使用 Java 将 Unix 时间中的日期时间作为字节数组获取，其大小为 4 个字节
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Getting Date Time in Unix Time as Byte Array which size is 4 bytes with Java
提问by Figen Güng?r
How Can I get the date time in unix time as byte array which should fill 4 bytes space in Java?
如何将 unix 时间中的日期时间作为字节数组获取，该数组应该在 Java 中填充 4 个字节的空间？
Something like that:
类似的东西：
byte[] productionDate = new byte[] { (byte) 0xC8, (byte) 0x34,
(byte) 0x94, 0x54 };
回答by Jesper
First: Unix timeis a number of secondssince 01-01-1970 00:00:00 UTC. Java's System.currentTimeMillis()returns millisecondssince 01-01-1970 00:00:00 UTC. So you will have to divide by 1000 to get Unix time:
第一：Unix时间是一个数字的秒因为01-01-1970 00:00:00 UTC。JavaSystem.currentTimeMillis()返回自 01-01-1970 00:00:00 UTC 以来的毫秒数。所以你必须除以 1000 才能得到 Unix 时间：
int unixTime = (int)(System.currentTimeMillis() / 1000);
Then you'll have to get the four bytes in the intout. You can do that with the bit shift operator>>(shift right). I'll assume you want them in big endianorder:
然后你必须在输出中获取四个字节int。您可以使用位移运算符>>（右移）来做到这一点。我假设您希望它们按大端顺序排列：
byte[] productionDate = new byte[]{
(byte) (unixTime >> 24),
(byte) (unixTime >> 16),
(byte) (unixTime >> 8),
(byte) unixTime
};
回答by Peter Lawrey
You can use ByteBufferto do the byte manipulation.
您可以使用ByteBuffer进行字节操作。
int dateInSec = (int) (System.currentTimeMillis() / 1000);
byte[] bytes = ByteBuffer.allocate(4).putInt(dateInSec).array();
You may wish to set the byte order to little endian as the default is big endian.
您可能希望将字节顺序设置为小端，因为默认是大端。
To decode it you can do
要解码它，你可以做
int dateInSec = ByteBuffer.wrap(bytes).getInt();
