我想动态生成基于当前一天的文本字符串。所以,例如,如果它是第1天,那么我想我的代码生成=“它的<动态>1*<动态字符串>st</动态字符串>*</动态>”。

总共有12天,所以我做了以下事情:

我设置了一个for循环,循环12天。 在我的html中,我已经给了我的元素一个唯一的id来瞄准它,如下所示: <h1 id="dynamicTitle" class="CustomFont leftHeading shadow">On The <span></span> <em>of rest of generic text</em></h1> 然后,在我的for循环中,我有以下代码: $ (" # dynamicTitle跨度”). html(我); Var day = i; If (day == 1) { Day = I +“st”; } else if (day == 2) { 日= I + "nd" } else if (day == 3) { Day = I + "rd" }

更新

这是请求的整个for循环:

$(document).ready(function () {
    for (i = 1; i <= 12; i++) {
        var classy = "";
        if (daysTilDate(i + 19) > 0) {
            classy = "future";
            $("#Day" + i).addClass(classy);
            $("#mainHeading").html("");
            $("#title").html("");
            $("#description").html("");
        } else if (daysTilDate(i + 19) < 0) {
            classy = "past";
            $("#Day" + i).addClass(classy);
            $("#title").html("");
            $("#description").html("");
            $("#mainHeading").html("");
            $(".cta").css('display', 'none');
            $("#Day" + i + " .prizeLink").attr("href", "" + i + ".html");
        } else {
            classy = "current";
            $("#Day" + i).addClass(classy);
            $("#title").html(headings[i - 1]);
            $("#description").html(descriptions[i - 1]);
            $(".cta").css('display', 'block');
            $("#dynamicImage").attr("src", ".." + i + ".jpg");
            $("#mainHeading").html("");
            $(".claimPrize").attr("href", "" + i + ".html");
            $("#dynamicTitle span").html(i);
            var day = i;
            if (day == 1) {
                day = i + "st";
            } else if (day == 2) {
                day = i + "nd"
            } else if (day == 3) {
                day = i + "rd"
            } else if (day) {
            }
        }
    }

当前回答

Intl。PluralRules,标准方法。

我只是想在这里放弃做这个的规范方法,因为似乎没有人知道它。不要白费力气。

如果你想让你的代码

自我记录 易于本地化 用现代标准

-这才是正确的方法。

const english_ordinal_rules = new Intl。PluralRules("en", {type: "ordinal"}); Const后缀= { 一:“圣”, 二:“和”, 一些:“路”, 其他:“th” }; 函数序数(number/*: number */) { Const category = english_ordinal_rules.select(number); Const suffix =后缀[类别]; 返回(数字+后缀); } // ->字符串 const test =数组(201) .fill () .map((_, index) => index - 100) . map(顺序) . join (" "); console.log(测试);

Intl。PluralRules构造函数(ECMA-402草案) 统一码的六个多元类别


Code-golf

虽然我不建议用你的代码打高尔夫,杀死可读性,但我为那些高尔夫球手想出了一个(92字节):

n=>n+{e:"st",o:"nd",w:"rd",h:"th"}[new Intl.PluralRules("en",{type:"ordinal"}).select(n)[2]]

其他回答

规则如下:

St用于以1结尾的数字(例如1st,发音为first) Nd与以2结尾的数字连用(例如92nd,发音为ninety-second) Rd与以3结尾的数字连用(例如33rd,发音为33third) 作为上述规则的例外,所有以11、12或13结尾的“青少年”数字都使用-th(例如,11,发音为111,112, 读作一百零十二位) Th用于所有其他数字(例如:9th,发音为ninth)。

以下JavaScript代码(在14年6月重写)实现了这一点:

function ordinal_suffix_of(i) {
    var j = i % 10,
        k = i % 100;
    if (j == 1 && k != 11) {
        return i + "st";
    }
    if (j == 2 && k != 12) {
        return i + "nd";
    }
    if (j == 3 && k != 13) {
        return i + "rd";
    }
    return i + "th";
}

0-115之间数字的示例输出:

  0  0th
  1  1st
  2  2nd
  3  3rd
  4  4th
  5  5th
  6  6th
  7  7th
  8  8th
  9  9th
 10  10th
 11  11th
 12  12th
 13  13th
 14  14th
 15  15th
 16  16th
 17  17th
 18  18th
 19  19th
 20  20th
 21  21st
 22  22nd
 23  23rd
 24  24th
 25  25th
 26  26th
 27  27th
 28  28th
 29  29th
 30  30th
 31  31st
 32  32nd
 33  33rd
 34  34th
 35  35th
 36  36th
 37  37th
 38  38th
 39  39th
 40  40th
 41  41st
 42  42nd
 43  43rd
 44  44th
 45  45th
 46  46th
 47  47th
 48  48th
 49  49th
 50  50th
 51  51st
 52  52nd
 53  53rd
 54  54th
 55  55th
 56  56th
 57  57th
 58  58th
 59  59th
 60  60th
 61  61st
 62  62nd
 63  63rd
 64  64th
 65  65th
 66  66th
 67  67th
 68  68th
 69  69th
 70  70th
 71  71st
 72  72nd
 73  73rd
 74  74th
 75  75th
 76  76th
 77  77th
 78  78th
 79  79th
 80  80th
 81  81st
 82  82nd
 83  83rd
 84  84th
 85  85th
 86  86th
 87  87th
 88  88th
 89  89th
 90  90th
 91  91st
 92  92nd
 93  93rd
 94  94th
 95  95th
 96  96th
 97  97th
 98  98th
 99  99th
100  100th
101  101st
102  102nd
103  103rd
104  104th
105  105th
106  106th
107  107th
108  108th
109  109th
110  110th
111  111th
112  112th
113  113th
114  114th
115  115th

你只有12天?我想让它只是一个简单的查找数组:

var suffixes = ['','st','nd','rd','th','th','th','th','th','th','th','th','th'];

然后

var i = 2;
var day = i + suffixes[i]; // result: '2nd'

or

var i = 8;
var day = i + suffixes[i]; // result: '8th'

我强烈推荐这本书,它超级简单易懂。希望有帮助?

它避免使用负整数,即小于1的数字并返回false 如果输入为0,则返回0

function numberToOrdinal(n) {

  let result;

  if(n < 0){
    return false;
  }else if(n === 0){
    result = "0";
  }else if(n > 0){

    let nToString = n.toString();
    let lastStringIndex = nToString.length-1;
    let lastStringElement = nToString[lastStringIndex];

    if( lastStringElement == "1" && n % 100 !== 11 ){
      result = nToString + "st";
    }else if( lastStringElement == "2" && n % 100 !== 12 ){
      result = nToString + "nd";
    }else if( lastStringElement == "3" && n % 100 !== 13 ){
      result = nToString + "rd";
    }else{
      result = nToString + "th";
    }

  }

  return result;
}

console.log(numberToOrdinal(-111));
console.log(numberToOrdinal(0));
console.log(numberToOrdinal(11));
console.log(numberToOrdinal(15));
console.log(numberToOrdinal(21));
console.log(numberToOrdinal(32));
console.log(numberToOrdinal(43));
console.log(numberToOrdinal(70));
console.log(numberToOrdinal(111));
console.log(numberToOrdinal(300));
console.log(numberToOrdinal(101));

输出

false
0
11th
15th
21st
32nd
43rd
70th
111th
300th
101st

我为更大的数字和所有测试用例写了这个函数

function numberToOrdinal(num) {
    if (num === 0) {
        return '0'
    };
    let i = num.toString(), j = i.slice(i.length - 2), k = i.slice(i.length - 1);
    if (j >= 10 && j <= 20) {
        return (i + 'th')
    } else if (j > 20 && j < 100) {
        if (k == 1) {
            return (i + 'st')
        } else if (k == 2) {
            return (i + 'nd')
        } else if (k == 3) {
            return (i + 'rd')
        } else {
            return (i + 'th')
        }
    } else if (j == 1) {
        return (i + 'st')
    } else if (j == 2) {
        return (i + 'nd')
    } else if (j == 3) {
        return (i + 'rd')
    } else {
        return (i + 'th')
    }
}

Intl。PluralRules,标准方法。

我只是想在这里放弃做这个的规范方法,因为似乎没有人知道它。不要白费力气。

如果你想让你的代码

自我记录 易于本地化 用现代标准

-这才是正确的方法。

const english_ordinal_rules = new Intl。PluralRules("en", {type: "ordinal"}); Const后缀= { 一:“圣”, 二:“和”, 一些:“路”, 其他:“th” }; 函数序数(number/*: number */) { Const category = english_ordinal_rules.select(number); Const suffix =后缀[类别]; 返回(数字+后缀); } // ->字符串 const test =数组(201) .fill () .map((_, index) => index - 100) . map(顺序) . join (" "); console.log(测试);

Intl。PluralRules构造函数(ECMA-402草案) 统一码的六个多元类别


Code-golf

虽然我不建议用你的代码打高尔夫,杀死可读性,但我为那些高尔夫球手想出了一个(92字节):

n=>n+{e:"st",o:"nd",w:"rd",h:"th"}[new Intl.PluralRules("en",{type:"ordinal"}).select(n)[2]]