Java：扫描字符串中的模式
这大概是一个快手吧。为什么这段代码不返回任何内容？
import java.util.Scanner;
public class MainClass {
public static void main(String[] args) {
try {
Scanner sc = new Scanner("asda ASA adad");
String pattern = "[A-Z]+";
while ((sc.hasNext(pattern))) {
System.out.println(sc.next(pattern));
}
sc.close();
} catch (Exception e) {
e.printStackTrace();
}
}
}
This is probably a quicky. Why does this code not return anything?
import java.util.Scanner;
public class MainClass {
public static void main(String[] args) {
try {
Scanner sc = new Scanner("asda ASA adad");
String pattern = "[A-Z]+";
while ((sc.hasNext(pattern))) {
System.out.println(sc.next(pattern));
}
sc.close();
} catch (Exception e) {
e.printStackTrace();
}
}
}
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评论（3）
hasNext(String pattern)仅当下一个标记与该模式匹配时才返回true。在您的情况下，“asda”是下一个标记，它与“[AZ]+”不匹配。该文档很清楚，“扫描仪不会前进超过任何输入”。如果您将模式更改为
"[A-Za-z]+"，那么您将获得三个标记，这可能正是您想要的。如果实际上您只想获取与
"[AZ]+"匹配的标记，那么您可以执行以下任一操作：useDelimiter("[^AZ] +")，然后只需调用next()skip("[^AZ]+")findInLine("[AZ]+ ")提示：如果性能至关重要，您需要使用这些方法的预编译
Pattern重载。提示：请记住，
"Xooo ABC"有两个"[AZ]+"匹配项。如果这不是您想要的，那么正则表达式必须更复杂一些。或者您始终可以简单地丢弃不匹配的标记。hasNext(String pattern)only returnstrueif the next token matches the pattern. In your case,"asda"is the next token, and that does NOT match"[A-Z]+". The documentation is clear in that "[the] scanner does not advance past any input".If you change the pattern to
"[A-Za-z]+", then you'd get three tokens, which may be what you intended.If in fact you only want to get tokens that match
"[A-Z]+", then you can do any of the following:useDelimiter("[^A-Z]+"), then simply invokenext()skip("[^A-Z]+")findInLine("[A-Z]+")Tip: if performance is critical, you'd want to use the precompiled
Patternoverloads of these methods.Tip: do keep in mind that
"Xooo ABC"has two"[A-Z]+"matches. If this is not what you want, then the regex will have to be a bit more complicated. Or you can always simply discard non-matching tokens.更改
为
Change
to
如果您希望打印出分隔符包围的所有单词，您可能需要安全起见并完全排除该模式。这样，您就不会遇到包含不在您的模式中的字符的单词，这会导致您的程序退出该循环（就像它当前正在做的那样）。例如：
If you are looking to print out all words surrounded by your delimiter you might want to be safe and exclude the pattern altogether. That way you don't come across a word that contains a character not in your pattern which would cause your program exit that loop (as it's currently doing). For example: