JavaScript函数格式字符串

Question

所以，我有几个测试用例字符串：

测试：00 -44 48 5555 8361' 预计：004-448-555-583-61

测试：0 - 22 1985--324' 预计：022-198-53-24

测试： '555372654' 预计：555-372-654

任何建议将是有益的。我一直在这里待了几个小时，试图找出为什么一些预期的结果有效，而另一些则没有。给定表示电话号码的字符串S，字符串将需要重新格式化。字符串S由N个字符组成：数字，空格和/或破折号。它至少包含两位数字。字符串S中的空格和破折号可以忽略。我想重新格式化给定的电话号码，以便数字按长度为三的块分组，并用单个破折号分隔。如果有必要，最后一个块或最后两个块的长度可以是两个。

var s = "00-44 48 5555 8361"; 
 
var n = s.replace(/\D/g,''); 
 
var temp = formatNumber(n); 
 
var final = temp.replace(/\-$/,''); 
 
alert(final); 
 

 
function formatNumber(S){ 
 
    var len = S.length; 
 
    var max = len - 2 * ((3-len) % 3); 
 
    var dash = "-"; 
 
    var result; 
 
    for(i = 0; i < max; i+=3){ 
 
    \t result += S.substring(i, i+3).concat(dash); 
 
    } 
 
    for(i = max; i < len; i+=2){ 
 
    \t result += S.substring(i, i+2).concat(dash); 
 
    } 
 
\t return result.toString(); 
 
}

Answer 1

在行var max = len - 2 * ((3-len) % 3);，实际上是增加了len未从其减去按预期（因为len >= 3从而3 - len <= 0从而(3 - len) % 3 <= 0因此通过-2倍增时，它的结果将是肯定的）。所以max总是等于或大于len的长度。

试试这个：

max = len - (len % 3? ((len + 1) % 3? 4: 2): 0); 

可以通过解释：

var max = len; 
if(len % 3 === 0) {    // if len is divisible by 3 (if len === 0, 3, 6, 9, ...) 
    max -= 0;      // substract 0 (leave max === len) 
} else {       // if len is not divisible by 3 
    if((len + 1) % 3 === 0) {  // if len + 1 is divisible by 3 (if len === 2, 5, 8, 11, ...) 
     max -= 2;     // substract 2 (only one part containing 2 digits) 
    } else {      // if len === 1, 4, 7, 10, ... 
     max -= 4;     // substract 4 (two parts containing 2 digits) 
    } 
} 

function formatNumber(s) { 
 
    var l = s.length, 
 
     max = l - (l % 3? ((l + 1) % 3? 4: 2): 0); 
 
    var result = ""; 
 
    for(var i = 0; i < max; i+=3) { 
 
    result += s.slice(i, i + 3) + "-";   // no need for concat just add "-" to the result of slice 
 
    } 
 
    for(var i = max; i < l; i+=2) { 
 
    result += s.slice(i, i + 2) + "-"; 
 
    } 
 
    return result.slice(0, -1);     // the same as result.slice(0, result.length - 1) to remove the last '-' 
 
} 
 

 
console.log(formatNumber("123456789")); 
 
console.log(formatNumber("12345678")); 
 
console.log(formatNumber("1234567890"));

Answer 2

看来，你总是需要这样的格式：(\d{3}-)+。

function formatter(input) { 
 
    return input 
 
    
 
    // rm ws 
 
    .replace(/\s+/g, '') 
 
    
 
    // rm dashes 
 
    .replace(/-/g, '') 
 
    
 
    // split by 3 chars 
 
    .split(/(.{3})/) 
 
    
 
    .filter(g => g) 
 
    
 
    .join('-') 
 
    ; 
 
} 
 

 

 
var raw = [ 
 
    "00-44 48 5555 8361", 
 
    "555372654", 
 
    "0 - 22 1985--324" 
 
]; 
 

 
console.log('Formatted', raw.map(formatter));

Answer 3

你可以尝试：

var phones = [ 
 
    '00-44 48 5555 8361', 
 
    '0 - 22 1985--324', 
 
    '555372654' 
 
]; 
 

 
phones = phones.map(function(phone) { 
 
    var num = phone.replace(/\D/g, ''); 
 
    return num.replace(/(...?)(?!.?$)/g, '$1-'); 
 
}); 
 

 
console.log(phones);

Answer 4

var n = '00-44 48 5555 8361'; 

console.log(formatPhone(n, 3)); 

function formatPhone(num, spacing) { 
    num = (num+'').replace(/[^\d]/g, '').split(''); 

    return num.map(function (currentValue, index) { 
     // Plus 1 because it's 0-indexed. 
     return (index + 1) % spacing === 0 
       ? currentValue + '-' 
       : currentValue 
     ; 
    }).join(''); 
} 

https://jsfiddle.net/efpy3vLn/

Answer 5

你可以做任何事情，只需几个正则表达式替换：

// Initial string 
var s0 = "00-44 48 5555 8361"; 

// Remove everything that's not a number 
var s1 = s0.replace(/[^0-9]/g,''); 

// A dash every three numbers 
var s2 = s1.replace(/(...)/g,'$1\-'); 

// Remove the traling slash if any 
var s3 = s2.replace(/\-$/,'');