问题描述
在Java for String类中有一个名为matches的方法，如何使用此方法检查我的字符串是否只使用正则表达式的数字。我尝试了下面的例子，但是他们两个都把我弄错了。
In Java for String class there is a method called matches, how to use this method to check if my string is having only digits using regular expression. I tried with below examples, but both of them returned me false as result.
String regex = "[0-9]";
String data = "23343453";
System.out.println(data.matches(regex));
String regex = "^[0-9]";
String data = "23343453";
System.out.println(data.matches(regex));
推荐答案
尝试
String regex = "[0-9]+";
或
String regex = "\\d+";
根据Java 正则表达式， + 表示一次或多次， \d 表示一位数。
As per Java regular expressions, the + means "one or more times" and \d means "a digit".
注意：双反斜杠是转义序列以获得单个反斜杠 - 因此，java字符串中的 \\d 为您提供实际结果： \d
Note: the "double backslash" is an escape sequence to get a single backslash - therefore, \\d in a java String gives you the actual result: \d
参考文献：
编辑由于在其他答案中有些困惑，我正在编写一个测试用例，并会详细解释更多的事情。
due to some confusion in other answers, I am writing a test case and will explain some more things in detail.
首先，如果您对此解决方案的正确性有疑问（或其他），请运行此测试c ase：
Firstly, if you are in doubt about the correctness of this solution (or others), please run this test case:
String regex = "\\d+";
// positive test cases, should all be "true"
System.out.println("1".matches(regex));
System.out.println("12345".matches(regex));
System.out.println("123456789".matches(regex));
// negative test cases, should all be "false"
System.out.println("".matches(regex));
System.out.println("foo".matches(regex));
System.out.println("aa123bb".matches(regex));
问题1：是否需要添加 ^ 和 $ 到正则表达式，所以它不匹配aa123bb？
Question 1: Isn't it necessary to add ^ and $ to the regex, so it won't match "aa123bb" ?
否。在java中，匹配方法（在问题中指定）匹配完整的字符串，而不是片段。换句话说，没有必要使用 ^ \\d + $ （即使它也是正确的）。请参阅上一个负面测试用例。
No. In java, the matches method (which was specified in the question) matches a complete string, not fragments. In other words, it is not necessary to use ^\\d+$ (even though it is also correct). Please see the last negative test case.
请注意，如果您使用在线正则表达式检查程序，那么这可能会有所不同。要在Java中匹配字符串的片段，您可以使用 find 方法，详细说明如下：
Please note that if you use an online "regex checker" then this may behave differently. To match fragments of a string in Java, you can use the find method instead, described in detail here:
问题2：此正则表达式是否也与空字符串匹配，？
否。正则表达式 \\\\ * 将匹配空字符串，但 \\d + 没有。星号 * 表示零或更多，而加号 + 表示一个或多个。请参阅第一个否定测试用例。
No. A regex \\d* would match the empty string, but \\d+ does not. The star * means zero or more, whereas the plus + means one or more. Please see the first negative test case.
问题3：编译正则表达式模式不是更快吗？
是。编译一次正则表达式模式确实更快，而不是每次调用匹配，因此，如果性能影响很重要，那么 Pattern 可以像这样编译和使用：
Yes. It is indeed faster to compile a regex Pattern once, rather than on every invocation of matches, and so if performance implications are important then a Pattern can be compiled and used like this:
Pattern pattern = Pattern.compile(regex);
System.out.println(pattern.matcher("1").matches());
System.out.println(pattern.matcher("12345").matches());
System.out.println(pattern.matcher("123456789").matches());
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