正则表达式检查字符串只包含十六进制字符 [英] Regex to check string contains only Hex characters

问题描述

我以前从未使用过正则表达式，我发现它们对于处理字符串非常有用.我看到了一些教程(例如)，但我仍然无法理解如何制作一个简单的 Java 正则表达式检查字符串中的十六进制字符.

用户将在文本框中输入类似:0123456789ABCDEF 并且我想知道输入是否正确，否则如果返回 false 时类似 XTYSPG456789ABCDEF.

是否可以使用正则表达式来做到这一点，还是我误解了它们的工作原理?

解决方案

是的，您可以使用正则表达式做到这一点:

<前>^[0-9A-F]+$

说明:

<前>^ 行首.[0-9A-F] 字符类:0 到 9 或 A 到 F 中的任何字符.+ 量词:上述一项或多项.$ 行尾.

<小时>

要在 Java 中使用此正则表达式，您可以例如在字符串上调用 matches 方法:

boolean isHex = s.matches("[0-9A-F]+");

请注意，matches 只会找到完全匹配的内容，因此在这种情况下您不需要行首和行尾锚点.在线查看它的运行情况:ideone

您可能还希望允许大写和小写 A-F，在这种情况下，您可以使用此正则表达式:

<前>^[0-9A-Fa-f]+$

I have never done regex before, and I have seen they are very useful for working with strings. I saw a few tutorials (for example) but I still cannot understand how to make a simple Java regex check for hexadecimal characters in a string.

The user will input in the text box something like: 0123456789ABCDEF and I would like to know that the input was correct otherwise if something like XTYSPG456789ABCDEF when return false.

Is it possible to do that with a regex or did I misunderstand how they work?

解决方案

Yes, you can do that with a regular expression:

^[0-9A-F]+$

Explanation:

^            Start of line.
[0-9A-F]     Character class: Any character in 0 to 9, or in A to F.
+            Quantifier: One or more of the above.
$            End of line.

---

To use this regular expression in Java you can for example call the matches method on a String:

boolean isHex = s.matches("[0-9A-F]+");

Note that matches finds only an exact match so you don't need the start and end of line anchors in this case. See it working online: ideone

You may also want to allow both upper and lowercase A-F, in which case you can use this regular expression:

^[0-9A-Fa-f]+$

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