JavaScript:对数组排序

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未排序的数组[input]:

["> 30 days", "< 7 days", "< 30 days", "< 10 days"];

数组中元素的格式始终类似于:</>X天

The format of an elements in array will always be like : </> X days

要求:

上面提到的数组应按照 大然后(>) 和 小然后符号(<)进行排序，并牢记天数(应减少天数).

Above mentioned array should be sorted as per the greater then (>) and lesser then symbol (<) and also keep the number of days in mind (less number of days should come first).

期望的数组[输出]:

["< 7 days", "< 10 days", "< 30 days", "> 30 days"];

到目前为止已尝试:

我尝试了 Array.sort() 函数，但未获得预期的输出.

I tried Array.sort() function but did not get expected output.

var arr = ["> 30 days", "< 7 days", "< 30 days", "< 10 days"]; var sortedArr = arr.sort(); console.log(sortedArr); // ["< 30 days", "< 10 days", "< 7 days", "> 30 days"]

推荐答案

您可以按数字排序，如果有比较符号可用，则为相同的数值获取两个偏移的增量，以反映比较的顺序.

You could sort by numbers and if a comparison sign is available, then take for the same numerical value the delta of both offsets, which reflects the order of comparison.

var array = ["> 30 days", "< 7 days", "30 days", "< 10 days", "> 10 days", "< 11 days", "42 days", ">= 42 days", "> 42 days", "<= 42 days", "< 42 days"]; array.sort(function (a, b) { function getV(s) { return { value: s.match(/\d+/), offset: { '<': -2, '<=': -1, null: 0, '>=': 1, '>': 2 }[s.match(/[<>]={0,1}(?=\s*\d)/)] }; } var aa = getV(a), bb = getV(b); return aa.value - bb.value || aa.offset - bb.offset; }); console.log(array);

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第一个/以前的解决方案

First/former solution

您可以解析字符串，并将其用于对数字应用更正值-然后进行相应的排序.

You could parse the string and use it for applying a correction value to the number - then sort accordingly.

{ '>': 0.1, '<': -0.1, null: 0 }[s.match(/[<>](?=\s*\d)/)] + +s.match(/\d+/)

具有校正值的对象

{ '>': 0.1, '<': -0.1, null: 0 }

使用正则表达式获取符号并将其用作对象的键

get the sign with a regular expression and use it as key for the object

s.match(/[<>](?=\s*\d)/)

然后添加字符串的数字部分

then add the number part of the string

+ +s.match(/\d+/)

var array = ["> 30 days", "< 7 days", "30 days", "< 10 days"]; array.sort(function (a, b) { function getV(s) { return { '>': 0.1, '<': -0.1, null: 0 }[s.match(/[<>](?=\s*\d)/)] + +s.match(/\d+/); } return getV(a) - getV(b); }); console.log(array);