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Java RegEx meta character (.) and ordinary dot?

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java

regex

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What is Dot in regex in Java?

JavaObject Oriented ProgrammingProgramming. The subexpression/metacharacter “.” matches any single character except a newline.

Does regex match dot space?

Yes, the dot regex matches whitespace characters when using Python's re module. What is this? The dot matches all characters in the string --including whitespaces. You can see that there are many whitespace characters ' ' among the matched characters.

How do I match a dot character in regex?

A . in regex is a metacharacter, it is used to match any character. To match a literal dot in a raw Python string ( r"" or r'' ), you need to escape it, so r"\." Unless the regular expression is stored inside a regular python string, in which case you need to use a double \ ( \\ ) instead.

Is dot a special character in Java?

Java Example to Split String by Dot The examples are pretty much similar to splitting String by any delimiter, with only a focus on using the correct regular expression because the dot is a special character in Java's regular expression API. That's all about how to split a String by dot in Java.


If you want the dot or other characters with a special meaning in regexes to be a normal character, you have to escape it with a backslash. Since regexes in Java are normal Java strings, you need to escape the backslash itself, so you need two backslashes e.g. \\.


Solutions proposed by the other members don't work for me.

But I found this :

to escape a dot in java regexp write [.]


Perl-style regular expressions (which the Java regex engine is more or less based upon) treat the following characters as special characters:

.^$|*+?()[{\ have special meaning outside of character classes,

]^-\ have special meaning inside of character classes ([...]).

So you need to escape those (and only those) symbols depending on context (or, in the case of character classes, place them in positions where they can't be misinterpreted).

Needlessly escaping other characters may work, but some regex engines will treat this as syntax errors, for example \_ will cause an error in .NET.

Some others will lead to false results, for example \< is interpreted as a literal < in Perl, but in egrep it means "word boundary".

So write -?\d+\.\d+\$ to match 1.50$, -2.00$ etc. and [(){}[\]] for a character class that matches all kinds of brackets/braces/parentheses.

If you need to transform a user input string into a regex-safe form, use java.util.regex.Pattern.quote.

Further reading: Jan Goyvaert's blog RegexGuru on escaping metacharacters


Escape special characters with a backslash. \., \*, \+, \\d, and so on. If you are unsure, you may escape any non-alphabetical character whether it is special or not. See the javadoc for java.util.regex.Pattern for further information.


Here is code you can directly copy paste :

String imageName = "picture1.jpg";
String [] imageNameArray = imageName.split("\\.");
for(int i =0; i< imageNameArray.length ; i++)
{
   system.out.println(imageNameArray[i]);
}

And what if mistakenly there are spaces left before or after "." in such cases? It's always best practice to consider those spaces also.

String imageName = "picture1  . jpg";
String [] imageNameArray = imageName.split("\\s*.\\s*");
    for(int i =0; i< imageNameArray.length ; i++)
    {
       system.out.println(imageNameArray[i]);
    }

Here, \\s* is there to consider the spaces and give you only required splitted strings.