给定这个函数,我想用随机颜色生成器替换颜色。
document.overlay = GPolyline.fromEncoded({
color: "#0000FF",
weight: 10,
points: encoded_points,
zoomFactor: 32,
levels: encoded_levels,
numLevels: 4
});
我该怎么做呢?
给定这个函数,我想用随机颜色生成器替换颜色。
document.overlay = GPolyline.fromEncoded({
color: "#0000FF",
weight: 10,
points: encoded_points,
zoomFactor: 32,
levels: encoded_levels,
numLevels: 4
});
我该怎么做呢?
当前回答
可能是最简单的
'#' + Math.random().toString(16).substring(9)
其他回答
下面是对这个问题的另一种看法。
我的目标是创造充满活力和独特的颜色。为了确保颜色的不同,我避免使用随机生成器,并从彩虹中选择“均匀间隔”的颜色。
这非常适合在谷歌map中创建具有最佳“唯一性”(即没有两个标记具有相似的颜色)的弹出式标记。
/**
* @param numOfSteps: Total number steps to get color, means total colors
* @param step: The step number, means the order of the color
*/
function rainbow(numOfSteps, step) {
// This function generates vibrant, "evenly spaced" colours (i.e. no clustering). This is ideal for creating easily distinguishable vibrant markers in Google Maps and other apps.
// Adam Cole, 2011-Sept-14
// HSV to RBG adapted from: http://mjijackson.com/2008/02/rgb-to-hsl-and-rgb-to-hsv-color-model-conversion-algorithms-in-javascript
var r, g, b;
var h = step / numOfSteps;
var i = ~~(h * 6);
var f = h * 6 - i;
var q = 1 - f;
switch(i % 6){
case 0: r = 1; g = f; b = 0; break;
case 1: r = q; g = 1; b = 0; break;
case 2: r = 0; g = 1; b = f; break;
case 3: r = 0; g = q; b = 1; break;
case 4: r = f; g = 0; b = 1; break;
case 5: r = 1; g = 0; b = q; break;
}
var c = "#" + ("00" + (~ ~(r * 255)).toString(16)).slice(-2) + ("00" + (~ ~(g * 255)).toString(16)).slice(-2) + ("00" + (~ ~(b * 255)).toString(16)).slice(-2);
return (c);
}
如果您希望看到这看起来像在行动,请参阅简单JavaScript彩虹颜色生成器谷歌地图标记。
有很多方法可以做到这一点。以下是我做的一些:
简短的一行代码,保证有效的颜色
'#'+(Math.random().toString(16)+'00000').slice(2,8)
生成6个随机十六进制数字(0-F)
function randColor() {
for (var i=0, col=''; i<6; i++) {
col += (Math.random()*16|0).toString(16);
}
return '#'+col;
}
// ES6 one-liner version
[..."000000"].map(()=>Math.random().toString(16)[2]).join("")
生成单独的HEX组件(00-FF)
function randColor2() {
var r = ('0'+(Math.random()*256|0).toString(16)).slice(-2),
g = ('0'+(Math.random()*256|0).toString(16)).slice(-2),
b = ('0'+(Math.random()*256|0).toString(16)).slice(-2);
return '#' +r+g+b;
}
过度设计的十六进制字符串(XORs 3输出一起形成颜色)
function randColor3() {
var str = Math.random().toString(16) + Math.random().toString(16),
sg = str.replace(/0./g,'').match(/.{1,6}/g),
col = parseInt(sg[0], 16) ^
parseInt(sg[1], 16) ^
parseInt(sg[2], 16);
return '#' + ("000000" + col.toString(16)).slice(-6);
}
我认为第一个回答是最简洁/有用的,但我只是写了一个初学者可能更容易理解的回答。
function randomHexColor(){
var hexColor=[]; //new Array()
hexColor[0] = "#"; //first value of array needs to be hash tag for hex color val, could also prepend this later
for (i = 1; i < 7; i++)
{
var x = Math.floor((Math.random()*16)); //Tricky: Hex has 16 numbers, but 0 is one of them
if (x >=10 && x <= 15) //hex:0123456789ABCDEF, this takes care of last 6
{
switch(x)
{
case 10: x="a"
break;
case 11: x="b"
break;
case 12: x="c"
break;
case 13: x="d"
break;
case 14: x="e"
break;
case 15: x="f"
break;
}
}
hexColor[i] = x;
}
var cString = hexColor.join(""); //this argument for join method ensures there will be no separation with a comma
return cString;
}
我喜欢这个:'#' + (Math.random().toString(16) + "000000").substring(2,8)
我怀疑还有什么能比这条更快或更短:
"#" + ((1 << 24) * Math.random() | 0).toString(16).padStart(6, "0")
挑战!