正则表达式排除包含java中特定单词的句子

Question

I am reading a file which contains lots of information like shown below:

    type dw_3 from u_dw within w_pg6p0012_01
    boolean visible = false
    integer x = 1797
    integer y = 388
    integer width = 887
    integer height = 112
    integer taborder = 0
    boolean bringtotop = true
    string dataobject = "d_pg6p0012_14"
    end type

    type dw_3 from u_dw within w_pg6p0012_01
    integer x = 1797
    integer y = 388
    integer width = 887
    integer height = 112
    integer taborder = 0
    boolean bringtotop = true
    string dataobject = "d_pg6p0012_14"
    end type

I made regex : (?i)type dw_\\d\\s+(.*?)\\s+within(.*?)\\s+(?!boolean visible = false)(.*) I want to extract all the strings which do not contain "boolean visible = false" but mine one is returning all. I also tried many similar posts on stack but the result is similar to mine, please suggest a way.

solution : (?i)type\\\\s+dw_(\\\\d+|\\\\w+)\\\\s+from\\\\s+.*?within\\\\s+.*?\\\\s+(string|integer)?\\\\s+.*\\\\s+.*\\\\s+.*\\\\s+.*?\\\\s+.*?\\\\s+.*?\\\\s*string\\\\s+dataobject\\\\s+=\\\\s+(.*?)\\\\s+end\\\\s+type")

This is working well on regex checker but when i tried it on java it keep on running without giving any output

Answer 1

It will be much easier (and more readable) if you make a regex to match "boolean visible = false" and then exclude those lines that contain a match for it.

Pattern pattern = Pattern.compile("boolean visible = false");

Files.lines(filepath)
     .filter(line -> !pattern.matcher(line).find())  // note the "!"
     .forEach(/* do stuff */);

Notes:

• Because we are using Files#lines(String) , it is not necessary to break apart separate lines in the regex. This is already done for us.
• The Matcher#find() method returns whether the given character sequence contains a match for the regex anywhere in it. I believe this is what you want.

---

EDIT:

Now, if you are just really intent on using a pure regex, then try this:

^((?!boolean visible = false).)+$

This will match an entire (non-empty) line if-and-only-if it does not contain "boolean visible = false" anywhere within it. No fancy backreferences / capture group semantics needed to extract the desired text.

See proof by unit tests here: https://regex101.com/r/dbzdMB/1

---

EDIT #2:

Alternatively, if all you are trying to do is to get the file text without any "boolean visible = false" , then you could simply replace every instance of that target string with the empty string.

Pattern pattern = Pattern.compile("boolean visible = false");
Matcher matcher = pattern.matcher(fileAsCharSequence);  // e.g. StringBuilder
String output = matcher.replaceAll("");

Answer 2

You can use this RegEx

(\s*boolean visible = false)|(.*)

DEMO

This basically defines 2 capture groups

1. First capture group (\\s*boolean visible = false) will catch boolean visible = false .

2. Second Capture group (.*) will capture everything else except all that's capture by first capture group.

Now when you're extracting it, just capture second group and ignore first one.

---

Edit

Here's an example for clarification:

In this example,

• getOriginalFileContents() method gets the content of the file as shown in the program.
• Notice how we're getting both the groups, but ignoring the first group and printing only the second one.

See the output, which is without that line boolean visible = false .

Output

 type dw_3 from u_dw within w_pg6p0012_01
 integer x = 1797
 integer y = 388
 integer width = 887
 integer height = 112
 integer taborder = 0
 boolean bringtotop = true
 string dataobject = "d_pg6p0012_14"
 end type


 type dw_3 from u_dw within w_pg6p0012_01
 integer x = 1797
 integer y = 388
 integer width = 887
 integer height = 112
 integer taborder = 0
 boolean bringtotop = true
 string dataobject = "d_pg6p0012_14"
 end type

Java Implementation

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class RegexTut3 {

    public static void main(String args[]) {
        String file = getOriginalFileContents();
        Pattern pattern = Pattern.compile("(\\s*boolean visible = false)|(.*)");
        Matcher matcher = pattern.matcher(file);
        while (matcher.find()) {
            //System.out.print(matcher.group(1)); //ignore this group
            if (matcher.group(2) != null) System.out.println(matcher.group(2));
        }
    }

    //this method just get's the file contents as displayed in the
    //question. 
    private static String getOriginalFileContents() {
        String s = "     type dw_3 from u_dw within w_pg6p0012_01\n" +
            "     boolean visible = false\n" +
            "     integer x = 1797\n" +
            "     integer y = 388\n" +
            "     integer width = 887\n" +
            "     integer height = 112\n" +
            "     integer taborder = 0\n" +
            "     boolean bringtotop = true\n" +
            "     string dataobject = \"d_pg6p0012_14\"\n" +
            "     end type\n" +
            "     \n" +
            "     type dw_3 from u_dw within w_pg6p0012_01\n" +
            "     integer x = 1797\n" +
            "     integer y = 388\n" +
            "     integer width = 887\n" +
            "     integer height = 112\n" +
            "     integer taborder = 0\n" +
            "     boolean bringtotop = true\n" +
            "     string dataobject = \"d_pg6p0012_14\"\n" +
            "     end type";

        return s;
    }
}

Answer 3

type dw_\d\s+(.*?)\s+within(.*)\n(?!\s*boolean visible = false\s*)[\s\S]*?\s+end type

Try this.See demo.

https://regex101.com/r/Heex8W/1