You can use replaceAll () with wildcards

Question

Good morning. I understand that there are many questions regarding replacement and replaceAll() , but I have not seen this.

What needs to be done is to parse the line (containing the actual html for the point), after I see the second <p> instance in the line, I want to delete everything that starts with and ends with; until I see the next </p>

To do the second part, I was hoping to use something along the lines of s.replaceAll("&*;","")

This does not work, but hopefully it gets my point of view that I am looking to replace everything that starts with and ends with:

Answer 1

You should probably leave the parsing DOM parser ( see this question ). I can almost guarantee that you will need to do this to find the text in the <p> tags.

For the replacement logic, String.replaceAll uses regular expressions that can fulfill the desired match.

The "wildcard" in the regular expressions you want is the expression .* . Using your example:

 String ampStr = "This &escape;String"; String removed = ampStr.replaceAll("&.*;"); System.out.println(removed); 

This outputs This String . This is due to the fact that . represents any character, and * means "this character is 0 or more times." So. .* Basically means "any number of characters." However, feeding him:

 "This &escape;String &anotherescape;Extended" 

probably won't do what you want and This Extended will output. To fix this, you specify exactly what you want to look for instead of a character . . This is done using [^;] , which means "any character that is not a semicolon:

 String removed = ampStr.replaceAll("&[^;]*;"); 

This gives performance advantages over &.*?; for non-matching lines, so I highly recommend using this version, especially since not all HTML files will contain the &abc; token , and the version &.*?; can have tremendous neck performance as a result.

Answer 2

Do you want to:

 s.replaceAll("&.*?;",""); 

But do you really want to parse HTML this way? You might be better off using an XML parser.