一个把中文数字转换成整数数值的函数
Dieses Skript sollte nicht direkt installiert werden. Es handelt sich hier um eine Bibliothek für andere Skripte, welche über folgenden Befehl in den Metadaten eines Skriptes eingebunden wird // @require https://update.greatest.deepsurf.us/scripts/450829/1164707/numberDigit.js
//来自红烧鱼i https://blog.csdn.net/q1424966670/article/details/118338460
var map = {
"〇": 0,
"零": 0,
"一": 1,
"壹": 1,
"二": 2,
"贰": 2,
"两": 2,
"三": 3,
"叁": 3,
"四": 4,
"肆": 4,
"五": 5,
"伍": 5,
"六": 6,
"陆": 6,
"七": 7,
"柒": 7,
"八": 8,
"捌": 8,
"九": 9,
"玖": 9,
"十": 10,
"拾": 10,
"百": 100,
"佰": 100,
"千": 1000,
"仟": 1000,
"万": 10000,
"十万": 100000,
"百万": 1000000,
"千万": 10000000,
"亿": 100000000
};
// 解析失败返回-1，成功返回转换后的数字，不支持负数
function numberDigit(chinese_number) {
var len = chinese_number.length;
if (len == 0) return -1;
if (len == 1) return (map[chinese_number] <= 10) ? map[chinese_number] : -1;
var summary = 0;
if (map[chinese_number[0]] == 10) {
chinese_number = "一" + chinese_number;
len++;
}
if (len >= 3 && map[chinese_number[len - 1]] < 10) {
var last_second_num = map[chinese_number[len - 2]];
if (last_second_num == 100 || last_second_num == 1000 || last_second_num == 10000 || last_second_num == 100000000) {
for (var key in map) {
if (map[key] == last_second_num / 10) {
chinese_number += key;
len += key.length;
break;
}
}
}
}
if (chinese_number.match(/亿/g) && chinese_number.match(/亿/g).length > 1) return -1;
var splited = chinese_number.split("亿");
var rest;
if (splited.length == 2) {
rest = splited[1] == "" ? 0 : numberDigit(splited[1]);
return summary + numberDigit(splited[0]) * 100000000 + rest;
}
splited = chinese_number.split("万");
if (splited.length == 2) {
rest = splited[1] == "" ? 0 : numberDigit(splited[1]);
return summary + numberDigit(splited[0]) * 10000 + rest;
}
var i = 0;
while (i < len) {
var first_char_num = map[chinese_number[i]];
var second_char_num = map[chinese_number[i + 1]];
if (second_char_num > 9) summary += first_char_num * second_char_num;
i++;
if (i == len) summary += first_char_num <= 9 ? first_char_num : 0;
}
return summary;
}