我想把一个非常大的字符串(比如10,000个字符)分割成n大小的块。
就性能而言,最好的方法是什么?
例如: "1234567890"除以2将变成["12","34","56","78","90"]。
使用string。prototype。match可以实现这样的事情吗如果可以,从性能来看,这是最好的方式吗?
我想把一个非常大的字符串(比如10,000个字符)分割成n大小的块。
就性能而言,最好的方法是什么?
例如: "1234567890"除以2将变成["12","34","56","78","90"]。
使用string。prototype。match可以实现这样的事情吗如果可以,从性能来看,这是最好的方式吗?
当前回答
我会用正则表达式…
var chunkStr = function(str, chunkLength) {
return str.match(new RegExp('[\\s\\S]{1,' + +chunkLength + '}', 'g'));
}
其他回答
var l = str.length, lc = 0, chunks = [], c = 0, chunkSize = 2;
for (; lc < l; c++) {
chunks[c] = str.slice(lc, lc += chunkSize);
}
window.format = function(b, a) {
if (!b || isNaN(+a)) return a;
var a = b.charAt(0) == "-" ? -a : +a,
j = a < 0 ? a = -a : 0,
e = b.match(/[^\d\-\+#]/g),
h = e && e[e.length - 1] || ".",
e = e && e[1] && e[0] || ",",
b = b.split(h),
a = a.toFixed(b[1] && b[1].length),
a = +a + "",
d = b[1] && b[1].lastIndexOf("0"),
c = a.split(".");
if (!c[1] || c[1] && c[1].length <= d) a = (+a).toFixed(d + 1);
d = b[0].split(e);
b[0] = d.join("");
var f = b[0] && b[0].indexOf("0");
if (f > -1)
for (; c[0].length < b[0].length - f;) c[0] = "0" + c[0];
else +c[0] == 0 && (c[0] = "");
a = a.split(".");
a[0] = c[0];
if (c = d[1] && d[d.length -
1].length) {
for (var d = a[0], f = "", k = d.length % c, g = 0, i = d.length; g < i; g++) f += d.charAt(g), !((g - k + 1) % c) && g < i - c && (f += e);
a[0] = f
}
a[1] = b[1] && a[1] ? h + a[1] : "";
return (j ? "-" : "") + a[0] + a[1]
};
var str="1234567890";
var formatstr=format( "##,###.", str);
alert(formatstr);
This will split the string in reverse order with comma separated after 3 char's. If you want you can change the position.
function chunkString(str, length = 10) {
let result = [],
offset = 0;
if (str.length <= length) return result.push(str) && result;
while (offset < str.length) {
result.push(str.substr(offset, length));
offset += length;
}
return result;
}
var str = "123456789";
var chunks = [];
var chunkSize = 2;
while (str) {
if (str.length < chunkSize) {
chunks.push(str);
break;
}
else {
chunks.push(str.substr(0, chunkSize));
str = str.substr(chunkSize);
}
}
alert(chunks); // chunks == 12,34,56,78,9
我已经写了一个扩展函数,所以块长度也可以是一个数字数组,比如[1,3]
String.prototype.chunkString = function(len) {
var _ret;
if (this.length < 1) {
return [];
}
if (typeof len === 'number' && len > 0) {
var _size = Math.ceil(this.length / len), _offset = 0;
_ret = new Array(_size);
for (var _i = 0; _i < _size; _i++) {
_ret[_i] = this.substring(_offset, _offset = _offset + len);
}
}
else if (typeof len === 'object' && len.length) {
var n = 0, l = this.length, chunk, that = this;
_ret = [];
do {
len.forEach(function(o) {
chunk = that.substring(n, n + o);
if (chunk !== '') {
_ret.push(chunk);
n += chunk.length;
}
});
if (n === 0) {
return undefined; // prevent an endless loop when len = [0]
}
} while (n < l);
}
return _ret;
};
的代码
"1234567890123".chunkString([1,3])
将返回:
[ '1', '234', '5', '678', '9', '012', '3' ]