在javascript中添加带有前导零的数字字符串答案

【问题标题】：Addition of number string in javascript with leading zeros在javascript中添加带有前导零的数字字符串
【发布时间】：2020-08-11 08:54:57
【问题描述】：

以下是我的代码，除了前导零之外，它在大多数情况下都可以正常工作。它应该保留尾随零，例如 -001 + 1 = 002

代码-

function incrementString (str) {
  if(str === '') return "1";
  
  if(!str.slice(-1).match(/\d/)) return `${str}1`;
  
  const replacer = x => {
    // Check if number 
    return (parseInt(x) + 1).toString();
  }
  
  return str.replace(/\d+/g, replacer )
}

// Return foobar2 which is correct
console.log(incrementString("foobar1"))

// Return foobar100 which is correct
console.log(incrementString("foobar099"))

// Return foobar2 which is incorrect, is should be foobar002
console.log(incrementString("foobar001"))

// Return foobar1 which is incorrect, is should be foobar001
console.log(incrementString("foobar000"))

// Return foobar101 which is incorrect, is should be foobar0101
console.log(incrementString("foobar0100"))

【问题讨论】：

标签： javascript html regex reactjs ecmascript-6

【解决方案1】：

您可以使用这个正则表达式解决方案：

function incrementString (str) {
  if(str === '') return "1";
  
  if(!str.slice(-1).match(/\d/)) return `${str}1`;
  
  const replacer = (m, g1, g2) => {
    // Check if number 
    var nn = (g1?g1:"") + (parseInt(g2) + 1).toString()
    return nn.slice(-1 * m.length)

  }
  
  return str.replace(/(0*)(\d+)/g, replacer )
}

// Return foobar2
console.log(incrementString("foobar1"))

// Return foobar100
console.log(incrementString("foobar099"))

// Return foobar002
console.log(incrementString("foobar001"))

// Return foobar001
console.log(incrementString("foobar000"))

// Return foobar0101
console.log(incrementString("foobar0100"))

// Return foobar01000
console.log(incrementString("foobar00999"))

// Return foobar010
console.log(incrementString("foobar009"))

【讨论】：

感谢/(0*)(\d+)/...我错过了这部分。
这是错误的顺便说一句 - incrementString("foobar099") 它正在制作 0100
foobar0100 变为 foobar0101，foobar0999 变为 foobar1000 表示 0 应保持到以下数字不超过其递增限制的时间 -> foobar0000100 变为 foobar0000101 & foobar0000999 变为 foobar0001000
我希望我为你提供了一个合理的解释。老实说，我没有考虑零，而是增加，但这是问题陈述:(
Expected: 'foobar01000', instead got: 'foobar001000', Expected: '010', instead got: '0010'

【解决方案2】：

一切似乎都很完美，您只需要处理replacer 函数中前导零的正则表达式部分。下面是相同的更新代码。

function incrementString(str) {
  if (str === '')
    return "1";

  if (!str.slice(-1).match(/\d/)) {
    return `${str}1`;
  }

  const replacer = x => {
    var leadingZerosMatched = x.match(/^0+/);
    var incrementedNumber = (parseInt(x) + 1).toString();
    var leadingZeroes;
    if (leadingZerosMatched && incrementedNumber.length < x.length) {
      leadingZeroes = leadingZerosMatched[0];
      if(leadingZeroes.length === x.length) {
        leadingZeroes = leadingZeroes.slice(0, leadingZeroes.length-1)
      }
    }
    return leadingZeroes ? leadingZeroes + incrementedNumber : incrementedNumber;
  }
  return str.replace(/\d+/g, replacer)
}

【讨论】：

【解决方案3】：

您可以将字符串与数字分开，并在增量后使用padStart 以保留前导0：

const incrementString = (str) => {
  const [chars, nums] = str.split(/(\d+)/)

  return [
    ...chars, 
    String(Number(nums) + 1)
    .padStart(nums.length, '0')
  ].join('')
}

console.log(incrementString("foobar1"))

console.log(incrementString("foobar099"))

console.log(incrementString("foobar001"))

console.log(incrementString("foobar000"))

console.log(incrementString("foobar0100"))

【讨论】：

【解决方案4】：

function incrementString (str) {
  let [
    openingPartial,
    terminatingInt
  ] = str.split(/(\d*)$/);

  if (terminatingInt) {
    const incrementedInt = String(parseInt(terminatingInt, 10) + 1);
    const leadingZeroCount = (terminatingInt.length - incrementedInt.length);

    if (leadingZeroCount >= 1) {

      terminatingInt = Array(leadingZeroCount).fill("0").concat(incrementedInt).join('');
    } else {
      terminatingInt = incrementedInt;
    }
  } else {
    terminatingInt = '1';
  }
  return `${ (openingPartial || '') }${ terminatingInt }`;
}

// Should return 'foo_003_bar1'.
console.log(incrementString("foo_003_bar"));

// Should return 'foo_003_bar_01'.
console.log(incrementString("foo_003_bar_00"));

// Should return 'foobar1'.
console.log(incrementString("foobar"));

// Should return 'foobar2'.
console.log(incrementString("foobar1"));

// Should return 'foobar100'.
console.log(incrementString("foobar099"));

// Should return 'foobar002'.
console.log(incrementString("foobar001"));

// Should return 'foobar001'.
console.log(incrementString("foobar000"));

// Should return 'foobar0101'.
console.log(incrementString("foobar0100"));

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【讨论】：