带有正则表达式分隔符的 Java Scanner
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Java Scanner with regex delimiter
提问by Sawyer
Why does the following code return false?
为什么下面的代码返回false？
Scanner sc = new Scanner("-v ");
sc.useDelimiter("-[a-zA-Z]\s+");
System.out.println(sc.hasNext());
The weird thing is -[a-zA-Z]//s+will return true.
奇怪的是-[a-zA-Z]//s+会返回true。
I also can't understand why this returns true:
我也不明白为什么这会返回 true：
Scanner sc = new Scanner(" -v");
sc.useDelimiter("-[a-zA-Z]\s+");
System.out.println(sc.hasNext());
回答by John Kugelman
A scanner is used to break up a string into tokens. Delimiters are the separatorsbetween tokens. The delimiters are what aren'tmatched by the scanner; they're discarded. You're telling the scanner that -[a-zA-Z]\\s+is a delimiter and since -vmatches that regex it skips it.
扫描器用于将字符串分解为标记。分隔符是分隔标记之间。分隔符是扫描仪不匹配的内容；他们被丢弃了。您告诉扫描仪这-[a-zA-Z]\\s+是一个分隔符，并且由于-v匹配该正则表达式，因此它会跳过它。
If you're looking for a string that matches the regex, use String.matches().
如果您正在寻找与正则表达式匹配的字符串，请使用String.matches().
If your goal really is to split a string into tokens then you might also consider String.split(), which is sometimes more convenient to use.
如果您的目标确实是将字符串拆分为标记，那么您也可以考虑String.split()，有时使用起来更方便。
回答by Sawyer
Thanks John Kugelman, I think you're right.
谢谢约翰·库格曼，我认为你是对的。
Scanner can use customized delimiter to split input into tokens. The default delimiter is a whitespace.
Scanner 可以使用自定义分隔符将输入拆分为令牌。默认分隔符是空格。
When delimiter doesn't match any input, it'll result all the input as one token:
当分隔符不匹配任何输入时，它会将所有输入作为一个标记：
Scanner sc = new Scanner("-v");
sc.useDelimiter( "-[a-zA-Z]\s+");
if(sc.hasNext())
System. out.println(sc.next());
In the code above, the delimiter actually doesn't get any match, all the input "-v" will be the single token. hasNext() means has next token.
在上面的代码中，分隔符实际上没有得到任何匹配，所有输入“-v”都将是单个标记。 hasNext() 表示有下一个标记。
Scanner sc = new Scanner( "-v ");
sc.useDelimiter( "-[a-zA-Z]\s+");
if(sc.hasNext())
System. out.println(sc.next());
this will match the delimiter, and the split ended up with 0 token, so the hasNext() is false.
这将匹配分隔符，并且拆分以 0 标记结束，因此 hasNext() 为 false。
