如何使用正则表达式在 Javascript 中 10 以下的任何数字前添加零
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How to prepend a zero in front of any number below 10 in Javascript using Regexp
提问by Cybrix
Good day,
再会，
Is there a Regex that I could use to prepend a 0 before any number that is below 10?
是否有一个正则表达式可以用来在任何小于 10 的数字之前添加 0？
I am not looking for a date parsing library, ternary or if/else solutions. (hopefully)
我不是在寻找日期解析库、三元或 if/else 解决方案。（希望）
var currentDate = new Date(),
stringDate = currentDate.getFullYear() + "-" + currentDate.getMonth() + "-" + currentDate.getDate() + " " + currentDate.getHours() + ":" + currentDate.getMinutes() + ":" + currentDate.getSeconds();
alert( stringDate ); //2011-10-17 10:3:7
I would like a RegExp that I could apply to stringDate to get 2011-10-17 10:03:07
我想要一个可以应用于 stringDate 的 RegExp 以获得 2011-10-17 10:03:07
Thank you very much!
非常感谢你！
回答by Alex Peattie
Just add the leading 0 every time, then use slice(-2)to get the last two characters, like so:
每次只需添加前导 0，然后使用slice(-2)获取最后两个字符，如下所示：
('0' + currentDate.getHours()).slice(-2)
回答by Davan Etelamaki
The following function will allow you to declare a minimum length for a number along or within a string and will pad it with zeros to make it the appropriate length.
以下函数将允许您为字符串中或字符串中的数字声明最小长度，并用零填充它以使其成为适当的长度。
var PrependZeros = function (str, len, seperator) {
if(typeof str === 'number' || Number(str)){
str = str.toString();
return (len - str.length > 0) ? new Array(len + 1 - str.length).join('0') + str: str;
}
else{
for(var i = 0,spl = str.split(seperator || ' '); i < spl.length; spl[i] = (Number(spl[i])&& spl[i].length < len)?PrependZeros(spl[i],len):spl[i],str = (i == spl.length -1)?spl.join(seperator || ' '):str,i++);
return str;
}
};
For those wanting a less cryptic version
对于那些想要一个不那么神秘的版本的人
var PrependZeros = function (str, len, seperator) {
if (typeof str === 'number' || Number(str)) {
str = str.toString();
return (len - str.length > 0) ? new Array(len + 1 - str.length).join('0') + str : str;
}
else {
var spl = str.split(seperator || ' ')
for (var i = 0 ; i < spl.length; i++) {
if (Number(spl[i]) && spl[i].length < len) {
spl[i] = PrependZeros(spl[i], len)
}
}
return spl.join(seperator || ' ');
}
};
Examples:
例子：
PrependZeros("1:2:3",2,":"); // "01:02:03"
PrependZeros(1,2); // "01"
PrependZeros(123,2); // "123"
PrependZeros("1 2 3",3); // "001 002 003"
PrependZeros("5-10-2012",2,"-"); //"05-10-2012"
回答by Alex Turpin
You don't need regex for that. You can make a simple pad function yourself:
你不需要正则表达式。你可以自己做一个简单的pad函数：
function pad(n) {
if (n < 10)
return "0" + n;
return n;
}
alert(pad(8));
alert(pad(11));
回答by Toto
How about:
怎么样：
x.replace(/^(\d)$/, "0");
回答by John
回答by Timo
%0{length} => %05 => 1=00001, 2=00002,... 55=00055,...
%0{长度} => %05 => 1=00001, 2=00002,... 55=00055,...
回答by Andreas
That should do the work (it has an ternary operator in it, but also an regex^^)
这应该可以完成工作（它有一个三元运算符，还有一个正则表达式^^）
stringDate.replace(/\d+/g, function(m) {
return parseInt(m, 10) < 10 ? "0" + m : m;
});
回答by Ya Zhuang
I prefer Alex's way, but if you reallywant the regex way, you can try:
我更喜欢 Alex 的方式，但如果你真的想要正则表达式的方式，你可以尝试：
"2011-10-10 10:2:27".replace(/:(?=[^0](?::|$))/g, ":0");
回答by John Strickler
Are you ready for your one-liner do-it-all regex solution?
您准备好使用单行全能正则表达式解决方案了吗？
Pass in your entire string... and this will match single-digits and pad them with a 0.
传入整个字符串……这将匹配单个数字并用 0 填充它们。
"2011-10-17 10:3:7"
.replace(/(^|[^0-9])([0-9])(?=($|[^0-9]))/ig, function (##代码##, ) {
return ##代码##[0] + '0' + ##代码##[1];
});
//returns "2011-10-17 10:03:07"
This would be made so much easier if JavaScript supported lookbehind-assertions.
如果 JavaScript 支持后视断言，这将变得更加容易。
