Java 正则表达式某些字符可以存在或不存在，但之后什么都没有
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regex certain character can exist or not but nothing after that
提问by mysticfalls
I'm new to regex and I'm trying to do a search on a couple of string.
我是 regex 的新手，我正在尝试对几个字符串进行搜索。
I wanted to check if a certain character, in this case its ":" (without the quote) exist on the strings.
我想检查字符串中是否存在某个字符，在这种情况下是它的“：”（不带引号）。
If : does not exist in the string it would still match, but if : exist there should be nothing after that only space and new line will be allowed.
如果 : 不存在于字符串中，它仍然会匹配，但如果 : 存在，则之后应该没有任何内容，只允许空格和新行。
I have this pattern, but it does not seem to work as I want it.
我有这个模式，但它似乎并没有像我想要的那样工作。
(.*)(:?\s*\n*)
Thank you.
谢谢你。
采纳答案by Vector Gorgoth
If I understand your question correctly, ^[^:]*(:\s*)?$
如果我正确理解你的问题， ^[^:]*(:\s*)?$
Let's break this down a bit:
让我们把它分解一下：
^Starting anchor; without this, the match can restart itself every time it sees another colon, or non-whitespace following a colon.
^启动锚；如果没有这个，匹配可以在每次看到另一个冒号或冒号后面的非空白时重新启动。
[^:]*Match any number of characters that AREN'T colon characters; this way, if the entire string is non-colon characters, the string is treated as a valid match.
[^:]*匹配任意数量的非冒号字符；这样，如果整个字符串都是非冒号字符，则该字符串将被视为有效匹配。
(:\s*)?If at any point we dosee a colon, all following characters must be white space until the end of the string; the grouping parens and following ?act to make this an all-or-nothing conditional statement.
(:\s*)?如果在任何时候我们确实看到了一个冒号，那么所有后面的字符必须是空格，直到字符串的末尾；分组括号和以下?行为使其成为全有或全无的条件语句。
$Ending anchor; without this, the regex won't know that if it sees a colon the following whitespace MUST persist until the end of the string.
$结束锚；没有这个，正则表达式不会知道如果它看到一个冒号，后面的空格必须持续到字符串的末尾。
回答by arnaudbey
here is a pattern which should work
这是一个应该工作的模式
/^([^:]*|([^:]*:\s*))$/
you can use the pipe to manage alternatives
您可以使用管道来管理替代品
