在 Java 中为当前时间添加一个大时间戳 [英] Adding a large timestamp to the current time in Java

问题描述

我正在开发一个支持 Google 两步验证的应用程序.此应用程序还支持信任此设备 30 天"的功能.

I am working on an application that supports Google 2-step verification. This application also supports a feature to 'trust this device for 30 days'.

我使用数据库来保存所有这些信息，例如 IP 地址和过期时间.现在，当我填写时间戳 System.currentTimeMillis() + 30 * 24 * 60 * 60 * 1000 以将当前时间添加 30 天时，它会在数据库中插入一个早于当前时间的时间戳.

I use a database to save all this information such as IP-addresses and expire times. Now when I fill in the timestamp System.currentTimeMillis() + 30 * 24 * 60 * 60 * 1000 to add 30 days to the current time, it inserts a timestamp earlier than the current time into the database.

例如:当前时间 = 1483223733000 (2016-31-12 11:36 PM UTC+1).现在，当我添加 30 天(即 2592000000 毫秒)时，它的日期类似于 1481520984841 (2016-12-12 6:36 AM UTC+1)不是提前 30 天，而是提前 19 天.

For example: the current time = 1483223733000 (2016-31-12 11:36 PM UTC+1). Now when I add 30 days (which is 2592000000 milliseconds, it comes to a date similar to 1481520984841 (2016-12-12 6:36 AM UTC+1) which is not 30 days ahead, but rather about 19 days back in time.

推荐答案

这个问题与 32 位整数溢出有关.由于整数的最大值是 2147483647，以毫秒为单位的 30 天对于整数来说太大了，并且会导致像 -1702967296 这样的整数(大约是 -19以毫秒为单位的天数.)

This problem had to do with a 32-bit integer overflow. Since the maximum value for an integer is 2147483647, 30 days in milliseconds would be too large for an integer and would result in an integer like -1702967296 (Which is about -19 days in milliseconds.)

为了解决这个问题，我使用了 long 而不是 int.所以现在我这样做:System.currentTimeMillis() + 30L * 24 * 60 * 60 * 1000;

To solve this problem, I use a long instead of an int. So now I do: System.currentTimeMillis() + 30L * 24 * 60 * 60 * 1000;

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