javascript - 这个函数可以用正则表达式重写吗？

Question

我想重新格式化并验证用户是否提供了有效的比利时企业编号。因为输入可以是下面所有的例子:

• BE 0123.321.123
• BE0123.321.123
• BE0123 321 123
• 0123.321.123
• 123.321.123
• 123321123

我编写了一个函数来验证输入并将其重新格式化为“显示”版本 (BE 0123.123.123) 和“代码”版本 (123123123)。这个函数看起来像这样。

formatAndValidateEnterpriseNumber = enterpriseNumber => {
    if(enterpriseNumber === undefined || !enterpriseNumber || (enterpriseNumber || '').length < 3) return { isValid: false, error: 'Please fill in your enterprise number' };

        //Remove space, dots, ...
        enterpriseNumber = enterpriseNumber.toUpperCase();
        enterpriseNumber = enterpriseNumber.replace(/[. ,:-]+/g, '');

        //Check for double country code
        const reDouble = /^[a-zA-Z]{4}/;
        if (reDouble.test(enterpriseNumber)) enterpriseNumber = enterpriseNumber.substring(2);

        if (enterpriseNumber.length < 9 || enterpriseNumber.length > 12) return { isValid: false, error: 'The length of the provided number is incorrect' };

        //Check country code
        const reBE = /^[a-zA-Z]{2}/;
        if (reBE.test(enterpriseNumber)) {
            //Check if country code = BE
            if (enterpriseNumber.slice(0, 2) !== 'BE') return { isValid: false, error: 'Please fill in a Belgian enterprise number' };
            // Remove country code
            else enterpriseNumber = enterpriseNumber.substring(2);
        }

        //Check if first digit is 0
        if (enterpriseNumber.length === 10 && enterpriseNumber.startsWith('0')) enterpriseNumber = enterpriseNumber.substring(1);

        //Check if enterpriseNumber is valid with modulo test
        if (parseInt(97 - (enterpriseNumber.slice(0, 7) % 97), 10) !== parseInt(enterpriseNumber.slice(7, 9), 10))
            return { isValid: false, error: 'The provided number is invalid'}

      return {
            isValid: true,
            enterpriseNumber: enterpriseNumber,
            displayEnterpriseNumber: `BE 0${enterpriseNumber.substring(0, 3)}.${enterpriseNumber.substring(3, 6)}.${enterpriseNumber.substring(6, 9)}`
      };
};

我认为这很困惑，我想知道是否可以通过重新格式化和验证用户输入的一个/两个正则表达式测试来改进这一点？

第二个问题:有时对于帐户或信用卡号，输入字段中已经有那些下划线和线 (-)，并且在输入时重新格式化数字。这种方法叫什么？可以针对比利时企业编号等特定事物执行此方法吗？

Answer 1

是的，你可以:

^(?:BE)?\s*[0-1]?(\d[. ]*){9}$

这个正则表达式应该可以做到!

This source (in Dutch) states what an enterprise number is for Belgium:

It has the country code: BE followed by a 0 or 1 and then followed by 9 digits.

https://regex101.com/r/4SRHxi/4

解释:

• ^:字符串必须以给定的正则表达式开头
• (?:BE)?:查找带有 BE 但 ? 表示它匹配零或 一次 - ?: 表示查找但不捕获
• \s*:搜索匹配零次或无限次的空格
• [0-1]?:检查零或一是否出现零次或一次
• ((\d[. ]*){9}):检查剩余字符串后是否有 9 位数字，与填充了多少点或空格无关。每次迭代都被捕获为第一个捕获组。当我们稍后更换时，这变得很重要。
• $:字符串必须结束

这将检查输入是否有效。

将它编辑成 code 版本很简单:

«code».replace(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/g, function(){
      return arguments[1].replace(/\D/g, "");
});

g 或全局修饰符将确保删除所有不需要的字符。通过使用其中包含替换的函数来替换所有非数字字符。此函数将输出我们想要的结果。

document.querySelector("pre").textContent.split("\n").forEach(function(element){
  if (element.match(/^(?:BE)?\s*[0-1]?(\d[. ]*){9}$/))
  {
     console.log(element.replace(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/g, function(){
      return arguments[1].replace(/\D/g, "");
     }));
  }
  else
  {
    console.log(`REJECTED: ${element}`);
  }

});

<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
BE 0123  321  123
BE 01 23 32 11 23
BE 0123 32 11 23
1123.321.123
123.321.123
123321123
AAA3434343A
BE  1233 445 4545 442
</pre>

现在很容易将字符串重建为正确的用户友好方式:

document.querySelector("pre").textContent.split("\n").forEach(function(element) {
  if (element.match(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/)) {
    var stripped = element.replace(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/g, function(){
          return arguments[1].replace(/\D/g, "");
    });

    //with the modulo check from your code added back in.
    if (97 - (parseInt(stripped.slice(0, 7), 10) % 97) == parseInt(stripped.slice(7, 9), 10)) {
      //use a literal string
      //use substring to put the dots between the sections of three numbers.
      var humanReadable = `BE 0${stripped.substring(0,3)}.${stripped.substring(3,6)}.${stripped.substring(6,9)}`;
      console.log(`CODE: ${stripped}`, `UI: ${humanReadable}`);
    }
  }

});

<pre>
BE 0123.321.123
BE0123.321.123
BE0123 321 123
0123.321.123
123.321.123
123321123
844256524
</pre>

---

第二个问题 是的，这可以完成，但是它需要您为其编写自己的代码。

简单版:

document.querySelector("div.enterprisenumber > input").addEventListener("keydown", function(e) {
  let value = this.value;

  //prevent the input from going back to 0
  if ( (value.length == 0 && (e.key == "Backspace" || e.key == "Delete"))) {
    e.preventDefault();
    return false;
  }
}, true);

document.querySelector("div.enterprisenumber > input").addEventListener("keyup", function(e) {
  //reset to a value without dots 
  let value = this.value.replace(/\./g, "");

  //strip the leading zero
  const valueWithout = value;
  //calculate how much iterations we need of a groups of three digits.
  const i = Math.floor(valueWithout.length / 3);
  let newValue = "";
  //check if backspace or delete are used to make sure the dot can be deleted.
  if (valueWithout.length < 9 && !(e.key == "Backspace" || e.key == "Delete")) {
    //only fire when higher than zero
    if (i > 0) {
      let t;
      //t is the index
      for (t = 0; t < i; t++) {
      //slice the correct portion of the string and append a dot, unless we are at the end of the groups
        newValue += valueWithout.slice(t * 3, t * 3 + 3) + (t == 2  ? "" : ".");
      }
      //append the remainder that is not a group of three.
      newValue += valueWithout.slice((t) * 3);
    } else {
      //return the value as is.
      newValue = value;
    }
    //set the new value to the input.
    this.value = newValue;
  }
}, true);

document.querySelector("div.enterprisenumber > input").addEventListener("blur", function(e) {
  let passed = false;
  if (this.value.match(/^(?:BE)?\s*[0-1]?((\d[. ]*){9})$/))
  {
    const value = this.value.replace(/\./g, "");
    //with modulo check
    if (97 - (parseInt(value.slice(0,7), 10) % 97) == value.slice(7, 9))
    {
      passed = true;
    }
  }
  document.querySelector(".enterprisenumber").classList[(passed ? "remove" : "add")]("error");
});

//if focus then focus input
document.querySelector("div.enterprisenumber").addEventListener("click", function(e) {
 if (e.target && e.target.nodeName != "SELECT")
 {
  this.querySelector("input").focus();
 }
});

* {
  box-sizing: border-box;
  font-family: tahoma;
  font-size: 10pt;
}

div.enterprisenumber {
  border: 1px solid #747474;
  width: 300px;
  padding: 0px;
  display: grid;
  grid-template-columns: 25px 40px auto;
  border-radius: 10px;
}

div.enterprisenumber.error {
  border: 1px solid #ff0000;
}

div.enterprisenumber>span {
  grid-column: 1;
  border: 0px;
  padding: 5px;
  background: linear-gradient(to right, rgba(0,0,0, 0.8) 33%, rgba(255,243,54, 0.8) 33%, rgba(255, 243, 54, 0.8) 66%, rgba(255, 15, 33, 0.8) 66%, rgba(255, 15, 33, 0.8) 100%);
  color: #ffffff;
  font-weight: bold;
  text-shadow: 1px 1px #000000;
  border-radius: 10px 10px 10px 10px;
}

div.enterprisenumber>select {
  grid-column: 2;
  border: 0px;
  padding: 5px;
}

div.enterprisenumber>input {
  grid-column: 3;
  border: 0px;
  padding: 5px;
  border-radius: 0px 10px 10px 0px;
}

Enter: 844256524
<div class="enterprisenumber">
  <span>BE</span><select><option value="0">0</option><option value="1">1</option><input value="" maxlength="11" />
</div>