Javascript 如何使此代码更具功能性和可读性?
如何使这些javascript语句看起来更具可读性。可以使用函数库ramda.js使代码看起来更好吗Javascript 如何使此代码更具功能性和可读性?,javascript,functional-programming,ramda.js,Javascript,Functional Programming,Ramda.js,如何使这些javascript语句看起来更具可读性。可以使用函数库ramda.js使代码看起来更好吗 var getTextSpace = function(len) { var tlength; if (len >= 1 && len <= 4) { tlength = 10; } else if (len === 5) {
var getTextSpace = function(len)
{
var tlength;
if (len >= 1 && len <= 4) {
tlength = 10;
} else if (len === 5) {
tlength = 14;
} else if (len === 6) {
tlength = 16;
} else if (len === 7) {
tlength = 18;
} else if (len >= 8 && len <= 10) {
tlength = 20;
} else if (len === 11) {
tlength = 22;
} else if (len === 12) {
tlength = 24;
} else if (len >= 13 && len <= 15) {
tlength = 26;
} else if (len === 16) {
tlength = 28;
} else if (len >= 17 && len <= 20) {
tlength = 32;
} else if (len >= 21 && len <= 34) {
tlength = tlength * 2;
} else if (len >= 35 && len <= 80) {
tlength = Math.round((len + len / 100 * 50));
}
else {
tlength = Math.round((len + len / 100 * 30));
}
return tlength;
};
也许
开关
是另一种选择。有一篇文章的主题与此相似
请向右看。您可以使用switch语句来避免所有其他if语句 此外,如果len始终为整数,则可以将TLENNGTS放入索引与len值匹配的数组中:
var getTextSpace = function(len) {
var tlengthArray = [10,10,10,10,14,16,18,20,20,20,22,24,26,26,26,28,32,32,32,32, len*2, Math.round((len + len / 100 * 50)), Math.round((len + len / 100 * 50))];
var tlength;
if (len >= 1 && len <=20) {
tlength = tlengthArray[len-1];
}
else if (len >= 21 && len <= 34) {
tlength = tlengthArray[20];
}
else if (len >= 35 && len <= 80) {
tlength = tlengthArray[21];
}
else {
tlength = tlengthArray[22];
}
return tlength;
}
var getTextSpace=function(len){
变量数组=[10,10,10,14,16,18,20,20,20,22,24,26,26,26,28,32,32,32,len*2,数学圆((len+len/100*50)),数学圆((len+len/100*50));
变长;
如果(len>=1&&len=21&&len=35&&len
函数getTextSpace(len){
//如果len在某个范围内,则返回该长度
变量映射=[
[1, 4, 10],
[5, 5, 14],
[6, 6, 16],
[7, 7, 18],
[8, 10, 20],
[11, 11, 22],
[12, 12, 24],
[13, 15, 26],
[16, 16, 28],
[17, 20, 32]
];
对于(变量i=0;i 如果(len>=范围[0]&&lenRamda可能会有所帮助。但最主要的是以可读的方式构造您的范围。下面的代码假设输入值是整数,您不需要测试其他数字类型。这些都可以完成,但是您需要比这里简单的之间的更复杂的东西。您需要iple函数或一种配置函数的方法,以确定开始和结束是包含的还是独占的
var getTextSpace = (function() {
// :: (Int, Int) -> (Int -> Bool)
var between = (begin, end) => R.both(R.gte(R.__, begin), R.lt(R.__, end));
return R.cond([
[between(1, 5), R.always(10)],
[between(5, 6), R.always(14)],
[between(6, 7), R.always(16)],
[between(7, 8), R.always(18)],
[between(8, 11), R.always(20)],
[between(11, 12), R.always(22)],
[between(12, 13), R.always(24)],
[between(13, 16), R.always(26)],
[between(16, 17), R.always(28)],
[between(17, 21), R.always(32)],
[between(21, 35), R.multiply(2)], // assuming original was typo
[between(35, 80), len => Math.round(len + len / 100 * 50)],
[R.T, len => Math.round(len + len / 100 * 30)]
]);
}());
(在最初的案例中似乎存在缺陷:
} else if (len >= 21 && len <= 34) {
tlength = tlength * 2;
}else if(len>=21&&len=21&&lenRamda非常实用,这意味着它的最佳用途是使用尽可能多的声明性和纯函数(泛型函数,可以在许多地方使用,而不仅仅是您的代码)。我的建议类似于以下代码:
var getTextSpace = function (len) {
var conds = [
{range: [1, 4], result: 10},
{range: [5, 5], result: 14},
{range: [6, 6], result: 16},
{range: [7, 7], result: 18},
{range: [8, 10], result: 20},
{range: [11, 11], result: 22},
{range: [12, 12], result: 24},
{range: [13, 15], result: 26},
{range: [16, 16], result: 28},
{range: [17, 20], result: 32},
{range: [21, 34], result: len * 2}, // You wrote tlength * 2 but it's not defined yet so I asumed you ment len * 2
{range: [35, 80], result: Math.round((len + len / 100 * 50))}
];
var test = function (obj) {
var rangeLens = R.lensProp('range');
var range = R.view(rangeLens, obj);
var lte = R.curry(R.lte)(range[0]);
var gte = R.curry(R.gte)(range[1]);
return R.both(lte, gte)(len);
}
var resultLens = R.lensProp('result');
var getResult = R.curry(R.view)(resultLens);
var chosen = R.find(test)(conds);
var defIfNotFound = R.defaultTo( {result: Math.round((len + len / 100 * 30))} );
return getResult(defIfNotFound(chosen));
};
我试着给每个函数一个名称来解释它的功能,并将它们分成许多部分,这几乎就像读一个句子,如果你想要一个普通的JS解决方案,这可能是一个替代方案
const isBetween = x => (s, e) =>
(Number(s) <= Number(x) && Number(x) <= Number(e))
? true : false
const getTextSpace = len => {
const lenIsBetween = isBetween(len)
return lenIsBetween(1,4)? 10
: lenIsBetween(5, 5) ? 14
: lenIsBetween(6, 6) ? 16
: lenIsBetween(7, 7) ? 18
: lenIsBetween(8, 10) ? 20
: lenIsBetween(11, 11) ? 22
: lenIsBetween(12, 12) ? 24
: lenIsBetween(13, 15) ? 26
: lenIsBetween(16, 16) ? 28
: lenIsBetween(17, 20) ? 32
: lenIsBetween(21, 34) ? len * 2
: lenIsBetween(35, 80) ? Math.round((len + len / 100 * 50))
: Math.round((len + len / 100 * 30))
}
const isBetween=x=>(s,e)=>
(数字)我投票结束这个问题,因为它可能属于“确定”,但你不会得到任何答案。但我会尝试回来。如果(len>=21&&len是的,这是一个错误。证明代码很难阅读。可能会很容易做一些事情:value.between(2,20)。然后(20)。between(21,22).然后(0).更大的(25).然后(25)).更少的(30).然后(函数(值){返回值*20})…)对不起,不知道你的意思。谢谢你的ramda实现,也谢谢你找到了Bug
var getTextSpace = function (len) {
var conds = [
{range: [1, 4], result: 10},
{range: [5, 5], result: 14},
{range: [6, 6], result: 16},
{range: [7, 7], result: 18},
{range: [8, 10], result: 20},
{range: [11, 11], result: 22},
{range: [12, 12], result: 24},
{range: [13, 15], result: 26},
{range: [16, 16], result: 28},
{range: [17, 20], result: 32},
{range: [21, 34], result: len * 2}, // You wrote tlength * 2 but it's not defined yet so I asumed you ment len * 2
{range: [35, 80], result: Math.round((len + len / 100 * 50))}
];
var test = function (obj) {
var rangeLens = R.lensProp('range');
var range = R.view(rangeLens, obj);
var lte = R.curry(R.lte)(range[0]);
var gte = R.curry(R.gte)(range[1]);
return R.both(lte, gte)(len);
}
var resultLens = R.lensProp('result');
var getResult = R.curry(R.view)(resultLens);
var chosen = R.find(test)(conds);
var defIfNotFound = R.defaultTo( {result: Math.round((len + len / 100 * 30))} );
return getResult(defIfNotFound(chosen));
};
const isBetween = x => (s, e) =>
(Number(s) <= Number(x) && Number(x) <= Number(e))
? true : false
const getTextSpace = len => {
const lenIsBetween = isBetween(len)
return lenIsBetween(1,4)? 10
: lenIsBetween(5, 5) ? 14
: lenIsBetween(6, 6) ? 16
: lenIsBetween(7, 7) ? 18
: lenIsBetween(8, 10) ? 20
: lenIsBetween(11, 11) ? 22
: lenIsBetween(12, 12) ? 24
: lenIsBetween(13, 15) ? 26
: lenIsBetween(16, 16) ? 28
: lenIsBetween(17, 20) ? 32
: lenIsBetween(21, 34) ? len * 2
: lenIsBetween(35, 80) ? Math.round((len + len / 100 * 50))
: Math.round((len + len / 100 * 30))
}