我将MAC地址存储为原始的48位数字,我想将其拆分并以xx:xx:xx:xx:xx:xx的标准十六进制格式打印。例如,原始编号81952921372024应为78:45:c4:26:89:4a。我的第一次尝试是,
var suspect = {mac: 2333752735057272};
console.log(
Number(suspect.mac & 0xFF).toString(16) + ":" +
Number((suspect.mac & 0xFF00) >> 8).toString(16) + ":" +
Number((suspect.mac & 0xFF0000) >> 16).toString(16) + ":" +
Number((suspect.mac & 0xFF000000) >> 24).toString(16) + ":" +
Number((suspect.mac & 0xFF00000000) >> 32).toString(16) + ":" +
Number((suspect.mac & 0xFF0000000000) >> 48).toString(16));
但是因为Javascript显然无法处理>进行移位操作时,32位整数,最后两个八位字节总是为0,
78:45:c4:26:0:0
答案 0 :(得分:13)
一种简单的方法如下:
var mac = 81952921372024;
mac.toString( 16 ) // "4a8926c44578"
.match( /.{1,2}/g ) // ["4a", "89", "26", "c4", "45", "78"]
.reverse() // ["78", "45", "c4", "26", "89", "4a"]
.join( ':' ) // "78:45:c4:26:89:4a"
> "78:45:c4:26:89:4a"
但是,如果您的整数很短(即00
),我建议仅针对病态情况添加其他mac = 150
组:
var mac = 81952921372024;
new Array( 6 ).join( '00' ) // '000000000000'
.match( /../g ) // [ '00', '00', '00', '00', '00', '00' ]
.concat(
mac.toString( 16 ) // "4a8926c44578"
.match( /.{1,2}/g ) // ["4a", "89", "26", "c4", "45", "78"]
) // ["00", "00", "00", "00", "00", "00", "4a", "89", "26", "c4", "45", "78"]
.reverse() // ["78", "45", "c4", "26", "89", "4a", "00", "00", "00", "00", "00", "00", ]
.slice( 0, 6 ) // ["78", "45", "c4", "26", "89", "4a" ]
.join( ':' ) // "78:45:c4:26:89:4a"
> "78:45:c4:26:89:4a"
答案 1 :(得分:1)
以下工作,
var hexValue = parseInt('44873434449413').toString(16);
var macaddress = [];
for (var i=0; i < hexValue.length; i=i+2) {
macaddress.push(hexValue.substr(i,2));
}
console.log(macaddress.join(':'));
输出:
28:cf:e9:1e:c6:05
编辑:
照顾尾随0的
str='3';
if (str.length < 12) { str = pad_after(str, 12, 0);}
var hexValue = parseInt(str).toString(16);
if (hexValue.length < 12) { hexValue = pad_before(hexValue, 12, 0);}
var macaddress = [];
for (var i=0; i < hexValue.length; i=i+2) {
macaddress.push(hexValue.substr(i,2));
}
console.log(macaddress.join(':'));
function pad_before(n, width, z) {
z = z || '0';
n = n + '';
return n.length >= width ? n : new Array(width - n.length + 1).join(z) + n;
}
function pad_after(n, width, z) {
z = z || '0';
n = n + '';
return n.length >= width ? n : n + new Array(width - n.length + 1).join(z);
}
输出:
00:45:d9:64:b8:00
答案 2 :(得分:0)
接受的答案没有通过我的单元测试(将mac地址转换为数字,然后转换回数字,并期望结果等于输入)。
@blueren表示接受的答案会返回相反的结果。
我的版本是:
(mac_num).toString(16).padStart(12, '0').match(/../g)
.reverse()
.slice(0, 6)
.reverse()
.join(':')
我的测试:
expect(mac.toString(mac.toLong('b4:ae:2b:db:1c:3d'))).to.deep.equal('b4:ae:2b:db:1c:3d');
expect(mac.toString(mac.toLong('00:00:00:db:1c:3d'))).to.deep.equal('00:00:00:db:1c:3d');
expect(mac.toString(mac.toLong('b4:ae:2b:00:00:00'))).to.deep.equal('b4:ae:2b:00:00:00');
expect(mac.toString(mac.toLong('00:00:00:00:00:00'))).to.deep.equal('00:00:00:00:00:00');
expect(mac.toString(mac.toLong('10:00:00:00:00:00'))).to.deep.equal('10:00:00:00:00:00');
expect(mac.toString(mac.toLong('00:00:01:00:00:00'))).to.deep.equal('00:00:01:00:00:00');
expect(mac.toString(mac.toLong('00:00:00:00:00:01'))).to.deep.equal('00:00:00:00:00:01');