[removed] Ordinal suffix for numbers with specific

问题:

I am trying to display numbers within a particular table with ordinal suffixes. The table always shows three numbers which come from an XML file. The numbers show ranks, so for example they may be 6th, 120th, 131st. The output is a table that would look like this:

<table>
    <tr>
        <td class='ordinal'>6</td>
        <td class='ordinal'>120</td>
        <td class='ordinal'>131</td>
    </tr>
</table>

I would ideally like to use javascript and I found a few very good solutions on stackoverflow, for example this one. However, I am struggling to apply the function to all numbers within the table, rather than putting in each number individually. I tried using a CSS class so that my function looks like this:

<script type="text/javascript">
$(function(){
    $(".ordinal").each(function(){
        var j = i % 10;
        if (j == 1 && i != 11) {
            return i + "st";
        }
        if (j == 2 && i != 12) {
            return i + "nd";
        }
        if (j == 3 && i != 13) {
            return i + "rd";
        }
        return i + "th";
        });
})
</script>

but it's not working, probably because I screwed up the code somewhere. Maybe somebody here can help me out and tell me where I went wrong?

Thank you very much for your help!

回答1:

My own suggestion, would be:

$(".ordinal").text(function (i, t) {
    i++;
    var str = i.toString().slice(-1),
        ord = '';
    switch (str) {
        case '1':
            ord = 'st';
            break;
        case '2':
            ord = 'nd';
            break;
        case '3':
            ord = 'rd';
            break;
        case '4':
        case '5':
        case '6':
        case '7':
        case '8':
        case '9':
        case '0':
            ord = 'th';
            break;
    }
    return i + ord;
});

JS Fiddle demo.

This effectively takes the incremented number (i++, in order to start from 1 not 0), converts it to a string, then looks at the last number of that string. This should work for any number, since the ordinal is based purely on that last number.

You could also extend the Number prototype to implement this functionality:

Number.prototype.ordinate = function(){
    var num = this + 1,
        last = num.toString().slice(-1),
        ord = '';
    switch (last) {
        case '1':
            ord = 'st';
            break;
        case '2':
            ord = 'nd';
            break;
        case '3':
            ord = 'rd';
            break;
        case '4':
        case '5':
        case '6':
        case '7':
        case '8':
        case '9':
        case '0':
            ord = 'th';
            break;
    }
    return num.toString() + ord;
};

$(".ordinal").text(function (i, t) {
    return i.ordinate();
});

JS Fiddle demo.

Edited to offer a slight alternative:

Number.prototype.ordinate = function(){
    var num = this,
        last = num.toString().slice(-1),
        ord = '';
    switch (last) {
        case '1':
            ord = 'st';
            break;
        case '2':
            ord = 'nd';
            break;
        case '3':
            ord = 'rd';
            break;
        default:
            ord = 'th';
            break;
    }
    return num.toString() + ord;
};

$(".ordinal").text(function (i,t) {
    return t.replace(/(\d+)/g, function(a){
        return parseInt(a, 10).ordinate();
    });
});

JS Fiddle demo.

This essentially iterates over each .ordinal element, replacing the numbers present with the (same) numbers with the ordinal suffix added to it.

---

Edited to address the problem, raised in the comments, below, that 11, 12 and 13 were receiving the ordinal suffix of st, nd and rd (respectively). This is now corrected to being th in all cases:

Number.prototype.ordinate = function(){
    var num = this,
        numStr = num.toString(),
        last = numStr.slice(-1),
        len = numStr.length,
        ord = '';
    switch (last) {
        case '1':
            ord = numStr.slice(-2) === '11' ? 'th' : 'st';
            break;
        case '2':
            ord = numStr.slice(-2) === '12' ? 'th' : 'nd';
            break;
        case '3':
            ord = numStr.slice(-2) === '13' ? 'th' : 'rd';
            break;
        default:
            ord = 'th';
            break;
    }
    return num.toString() + ord;
};

JS Fiddle demo.

References:

• slice().
• switch().
• text().
• toString().

回答2:

function nth(n){
    if(isNaN(n) || n%1) return n;   
    var s= n%100;
    if(s>3 && s<21) return n+'th';
    switch(s%10){
        case 1: return n+'st';
        case 2: return n+'nd';
        case 3: return n+'rd';
        default: return n+'th';
    }
}

You can take care of the teens in their own line, other integers follow the last digit rules.

回答3:

I created two approaches one using Prototype, the other as a plugin :

Number.prototype.between = function(n,m){ return this > n && this < m }
Number.prototype.ORDINATE_INDEX = ["th","st","nd","rd"];
Number.prototype.toOrdinate = function(){
    var
        nthMod = (this % 10),
        index =  nthMod > 3 || this.between(10,20) ? 0 : nthMod
    ;

    return this + this.ORDINATE_INDEX[index];
};

$(".ordinal").text(function (index, element) {
    return parseInt(element).toOrdinate();
});

This is the one as a Jquery Plugin :

(function($){
    var numberTool = new (function(){
        var private = {};

        private.ORDINATE_INDEX = ["th","st","nd","rd"];

        private.parseOrdinary = function(number)
        {
            var
                nthMod = (number % 10),
                index =  nthMod > 3 || private.between(number, 10,20) ? 0 : nthMod
            ;

            return number + private.ORDINATE_INDEX[index];
        }

        private.between = function(number, n,m){
            return number > n && number < m
        }

        this.isNumeric = function(number)
        {
            return !isNaN(parseFloat(number)) && isFinite(number);
        }

        this.toOrdinary = function(number)
        {
            return this.isNumeric(number) ? private.parseOrdinary(number) : number;
        }
    });


    $.fn.toOrdinary = function(){
        return this.each(function(){
            $element = $(this);
            $element.text(numberTool.toOrdinary($element.text()));
        }); 
    };
})(jQuery);

$(".ordinal").toOrdinary();
$(".ordinal").toOrdinary();
$(".ordinal").toOrdinary();

Tested on JSFIDDLE:

The prototype version example : http://jsfiddle.net/f8vQr/6/

The JQuery version example : http://jsfiddle.net/wCdKX/27/

回答4:

It's not working because you're returning the strings to $.each, not actually using them. Usage would depend on your HTML but here is an example of setting the .ordinal text to the value.

You were also missing the i parameter on the event handler and you can increment i to start from 1st instead of 0th.

jsFiddle

$(".ordinal").each(function (i) {
    i++;
    var j = i % 10,
        str;
    if (j == 1 && i != 11) {
        str = i + "st";
    } else if (j == 2 && i != 12) {
        str = i + "nd";
    } else if (j == 3 && i != 13) {
        str = i + "rd";
    } else {
        str = i + "th";
    }
    $(this).text(str);
});

If you have the numbers in your elements already then it would be best to not rely on the index and instead check the number, then append the string to the end.

jsFiddle

$(document).ready(function () {
    $(".ordinal").each(function (i) {
        var j = parseInt($('ordinal').text(), 10) % 10,
            str;
        if (j == 1 && i != 11) {
            str = "st";
        } else if (j == 2 && i != 12) {
            str = "nd";
        } else if (j == 3 && i != 13) {
            str = "rd";
        } else {
            str = "th";
        }
        var elem = $(this)
        elem.text(elem.text() + str);
    });
});

回答5:

Ordinal suffix in one line

var integerWithSuffix=originalInteger+(['st','nd','rd'][( originalInteger +'').match(/1?\d\b/)-1]||'th');

the concatenation of the original number and a string representing the ordinal derived from an array indexed by the result of a regex search on that number

http://jsfiddle.net/thisishardcoded/DbSMB/

回答6:

I would do something like this, based on David Thomas's answer:

Number.prototype.ordinate = function(){
    var num = this,
        ones = num % 10, //gets the last digit
        tens = num % 100, //gets the last two digits
        ord = ["st","nd","rd"][ tens > 10 && tens < 20 ? null : ones-1 ] || 'th';
    return num.toString() + ord;
};

It accomplishes the same thing. If a number's last 2 digits are within the 11-19 range OR the last digit is between 4-0 it defaults to 'th', otherwise it will pull a 'st', 'nd' or 'rd' out of the array based on the ones place.

I like the idea of creating a prototype function very much but I would definitely leave the incrementation of the index outside of the prototype function to make it more versatile:

$(".ordinal").text(function (i, t) {
    return (++i).ordinate();
});

JS Fiddle Demo

回答7:

function ordsfx(a){return["th","st","nd","rd"][(a=~~(a<0?-a:a)%100)>10&&a<14||(a%=10)>3?0:a]}

See annotated version at https://gist.github.com/furf/986113#file-annotated-js

Short, sweet, and efficient, just like utility functions should be. Works with any signed/unsigned integer/float. (Even though I can't imagine a need to ordinalize floats)

回答8:

This is what I use, and it works for any year, month, day (leap year) included:

// panelyear, panelmonth and panelday are passed as parameters

var PY01 = parseInt(panelyear); var PM01 = (parseInt(panelmonth) - 1); PD01 = parseInt(panelday);

var now = new Date(PY01, PM01, PD01);
var start = new Date(PY01, 0, 0);
var diff = (now - start) + ((start.getTimezoneOffset() - now.getTimezoneOffset()) * 60 * 1000);
var oneDay = 1000 * 60 * 60 * 24;
var day = Math.floor(diff / oneDay);

function getNumberWithOrdinal(n) { var s = ["th","st","nd","rd"], v = n % 100; return n + (s[(v - 20) % 10] || s[v] || s[0]); }

Use with

<script> document.write(getNumberWithOrdinal(day)); </script>