如何在java中生成一个4位数的随机长？
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how to generate a random long with 4 digits in java?
提问by Louise Lin
how to generate a long that is in the range of [0000,9999] (inclusive) by using the random class in java? The long has to be 4 digits.
java - 如何使用java中的随机类生成[0000,9999]（含）范围内的long？long 必须是 4 位数字。
采纳答案by Grogi
If you want to generate a number from range [0, 9999], you would use random.nextInt(10000).
如果要生成范围 [0, 9999] 中的数字，可以使用random.nextInt(10000).
Adding leading zeros is just formatting:
添加前导零只是格式化：
String id = String.format("%04d", random.nextInt(10000));
回答by Elliott Frisch
A Java intwill never have leading 0(s). You'll need a Stringfor that. You could use String.format(String, Object...)or (PrintStream.printf(String, Object...)) like
Javaint永远不会有前导0。你需要一个String。你可以使用String.format(String, Object...)或 ( PrintStream.printf(String, Object...)) 喜欢
Random rand = new Random();
System.out.printf("%04d%n", rand.nextInt(10000));
The format String%04dis for 0 filled 4 digits. And Random.nextInt(int)will be [0,10000) or[0,9999].
格式String%04d为 0 填充的 4 位数字。并且Random.nextInt(int)将是 [0,10000)或[0,9999]。
