正则表达式匹配应为1或2位数，但不是零，日期格式为单位数

var dateObj = {
  '01/01/2017' : true,
  '1/1/2016' : true,
  '1/1/16' : true,
  '1/1/116' : false,
  '01/11/2016' : true,
  '01.01.2016' : true,
  '01.01_2016' : false,
  '01-0-2016' : false,
  '01-01-2016' : true
};
var failedAttempts = [];
var date_val = "01/01/2017";
var re = /^(\d{1,2})[/.\-](\d{1,2})[/.\-](\d{4}|\d{2})$/;
for(let i in dateObj){
  let result = re.test(i);
  if(result != dateObj[i]){
    failedAttempts.push(i);
  }
}
if(failedAttempts.length > 0){
  console.error('Unit test fails');
  console.log(failedAttempts);
}else{
  console.log('Unit test pass');
}

'01-0-2016' : false consider this case returns true even it is in wrong format. I want to rewrite the Regex with the digit matches either 1 or 2 but not 0 in single digit.

'01 -0-2016':false认为这种情况即使格式错误也会返回true。我想重写正则表达式,数字匹配为1或2,但不是0,单位数。

4 个解决方案

#1

Restrict the digit matching patterns with negative (?!0+\b) lookaheads:

使用负(?!0+ \ b)前瞻限制数字匹配模式:

 /^(?!0+\b)(\d{1,2})[\/.-](?!0+\b)(\d{1,2})[\/.-](\d{4}|\d{2})$/
   ^^^^^^^^               ^^^^^^^^

See the regex demo

请参阅正则表达式演示

If you needn't restrict both day month parts, just remove the unnecessary lookahead.

如果您不需要限制两个月的零件,只需删除不必要的前瞻。

The (?!0+\b) pattern matches one or more zeros followed with a word boundary (that is, there cannot be any letter/digit/_ after it), and if the pattern is found, the match is failed.

(?!0+ \ b)模式匹配一个或多个零后跟一个单词边界(即,后面不能有任何字母/数字/ _),如果找到模式,则匹配失败。

#2

There could be several ways, I propose:

我建议可以有几种方法:

var re = /^(\d{1,2})[/.\-]([1-9]|[0-9]{2})[/.\-](\d{4}|\d{2})$/;

Let me know if it solves your issue.

如果它能解决您的问题,请告诉我。

#3

Here is a non-regex way of validating your input date strings using dd/mm/yyyy or /dd/mm/yy formats:

以下是使用dd / mm / yyyy或/ dd / mm / yy格式验证输入日期字符串的非正则表达方式:

for(let str in dateObj) {
   var parts = str.split(/[-.\/]/);
   var p = parts[0]*parts[1]*parts[2];
   if (isNaN(p) || p<10 || (p>99 && p<1000)) {
      console.log("Invalid date: ", str);
   }
}

Output:

Invalid date:  1/1/116
Invalid date:  01.01_2016
Invalid date:  01-0-2016

First we are splitting date string into 3 parts using delimiters - or . or /

首先,我们使用分隔符将日期字符串拆分为3个部分 - 或者。要么 /

Then we are multiplying all 3 parts and taking the product for validation

然后我们将所有3个部分相乘并将产品用于验证

Finally we are using isNaN(p) || p<10 || (p>99 && p<1000 conditions for validation. isNan will address all invalid date inputs, p<=0 will address day, mont or year being zero and last condition will address year part in 3 digits.

最后我们使用isNaN(p)|| p <10 || (p> 99 && p <1000条件用于验证.isNan将解决所有无效日期输入,p <= 0将解决日期,mont或年份为零,最后一个条件将解决年份部分的3位数。

#4

This should work for you:

这应该适合你:

const validateDate = (date) => /^(?:[1-2]\d|0?[1-9]|30|31)([.\/-])(?:0?[1-9]|1[0-2])\1(?:\d{4}|\d{2})$/.test(date)

const testData = [`01/01/2017`, `1/1/2016`, `1/1/16`, `01/11/2016`, `01.01.2016`, `01-01-2016`, `01.0.2016`, `01.01_2016`, `01-0-2016`, `1/1/116`, `41.23.23`]

for (const date of testData){
  console.log(`${date} ${validateDate(date)}`)
}

#1