Java正则表达式匹配具有特殊字符的精确单词

Question

I have list of keywords entered by the user and they may contains the special characters like $, #, @, ^, &, etc.

As per my requirement when ever i receive list of text messages i need to search for all the keywords in every message.

We need to match exact keyword .

CASE 1: Simple Keyword - Simple Message

I used \\b to match exact keyword and it works fine.

public static void main(String[] args) {
        String patternStr =  "(?i)\\bHello\\b";

        Pattern pattern = Pattern.compile(patternStr);

        List<String> strList = new ArrayList<String>();
        strList.add("HHello Message");
        strList.add("This is Hello Message ");
        strList.add("Now Hellos again.");

        for(String str : strList) {
            Matcher matcher = pattern.matcher(str);
            System.out.println(">> "+matcher.find());
        }
    }

OUTPUT as Expected

>> false
>> true
>> false

CASE 2 : Simple Keyword - Message with Special Character

Now, if i run above same code for following messages then it didn't work as expected.

List<String> strList = new ArrayList<String>();
strList.add("#Hello Message");
strList.add("This is Hello Message ");
strList.add("Now Hellos again.");

OUTPUT:

true
true
false

Expected OUTPUT

false
true
false

CASE 3 : Keyword & Message with Special Character

If i receive following messages and Keyword is #Hello . I wrote following code but it didn't work .

public static void main(String[] args) {
        String patternStr =  "(?i)\\b#Hello\\b";

        Pattern pattern = Pattern.compile(patternStr);

        List<String> strList = new ArrayList<String>();
        strList.add("HHello Message");
        strList.add("This is #Hello Message ");
        strList.add("Now Hellos again.");

        for(String str : strList) {
            Matcher matcher = pattern.matcher(str);
            System.out.println(">> "+matcher.find());
        }
    }

OUTPUT:

>> false
>> false
>> false

Expected OUTPUT:

>> false
>> true
>> false

How can i escape the special characters and resolve CASE 2 and CASE 3 .

Please help.

Answer 1

Case 2 seems the opposite as case 3, so I don't think you can combine the Pattern s.

For case 2, your Pattern could look like:

Pattern pattern = Pattern.compile("(\\s|^)Hello(\\s|$)", Pattern.CASE_INSENSITIVE);

In this case we surround the keyword by whitespace or beginning/end of input.

For case 3, your Pattern could look like:

Pattern pattern = Pattern.compile("[\\$#@\\^&]Hello(\\s|$)", Pattern.CASE_INSENSITIVE);

In this case, we precede the keyword with any of the special characters of your choice (note the escaped reserved characters $ and ^ ), then we accept whitespace or the end of input as the character following the keyword.

Answer 2

使用(?:^|\\s) （“文本或空白的开头”）代替第一个\\b和(?:$|\\s) （“文本结尾或空白”）而不是第二个\\b in你的正则表达式。

Answer 3

The problem comes from the way that "exact word" is defined. It is not just whitespace that can surround the word to make it a word. For example in most circumstances one would want an exact word match for 'Hello' to work with.

"hello there", "That young man just said hello to that other young man" and "I wish people would still answer the telephone by saying ahoy rather than Hello."

If you want the match to be only split on whitespace then I believe you will have to specify the whitespace condition. Assuming you also want to it to match at the end then I would propose something like this.

Pattern pattern = Pattern.compile("\(^\| \)" + escapeSearchString(patternString) + "\( \|$\)");

and then have a couple of methods like this

public String escapeSearchString(String patternString) {
    StringBuilder stringBuilder = new StringBuilder(patternString.length() * 3);
    for (char c : patternString.toCharArray()) {
        if (isEscapableCharacter(c)) {
            stringBuilder.append("\\");
        }
        stringBuilder.append(c);
    }
}

public boolean isEscapableCharacter(char c) {
    switch (c) {
        case '#':
        case '$':
        case '@':
        case '^':
        case '&':
            return true;
        default:
            return false;
    }
}

It would probably be better to iterate over a char[] for the escapable characters and load them from a config file.

Answer 4

Try maybe this way

String patternStr = "(?i)(?<=\\s|^)"+Pattern.quote(searchedStubstring)+"(?=\\s|$)";

(?<=...) and (?=...) is positive look behind and ahead so it will check if before your searchedStubstring will have

• white-space \\\\s or start of the input ^ before, and
• white-space \\\\s or end of the input & after it.

Also in case you would like to searched for special characters like $ + and others you need to escape them. To do this you can use Pattern.quote(searchedStubstring)

Answer 5

for example if your word want to have special char (for example here '#') at the begining and end of this you have to write the following:

Pattern p = Pattern.compile("(\\s|^|#)"+word+"(\\s|\\#|$)", Pattern.CASE_INSENSITIVE);

if you want exact match:

Pattern p = Pattern.compile("(\\s|^)"+word+"(\\s|$)", Pattern.CASE_INSENSITIVE);

with '|' is like OR so you can add as match special char's you want ..for example:

Pattern p = Pattern.compile("(\\s|^|#|:|-)"+word+"(\\s|\\#|\\,|\\.|$)", Pattern.CASE_INSENSITIVE);

char '^' means to detect the string at beginning of line and '$' means at end of line. see more here: Summary of regular-expression constructs