正如标题所说，我有一个字符串，我想把它分成n个字符的片段。
例如:
var str = 'abcdefghijkl';
当n=3时，它会变成
var arr = ['abc','def','ghi','jkl'];
有办法做到这一点吗?
当前回答
var b1 = "";
function myFunction(n) {
if(str.length>=3){
var a = str.substring(0,n);
b1 += a+ "\n"
str = str.substring(n,str.length)
myFunction(n)
}
else{
if(str.length>0){
b1 += str
}
console.log(b1)
}
}
myFunction(4)
其他回答
这里我们每隔n个字符就在一个字符串中穿插另一个字符串:
export const intersperseString = (n: number, intersperseWith: string, str: string): string => {
let ret = str.slice(0,n), remaining = str;
while (remaining) {
let v = remaining.slice(0, n);
remaining = remaining.slice(v.length);
ret += intersperseWith + v;
}
return ret;
};
如果我们像这样使用上面的语句:
console.log(splitString(3,'|', 'aagaegeage'));
我们得到:
亚美大陆煤层气有限公司|亚美大陆煤层气有限公司| aeg |坚毅不屈| e
这里我们做同样的事情，但是push到一个数组:
export const sperseString = (n: number, str: string): Array<string> => {
let ret = [], remaining = str;
while (remaining) {
let v = remaining.slice(0, n);
remaining = remaining.slice(v.length);
ret.push(v);
}
return ret;
};
然后运行它:
console.log(sperseString(5, 'foobarbaztruck'));
我们得到:
[“fooba”、“rbazt”、“ruck”]
如果有人知道简化上述代码的方法，请使用lmk，但它应该适用于字符串。
基于之前对这个问题的回答;下面的函数将分割一个字符串(str) n-number (size)的字符。
function chunk(str, size) {
return str.match(new RegExp('.{1,' + size + '}', 'g'));
}
Demo
(function() { function chunk(str, size) { return str.match(new RegExp('.{1,' + size + '}', 'g')); } var str = 'HELLO WORLD'; println('Simple binary representation:'); println(chunk(textToBin(str), 8).join('\n')); println('\nNow for something crazy:'); println(chunk(textToHex(str, 4), 8).map(function(h) { return '0x' + h }).join(' ')); // Utiliy functions, you can ignore these. function textToBin(text) { return textToBase(text, 2, 8); } function textToHex(t, w) { return pad(textToBase(t,16,2), roundUp(t.length, w)*2, '00'); } function pad(val, len, chr) { return (repeat(chr, len) + val).slice(-len); } function print(text) { document.getElementById('out').innerHTML += (text || ''); } function println(text) { print((text || '') + '\n'); } function repeat(chr, n) { return new Array(n + 1).join(chr); } function textToBase(text, radix, n) { return text.split('').reduce(function(result, chr) { return result + pad(chr.charCodeAt(0).toString(radix), n, '0'); }, ''); } function roundUp(numToRound, multiple) { if (multiple === 0) return numToRound; var remainder = numToRound % multiple; return remainder === 0 ? numToRound : numToRound + multiple - remainder; } }()); #out { white-space: pre; font-size: 0.8em; } <div id="out"></div>
我的解决方案(ES6语法):
const source = "8d7f66a9273fc766cd66d1d";
const target = [];
for (
const array = Array.from(source);
array.length;
target.push(array.splice(0,2).join(''), 2));
我们甚至可以这样创建一个函数:
function splitStringBySegmentLength(source, segmentLength) {
if (!segmentLength || segmentLength < 1) throw Error('Segment length must be defined and greater than/equal to 1');
const target = [];
for (
const array = Array.from(source);
array.length;
target.push(array.splice(0,segmentLength).join('')));
return target;
}
然后你可以以一种可重用的方式轻松地调用函数:
const source = "8d7f66a9273fc766cd66d1d";
const target = splitStringBySegmentLength(source, 2);
干杯
var b1 = "";
function myFunction(n) {
if(str.length>=3){
var a = str.substring(0,n);
b1 += a+ "\n"
str = str.substring(n,str.length)
myFunction(n)
}
else{
if(str.length>0){
b1 += str
}
console.log(b1)
}
}
myFunction(4)
如果你真的需要坚持使用.split和/或.raplace，那么使用/(?<=^(?:.{3})+)(?!$)/g
.split:
var arr = str.split( /(?<=^(?:.{3})+)(?!$)/ )
// [ 'abc', 'def', 'ghi', 'jkl' ]
.replace:
var replaced = str.replace( /(?<=^(?:.{3})+)(?!$)/g, ' || ' )
// 'abc || def || ghi || jkl'
/(?!$)/表示在字符串结束时不停止。没有它的:
var arr = str.split( /(?<=^(?:.{3})+)/ )
// [ 'abc', 'def', 'ghi', 'jkl' ] // is fine
var replaced = str.replace( /(?<=^(.{3})+)/g, ' || ')
// 'abc || def || ghi || jkl || ' // not fine
忽略组/(?:…)/是为了防止数组中存在重复项。没有它的:
var arr = str.split( /(?<=^(.{3})+)(?!$)/ )
// [ 'abc', 'abc', 'def', 'abc', 'ghi', 'abc', 'jkl' ] // not fine
var replaced = str.replace( /(?<=^(.{3})+)(?!$)/g, ' || ' )
// 'abc || def || ghi || jkl' // is fine
