Java中的英语单词
在这里，我要问一些奇怪的事情。
我想问一下，是否有任何方法/逻辑可以将整数值转换为包含数字英文单词的字符串值？
例如，用户输入22并获得二十二或二十二的输出。
谢谢
查看此代码，它可能是您正在寻找的。 例如，如果我们有：
System.out.println(convert(22)); 输出：
twenty two
编辑我已经复制了下面的代码，稍微清理了格式（主要方法在底部）：
import java.text.DecimalFormat; public class EnglishNumberToWords { private static final String[] tensNames = { "", " ten", " twenty", " thirty", " forty", " fifty", " sixty", " seventy", " eighty", " ninety" }; private static final String[] numNames = { "", " one", " two", " three", " four", " five", " six", " seven", " eight", " nine", " ten", " eleven", " twelve", " thirteen", " fourteen", " fifteen", " sixteen", " seventeen", " eighteen", " nineteen" }; private static String convertLessThanOneThousand(int number) { String soFar; if (number % 100 < 20) { soFar = numNames[number % 100]; number /= 100; } else { soFar = numNames[number % 10]; number /= 10; soFar = tensNames[number % 10] + soFar; number /= 10; } if (number == 0) return soFar; return numNames[number] + " hundred" + soFar; } public static String convert(long number) { // 0 to 999 999 999 999 if (number == 0) { return "zero"; } String snumber = Long.toString(number); // pad with "0" String mask = "000000000000"; DecimalFormat df = new DecimalFormat(mask); snumber = df.format(number); // XXXnnnnnnnnn int billions = Integer.parseInt(snumber.substring(0, 3)); // nnnXXXnnnnnn int millions = Integer.parseInt(snumber.substring(3, 6)); // nnnnnnXXXnnn int hundredThousands = Integer.parseInt(snumber.substring(6, 9)); // nnnnnnnnnXXX int thousands = Integer.parseInt(snumber.substring(9, 12)); String tradBillions; switch (billions) { case 0: tradBillions = ""; break; case 1: tradBillions = convertLessThanOneThousand(billions) + " billion "; break; default: tradBillions = convertLessThanOneThousand(billions) + " billion "; } String result = tradBillions; String tradMillions; switch (millions) { case 0: tradMillions = ""; break; case 1: tradMillions = convertLessThanOneThousand(millions) + " million "; break; default: tradMillions = convertLessThanOneThousand(millions) + " million "; } result = result + tradMillions; String tradHundredThousands; switch (hundredThousands) { case 0: tradHundredThousands = ""; break; case 1: tradHundredThousands = "one thousand "; break; default: tradHundredThousands = convertLessThanOneThousand(hundredThousands) + " thousand "; } result = result + tradHundredThousands; String tradThousand; tradThousand = convertLessThanOneThousand(thousands); result = result + tradThousand; // remove extra spaces! return result.replaceAll("^\\s+", "").replaceAll("\\b\\s{2,}\\b", " "); } public static void main(String[] args) { System.out.println(convert(22)); // "twenty two" } }
虽然接受的答案有效，但最好使用已经实现的function来执行此操作。 ICU4J包含一个类com.ibm.icu.text.RuleBasedNumberFormat ，可用于执行此操作。 它还支持英语以外的语言，以及反向操作，将文本字符串解析为整数值。
这是一个例子，假设我们在类路径中有ICU4J依赖 ：
import com.ibm.icu.text.RuleBasedNumberFormat; import java.util.Locale; RuleBasedNumberFormat nf = new RuleBasedNumberFormat (Locale.UK, RuleBasedNumberFormat.SPELLOUT); nf.format(24); // The result is "twenty-four"
我有一个非常简单的解决方案，可以将java中的整个整数范围转换为hinglish表示法
String h1[] = { "Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Tweleve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen" }; String h2[] = { "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety" }; String h3[] = { "Hundred", "Thousand", "Lakh", "Crore", "Arab" }; public String parseToHinglishNotation(int num) { String word = ""; if (num < 0) { word += "Minus "; num = -num; } if (num < 20) { word += h1[num]; } else if (num < 100) { int temp = num / 10 - 2; word += h2[temp]; temp = num % 10; if (temp > 0) { word += " " + parseToHinglishNotation(temp); } } else if (num < 1000) { int temp = num / 100; word += parseToHinglishNotation(temp) + " " + h3[0]; temp = num % 100; if (temp > 0) { word += " " + parseToHinglishNotation(temp); } } else if (num < 100000) { int temp = num / 1000; word += parseToHinglishNotation(temp) + " " + h3[1]; temp = num % 1000; if (temp > 0) { word += " " + parseToHinglishNotation(temp); } } else if (num < 10000000) { int temp = num / 100000; word += parseToHinglishNotation(temp) + " " + h3[2]; temp = num % 100000; if (temp > 0) { word += " " + parseToHinglishNotation(temp); } } else if (num < 1000000000) { int temp = num / 10000000; word += parseToHinglishNotation(temp) + " " + h3[3]; temp = num % 10000000; if (temp > 0) { word += " " + parseToHinglishNotation(temp); } } else if (num <= 2147483647) { int temp = num / 1000000000; word += parseToHinglishNotation(temp) + " " + h3[4]; temp = num % 1000000000; if (temp > 0) { word += " " + parseToHinglishNotation(temp); } } return word; }
我知道这篇文章有点陈旧，但我最近提出了自己的解决方案，并认为它可能值得分享。 主函数numToWords采用1到9999之间的任何Integer （包括）并输出其对应的英语单词，后跟括号中的数字String 。
示例：如果Integer x = 2614; ， numToText(x); 返回"Two Thousand Six Hundred Fourteen (2614)" ：
数字等效字存储在HashMaps中，取决于它们是1还是10的倍数（例如6 --> "Six" ， 60 --> "Sixty" ），而青少年则使用switch语句独立处理。
import java.util.HashMap; public class Converter{ public Converter(){ hm1.put(0, ""); hm1.put(1, "One"); hm1.put(2, "Two"); hm1.put(3, "Three"); hm1.put(4, "Four"); hm1.put(5, "Five"); hm1.put(6, "Six"); hm1.put(7, "Seven"); hm1.put(8, "Eight"); hm1.put(9, "Nine"); hm10.put(0, ""); hm10.put(1, ""); hm10.put(2, "Twenty "); hm10.put(3, "Thirty "); hm10.put(4, "Fourty "); hm10.put(5, "Fifty "); hm10.put(6, "Sixty "); hm10.put(7, "Seventy "); hm10.put(8, "Eighty "); hm10.put(9, "Ninety "); } public static String numToWords(Integer x){ // Obtain a char[] of the Integer to analyze each digit. String s = x.toString(); char[] cArray = s.toCharArray(); // Refer to the length of the Integer as its final index. int l = cArray.length-1; String retStr = ""; // The loop counts backwards from the right-most digit. for(int i = l; i >= 0; i--){ // Determine the numeric value at the left-most index, then move rightward. // This setup is attributed to the nuances of the English language. int j = Character.getNumericValue(cArray[li]); // if(i == 3){ retStr += hm1.get(j); retStr += " Thousand "; } else if(i == 2){ retStr += hm1.get(j); retStr += " Hundred "; } else if(i == 1){ //If the 10s digit is 1, the final word is 10 <= x <= 19. if(j == 1){ String tens = ""; // Check the 1s digit to determine x (10 <= x <= 19) switch(Character.getNumericValue(cArray[l])){ case 0: tens = "Ten"; break; case 1: tens = "Eleven"; break; case 2: tens = "Twelve"; break; case 3: tens = "Thirteen"; break; case 4: tens = "Fourteen"; break; case 5: tens = "Fifteen"; break; case 6: tens = "Sixteen"; break; case 7: tens = "Seventeen"; break; case 8: tens = "Eighteen"; break; case 9: tens = "Nineteen"; break; } i = -1; // Ensure it's the last word by indexing out of the loop. retStr += tens; } else { //If the 10s digit is 2 <= x <= 9, the 10s word is 20 <= x <= 90. retStr += hm10.get(j); } } else if(i == 0){ retStr += hm1.get(j); } } return retStr + " (" + x + ") "; } private HashMap hm1 = new HashMap(); private HashMap hm10 = new HashMap(); }