blocks|key|731347|text|将正则表达式更改为：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|731348|.*\\W%2B.*|code-block|syntax|javascript|731349|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Change your regex to:

<pre><code>.*\\W+.*
</code></pre>

blocks|key|947350|text|当我不得不做同样的事情时，我使用的正则表达式是"(\w)*“这就是我使用的。不确定大写字母w是否相同，但我也使用了括号。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|947351|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

When I had to do this same thing the regex I use is "(\w)*" Thats what I use. Not sure if capitol w is the same but I also used parenthesis.

blocks|key|731427|text|问题出在'*'上。'*'匹配零个或多个字符。您希望匹配至少一个非单词字符，因此必须使用'%2B'作为数量修饰符。因此匹配\W%2B+(大写W表示非单词)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|731428|entityMap^0|4|3|9|3|17|3|1M|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|G|$]]

The problem is the <code>'*'</code>. <code>'*'</code> matches ZERO or more characters. You want to match at least one non word character, so you must use <code>'+'</code> as the quantity modifier. Hence match <code>\W+</code> (Capital W there for NON word)

blocks|key|731681|text|方法的大小写错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|731682|匹配器被声明为m，但用作匹配器。|731683|重复应该是“一个或多个”%2B，而不是“零或多个”*这是正确的：|731684|String+term+=+"Hello-World";
boolean+found+=+false;
Pattern+p+=+Pattern.compile("\\W%2B");//<--+compile(+not+Compile(
Matcher+m+=+p.matcher(term);++//<--+matcher(+not+Matcher
if(m.find())+{++//<--+m+not+matcher
++++found+=+true;
}|code-block|syntax|javascript|731685|顺便说一句，如果你只需要：|731686|boolean+found+=+m.find();|731687|:)|731688|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|U|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|X|8|@]|9|@]|A|$I|J]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Methods are in the wrong case. 

The matcher was declared as m but used as matcher. 

The repetition should be "one or many" + instead of "zero or many " *
This works correctly:

<pre><code>String term = "Hello-World";
boolean found = false;
Pattern p = Pattern.compile("\\W+");//&lt;-- compile( not Compile(
Matcher m = p.matcher(term); //&lt;-- matcher( not Matcher
if(m.find()) { //&lt;-- m not matcher
 found = true;
}
</code></pre>

Btw, it would be enough if you just :

<pre><code>boolean found = m.find();
</code></pre>

:)

blocks|key|947652|text|您的表达式没有考虑到可能的非英文字母。它也比它需要的更复杂。除非您使用正则表达式是出于某些非必要的原因(如您的教授告诉您的那样)，否则您最好使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|947653|boolean+found+=+false;
for+(int+i=0;i<mystring.length();%2B%2Bi)+{
++if+(!Character.isLetterOrDigit(mystring.charAt(i)))+{
++++found=true;
++++break;
++}
}|code-block|syntax|javascript|947654|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Your expression does not take account of possible non-English letters. It's also more complicated than it needs to be. Unless you are using regexs for some reason other than need (such as your professor having told you to) you are much better off with:

<pre><code>boolean found = false;
for (int i=0;i&lt;mystring.length();++i) {
 if (!Character.isLetterOrDigit(mystring.charAt(i))) {
 found=true;
 break;
 }
}
</code></pre>

blocks|key|731757|text|现在是2016年或更晚，你应该考虑其他字母表中的国际字符串，而不仅仅是拉丁语。在这种情况下，经常引用的[%5Ea-zA-Z]将不匹配。现在在Java中有更好的方法：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|731758|[%5E\\p{IsAlphabetic}%5E\\p{IsDigit}]|code-block|syntax|javascript|731759|请参阅the+reference+(“Unicode脚本、块、类别和二进制属性的类”一节)。还有一个我觉得很有用的this+answer。|731760|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html|1|https://stackoverflow.com/a/29718900/2535335^0|1F|9|0|0|3|D|0|1L|B|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|M|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

It's 2016 or later and you should think about international strings from other alphabets than just Latin. The frequently cited <code>[^a-zA-Z]</code> will not match in that case. There are better ways in Java now:

<pre><code>[^\\p{IsAlphabetic}^\\p{IsDigit}]
</code></pre>

See <a href="https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html" rel="noreferrer">the reference</a> (section "Classes for Unicode scripts, blocks, categories and binary properties"). There's also <a href="https://stackoverflow.com/a/29718900/2535335">this answer</a> that I found helpful.

blocks|key|731764|text|if+(value.matches(".*[%5Ea-zA-Z0-9].*"))+{+//+tested,+seems+to+work.
++++System.out.println("match");
}+else+{
++++System.out.println("no+match");
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|731765|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>if (value.matches(".*[^a-zA-Z0-9].*")) { // tested, seems to work.
 System.out.println("match");
} else {
 System.out.println("no match");
}
</code></pre>

blocks|key|947745|text|如果您可以使用Apache+StringUtils，那么它就像下面这样简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|947746|StringUtils.isAlphanumeric(inp)|code-block|syntax|javascript|947747|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you are okay to use Apache StringUtils, then it's as simple as following 

<pre><code>StringUtils.isAlphanumeric(inp)
</code></pre>

This is my code to determine if a word contains any non-alphanumeric characters:

<pre><code> String term = "Hello-World";
 boolean found = false;
 Pattern p = Pattern.Compile("\\W*");
 Matcher m = p.Matcher(term);
 if(matcher.find())
 found = true;
</code></pre>

I am wondering if the regex expression is wrong. I know <code>"\W"</code> would matches any non-word characters. Any idea on what I am missing ??

Java regex: check if word has non alphanumeric characters

Java

这是我的代码，用于确定单词是否包含任何非字母数字字符：  String term = "Hello-World";  boolean found = false;  Pattern p = Pattern.Compile("\\W*");  Matcher m = p.Matcher(term);  if(matche...

问Java regex:检查word是否包含非字母数字字符
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