java 不允许在开始时使用空格的正则表达式
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regular expression for not allowing white space at start
提问by user679526
A regular expression to allow only alphabets and numbers and spaces only in between alphabets with a maximum size of 20.
一个正则表达式，只允许字母和数字以及字母之间的空格，最大大小为20。
([a-zA-Z0-9]+([\\s][a-zA-Z0-9]+)*){0,20}.
([a-zA-Z0-9]+([\\s][a-zA-Z0-9]+)*){0,20}.
This does not allow white space at start, but it is not checking the max size condition. How can I change this regular expression?
这在开始时不允许有空格，但它不会检查最大大小条件。如何更改此正则表达式？
回答by tripleee
You are specifying 20 repetitions of the entire pattern. I am guessing you probably mean something like
您正在指定整个模式的 20 次重复。我猜你的意思可能是
[a-zA-Z0-9][\sa-zA-Z0-9]{0,19}
If empty input should be allowed, wrap the whole thing in (...)?.
如果应该允许空输入，请将整个内容包装在(...)?.
回答by Code Jockey
All Sorts of ways to write this, and since you're using Java, why not use a Java regex "feature"? :D
各种各样的方法来写这个，既然你使用的是 Java，为什么不使用 Java 正则表达式“功能”？:D
String regexString = "(?<!\s+)[\w\s&&[^_]]{0,20}";
Broken down, this says:
分解，这说：
(?<!\s+) # not following one or more whitespace characters,
[ # match one of the following:
\w # word character (`a-z`, `A-Z`, `0-9`, and `_`)
\s # whitespace characters
&&[^_] # EXCEPT FOR `_`
] #
{0,20} # between 0 and 20 times
It will match a-z, A-Z, 0-9, and white space, even though the \w would otherwise include underscores, the extra part there says NOT underscores - I think it's unique to Java... anyways, that was fun!
它将匹配a-z、A-Z、0-9和空格，即使 \w 否则会包含下划线，那里的额外部分表示 NOT 下划线 - 我认为它是 Java 独有的......无论如何，这很有趣！
回答by FailedDev
The regex below :
下面的正则表达式：
boolean foundMatch = subjectString.matches("(?i)^(?=.{1,20}$)[a-z0-9]+(?:\s[a-z0-9]+)*$");
will match a string of 1 to 20 characters starting with an alphanumeric character followed by a single space and more alphanumeric characters. Note that the string must end with an alphanumeric character and not a space.
将匹配 1 到 20 个字符的字符串，以字母数字字符开头，后跟单个空格和更多字母数字字符。请注意，字符串必须以字母数字字符而不是空格结尾。
