正則表達式javascript

Question

是否有任何優雅的方式如何獲得在js /咖啡字格式的序數？ 事情是這樣的：

ordinalInWord(1) # => "first" 
ordinalInWord(2) # => "second" 
ordinalInWord(5) # => "fifth" 

Answer 1

恐怕序都是不正規足以避免他們每個人的打字了。

function ordinalInWord(cardinal) { 
    var ordinals = [ 'zeroth', 'first', 'second', 'third' /* and so on */ ]; 

    return ordinals[ cardinal ]; 
} 

如果您需要的功能，工作近20，你可以利用浮現的圖案：

function ordinalInWord(cardinal) { 
    var ordinals = [ 'zeroth', 'first', 'second', 'third' /* and so on up to "twentieth" */ ]; 
    var tens = { 
     20: 'twenty', 
     30: 'thirty', 
     40: 'forty' /* and so on up to 90 */ 
    }; 
    var ordinalTens = { 
     30: 'thirtieth', 
     40: 'fortieth', 
     50: 'fiftieth' /* and so on */ 
    }; 

    if(cardinal <= 20) {      
     return ordinals[ cardinal ]; 
    } 

    if(cardinal % 10 === 0) { 
     return ordinalTens[ cardinal ]; 
    } 

    return tens[ cardinal - (cardinal % 10) ] + ordinals[ cardinal % 10 ]; 
} 

演示：http://jsfiddle.net/AQCqK/

擴大該工作過去99不應該難。

Answer 2

下面是最高64位的解決方案。我無法想象你爲什麼需要它，但爲什麼不是問題。

var nth = function(i){ 
    var n = ["one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"]; 
    var s = ["zeroth", "first", "second", "third", "fourth", "fifth", "sixth", "seventh", "eighth", "ninth", "tenth", 
     "eleventh", "twelfth", "thirteenth", "fourteenth", "fifteenth", "sixteenth", "seventeenth", "eighteenth", "nineteenth"]; 
    var p = ["twent", "thirt", "fourt", "fift", "sixt", "sevent", "eight", "ninet"]; 
    var c = ["hundred", "thousand", "million", "billion", "trillion", "quadrillion", "quintillion"]; 
    var b = Math.floor(Math.log10(i)); 
    if(i<20) return s[i]; // Special case for sub-20 
    if(b==1){ // Between 21 and 99 
    if(i%10 === 0) return p[Math.floor(i/10)-2]+"ieth"; // On the tens, return p+"ieth" 
    return p[Math.floor(i/10)-2] + "y-" + s[i%10]; // Otherwise, return hyphenated 
    } 
    if(b==2){ // Between 100 and 999 
     var e = Math.floor(i/Math.pow(10,b)); // The first number 
     return n[e-1]+"-"+c[0]+" "+nth(i-(e*Math.pow(10,b))); 
    } 
    // Greater than 1000 we break into groups of 10^3 followed by a multiplier 
    var m = b%3 + 1; // Take the first m digits off 
    var cm = Math.floor(b/3); 
    var x = Math.floor(i/Math.pow(10,b-m+1)); 
    var numberToString = function(y){ // Converts a number less than 1000 to its string representation as a multiplier 
    if(y<20) return n[y-1]; 
    if(y<100) return p[Math.floor(y/10)-2] + "y-" + n[y%10-1]; 
    return n[Math.floor(y/100)-1] + " " + c[0] + " " + numberToString(y-(Math.floor(y/100)*100)); 
    } 
    return numberToString(x) + " " + c[cm] + " " + nth(i-(x*Math.pow(10, b-m+1))); 
}