如何使这些javascript语句看起来更具可读性。功能库ramda.js可以用来使这段代码看起来更好吗？
var getTextSpace = function(len)
{
var tlength;
if (len >= 1 && len <= 4) {
tlength = 10;
} else if (len === 5) {
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;
};
提前谢谢。
也许可以做一些允许这样做的事情？
value
.between(2,20).then(20)
.between(21,22).then(0)
.greater(25).then(25))
.less(30).then(function(value) {return value * 20 )})
答案 0 :(得分：6)
拉姆达可能会有所帮助。但主要的是以可读的方式构建范围。下面的代码假定输入值是整数，您不需要测试其他数字类型。这些可以做到，但是你需要比简单的between更复杂的东西。您需要多种功能或一种方法来配置一种功能，以确定开头和结尾各自是包含还是排他。
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)]
]);
}());
（在原案件中似乎存在错误：
} else if (len >= 21 && len <= 34) {
tlength = tlength * 2;
我认为是
} else if (len >= 21 && len <= 34) {
tlength = len * 2;
我在这里编写了相应的代码。）
您可以在 Ramda REPL 上看到这一点。
答案 1 :(得分：2)
let mapper = AWSDynamoDBObjectMapper.defaultDynamoDBObjectMapper()
let queryExpression = AWSDynamoDBQueryExpression()
/* this works
queryExpression.hashKeyValues = "1"
*/
// but I really want something like...
queryExpression.hashKeyValues = ["1", "5", "99"]
mapper.query(Person.self, expression: queryExpression) .continueWithBlock { (task) -> AnyObject! in
if (task.error != nil) {
print("Names from IDs lookup failed: \(task.error)")
} else if (task.result != nil) {
let people = task.result.items as! [Person]
for person in people {
print("---> \(person.name)")
}
} else {
print("Task result is nil")
}
return nil
}
function getTextSpace(len) {
// If len falls within a range then return that length
var map = [
[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]
];
for (var i = 0; i < map.length; i++) {
var range = map[i];
if (len >= range[0] && len <= range[1]) {
return range[2];
}
}
// We didn't find a range so return return calculation
// for the following ranges.
if (len >= 21 && len <= 34) {
return len * 2;
} else if (len >= 35 && len <= 80) {
return Math.round((len + len / 100 * 50));
}
// Return this calculation for everything else.
return Math.round((len + len / 100 * 30));
}
function test() {
var out = document.getElementById("out");
var text = "";
for (var i = 0; i < 100; i += 3) {
text += i + ": " + getTextSpace(i) + "\n";
}
out.innerHTML = text;
}
test();
答案 2 :(得分：1)
您可以使用switch语句来避免所有else if语句。
此外，如果len总是一个整数，你可以将tlength放在索引与len的值匹配的数组中：
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;
}
答案 3 :(得分：1)
Ramda非常实用，这意味着它的最佳用途是使用尽可能多的声明和纯函数（通用函数，可以在很多地方使用，而不仅仅是你的代码）。我的建议就像这段代码：
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));
};
我尝试给每个函数一个名称来解释它的作用，并将它们分成许多部分，这使得它几乎就像读一个句子
答案 4 :(得分：0)
也许switch是另一种选择。有一个帖子退出了类似的主题。
看起来正确here。
答案 5 :(得分：0)
如果您想要一个香草JS解决方案，这可能是另一种选择。
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))
}