr(0, 2), 16),
endGreen = parseInt(endColor.substr(2, 2), 16),
endBlue = parseInt(endColor.substr(4, 2), 16);
// calculate new color
let diffRed = endRed - startRed;
let diffGreen = endGreen - startGreen;
let diffBlue = endBlue - startBlue;
diffRed = ((diffRed * percent) + startRed);
diffGreen = ((diffGreen * percent) + startGreen);
diffBlue = ((diffBlue * percent) + startBlue);
let diffRedStr = diffRed.toString(16).split('.')[0];
let diffGreenStr = diffGreen.toString(16).split('.')[0];
let diffBlueStr = diffBlue.toString(16).split('.')[0];
// ensure 2 digits by color
if (diffRedStr.length === 1) diffRedStr = '0' + diffRedStr
if (diffGreenStr.length === 1) diffGreenStr = '0' + diffGreen
if (diffBlueStr.length === 1) diffBlueStr = '0' + diffBlue
return '#' + diffRedStr + diffGreenStr + diffBlueStr;
}
示例:
getGradientColor('#FF0000', '#00FF00', 0.4);
=>“#996600”
答案 11 :(得分:0)
答案 12 :(得分:0)
基于@desau 的回答和其他地方的一些代码,这里有一个 jQuery 分步演练:
function coloursBetween(fromColour, toColour, numberOfColours){
var colours = []; //holds output
var fromSplit = getRGBAValues(hexToRGBA(fromColour, 1.0)); //get raw values from hex
var toSplit = getRGBAValues(hexToRGBA(toColour, 1.0));
var fromRed = fromSplit[0]; //the red value as integer
var fromGreen = fromSplit[1];
var fromBlue = fromSplit[2];
var toRed = toSplit[0];
var toGreen = toSplit[1];
var toBlue = toSplit[2];
var difRed = toRed - fromRed; //difference between the two
var difGreen = toGreen - fromGreen;
var difBlue = toBlue - fromBlue;
var incrementPercentage = 1 / (numberOfColours-1); //how much to increment percentage by
for (var n = 0; n < numberOfColours; n++){
var percentage = n * incrementPercentage; //calculate percentage
var red = (difRed * percentage + fromRed).toFixed(0); //round em for legibility
var green = (difGreen * percentage + fromGreen).toFixed(0);
var blue = (difBlue * percentage + fromBlue).toFixed(0);
var colour = 'rgba(' + red + ',' + green + ',' + blue + ',1)'; //create string literal
colours.push(colour); //push home
}
return colours;
}
function getRGBAValues(string) {
var cleaned = string.substring(string.indexOf('(') +1, string.length-1);
var split = cleaned.split(",");
var intValues = [];
for(var index in split){
intValues.push(parseInt(split[index]));
}
return intValues;
}
function hexToRGBA(hex, alpha){
var c;
if(/^#([A-Fa-f0-9]{3}){1,2}$/.test(hex)){
c= hex.substring(1).split('');
if(c.length== 3){
c= [c[0], c[0], c[1], c[1], c[2], c[2]];
}
c= '0x'+c.join('');
return 'rgba('+[(c>>16)&255, (c>>8)&255, c&255].join(',')+','+alpha+')';
}
return rgba(0,0,0,1);
//throw new Error('Bad Hex');
}
共有三个功能:
coloursBetween(fromColour, toColour, numberOfColours)
getRGBAValues(string)
hexToRGBA(hex, alpha)
调用主函数 coloursBetween()
传入起始颜色和结束颜色,以及您想要返回的颜色总数。因此,如果您请求返回 10 种颜色,您会得到第一种颜色 + 8 种渐变颜色 + 最后一种颜色。
coloursBetween 函数首先将传入的十六进制颜色(例如#FFFFFF、#000000)转换为 rgba(例如 rgba(255,255,255,1) rgba(0,0,0,1)),然后减去红色、绿色和来自每个的蓝色值。
然后计算红色、绿色和蓝色之间的差异。在本例中,每种情况下都是 -255。计算增量并用于乘以红色、绿色和蓝色的新增量值。 Alpha 始终假定为 1(完全不透明)。然后将新值添加到颜色数组中,并在 for 循环完成后返回。
最后,像这样调用(从红色到蓝色):
var gradientColours = coloursBetween("#FF0000", "#0000FF", 5);
答案 13 :(得分:0)
这是一个可以满足您要求的脚本: https://gist.github.com/av01d/538b3fffc78fdc273894d173a83c563f
非常易于使用:
let colors;
colors = ColorSteps.getColorSteps('#000', 'rgba(255,0,0,0.1)', 10);
colors = ColorSteps.getColorSteps('red', 'blue', 5);
colors = ColorSteps.getColorSteps('hsl(180, 50%, 50%)', 'rgba(200,100,20,0.5)', 10);