Ant*_*lev 118 javascript jquery numbers
我想基于当天动态生成一串文本.所以,例如,如果它是第1天,那么我希望我的代码生成="它的<dynamic> 1*<dynamic string> st </ dynamic string>*</ dynamic>".
共有12天,所以我做了以下几点:
我已经建立了一个for循环,循环了12天.
在我的html中,我给了我的元素一个唯一的id来定位它,见下文:
<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 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"
}
UPDATE
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) {
}
}
}
Sal*_*n A 305
该规则如下:
- st用于以1结尾的数字(例如,1st,发音为first)
- nd用于以2结尾的数字(例如92nd,发音为ninety-second)
- rd用于以3结尾的数字(例如33,发音为第三十三)
- 作为上述规则的一个例外,所有以11,12或13结尾的"青少年"数字都使用-th(例如,第11,发音第11,第112,发音为100 [和]第12)
- th用于所有其他数字(例如,第9,发音第9).
以下JavaScript代码(在2014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
Fiz*_*han 201
来自Shopify
function getNumberWithOrdinal(n) {
var s=["th","st","nd","rd"],
v=n%100;
return n+(s[(v-20)%10]||s[v]||s[0]);
}
Кон*_*Ван 38
Intl.PluralRules，标准方法。我只想在这里放弃执行此操作的规范方式，因为似乎没有人知道。
如果你希望你的代码是
? 这是要走的路。
const english_ordinal_rules = new Intl.PluralRules("en", {type: "ordinal"});
const suffixes = {
one: "st",
two: "nd",
few: "rd",
other: "th"
};
function ordinal(number) {
const suffix = suffixes[english_ordinal_rules.select(number)];
return (number + suffix);
}
const test = Array(201)
.fill()
.map((_, index) => index - 100)
.map(ordinal)
.join(" ");
console.log(test);Tom*_*aas 35
function nth(n){return["st","nd","rd"][((n+90)%100-10)%10-1]||"th"}
(这是正整数,见下面的其他变化)
从带有后缀的数组开始["st", "nd", "rd"].我们想要将以1,2,3结尾的整数(但不是以11,12,13结尾)映射到索引0,1,2.
其他整数(包括那些以11,12,13结尾的整数)可以映射到其他任何东西 - 数组中找不到的索引将评估为undefined.这在javascript中是假的,并且使用逻辑或(|| "th")表达式将返回"th"这些整数,这正是我们想要的.
表达式((n + 90) % 100 - 10) % 10 - 1执行映射.打破它:
(n + 90) % 100:此表达式采用输入整数 - 10 mod 100,映射10到0,... 99到89,0到90,...,9到99.现在以11,12,13结尾的整数位于较低位置结束(映射到1,2,3).- 10:现在10映射到-10,19到-1,99到79,0到80,... 9到89.以11,12,13结尾的整数映射到负整数(-9,-8, -7).% 10:现在所有以1,2或3结尾的整数都映射到1,2,3.所有其他整数都映射到其他整数(11,12,13仍然映射到-9,-8,-7).- 1:减去一个给出1,2,3的最终映射到0,1,2.function nth(n){return["st","nd","rd"][((n+90)%100-10)%10-1]||"th"}
//test integers from 1 to 124
for(var r = [], i = 1; i < 125; i++) r.push(i + nth(i));
//output result
document.getElementById('result').innerHTML = r.join('<br>');<div id="result"></div>允许负整数:
function nth(n){return["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"}
在ES6胖箭头语法(匿名函数)中:
n=>["st","nd","rd"][(((n<0?-n:n)+90)%100-10)%10-1]||"th"
表达式是正整数的更短的替代方案
[,'st','nd','rd'][n%100>>3^1&&n%10]||'th'
请参阅此帖子以获得解释.
The*_*lis 18
您可以使用矩库本地数据函数。
代码：
moment.localeData().ordinal(1)
//1st
你只有12天？我很想把它变成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'
要么
var i = 8;
var day = i + suffixes[i]; // result: '8th'
小智 7
通过将数字拆分成数组并反转,我们可以使用array[0]和轻松检查数字的最后2位数array[1].
如果一个数字在十几岁,array[1] = 1它需要"th".
function getDaySuffix(num)
{
var array = ("" + num).split("").reverse(); // E.g. 123 = array("3","2","1")
if (array[1] != "1") { // Number is in the teens
switch (array[0]) {
case "1": return "st";
case "2": return "nd";
case "3": return "rd";
}
}
return "th";
}
小智 5
function getSuffix(n) {return n < 11 || n > 13 ? ['st', 'nd', 'rd', 'th'][Math.min((n - 1) % 10, 3)] : 'th'}
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