How can I get non-matching groups using a Matcher in Java? - java

I'm trying to write a java regex to catch some groups of words from a String using a Matcher.
Say i got this string: "Hello, we are #happy# to see you today".
I would like to get 2 group of matches, one having
Hello, we are
to see you today
and the other
happy
So far, I was only able to match the word between the #s using this Pattern:
Pattern p = Pattern.compile("#(.+?)#");
I've read about negative lookahead and lookaround, played a bit with it but without success.
I assume I should do some sort of negation of the regex so far, but I couldn't come up with anything.
Any help would be really appreciated, thank you.


From comment:
I may incur in a string where I got more than one instances of words wrapped by #, such as "#Hello# kind #stranger#"
From comment:
I need to apply some different style format to both the text inside and outside.
Since you need to apply different stylings, the code need to process each block of text separately, and needs to know if the text is inside or outside a #..# section.
Note, in the following code, it will silently skip the last #, if there is an odd number of them.
String input = ...
for (Matcher m = Pattern.compile("([^#]+)|#([^#]+)#").matcher(input); m.find(); ) {
if (m.start(1) != -1) {
String outsideText = m.group(1);
System.out.println("Outside: \"" + outsideText + "\"");
} else {
String insideText = m.group(2);
System.out.println("Inside: \"" + insideText + "\"");
}
}
Output for input = "Hello, we are #happy# to see you today"
Outside: "Hello, we are "
Inside: "happy"
Outside: " to see you today"
Output for input = "#Hello# kind #stranger#"
Inside: "Hello"
Outside: " kind "
Inside: "stranger"
Output for input = "This #text# has unpaired # characters"
Outside: "This "
Inside: "text"
Outside: " has unpaired "
Outside: " characters"


The best I could do is splitting in 3 groups, then merging the group 1 and 4 :
(^.*)(\#(.+?)\#)(.*)
Test it here
EDIT: Taking remarks from the comments :
(^[^\#]*)(?:\#(.+?)\#)([^\#]*)
Thanks to #Lino we don't capture the useless group with # anymore, and we capture anything except #, instead of any non whitespace character in the 1st and 2nd groups.
Test it here


Is this solution fine?
Pattern pattern =
Pattern.compile("([^#]+)|#([^#]*)#");
Matcher matcher =
pattern.matcher("Hello, we are #happy# to see you today");
List<String> notBetween = new ArrayList<>(); // not surrounded by #
List<String> between = new ArrayList<>(); // surrounded by #
while (matcher.find()) {
if (Objects.nonNull(matcher.group(1))) notBetween.add(matcher.group(1));
if (Objects.nonNull(matcher.group(2))) between.add(matcher.group(2));
}
System.out.println("Printing group 1");
for (String string :
notBetween) {
System.out.println(string);
}
System.out.println("Printing group 2");
for (String string :
between) {
System.out.println(string);
}

Related

Regex to remove line break within double quote in CSV

Hi I have a csv file with an error in it.so i want it to correct with regular expression, some of the fields contain line break, Example as below
"AHLR150","CDS","-1","MDCPBusinessRelationshipID",,,"Investigating","1600 Amphitheatre Pkwy
California",,"Mountain View",,"United States",,"California",,,"94043-1351","9958"
the above two lines should be in one line
"AHLR150","CDS","-1","MDCPBusinessRelationshipID",,,"Investigating","1600 Amphitheatre PkwyCalifornia",,"Mountain View",,"United States",,"California",,,"94043-1351","9958"
I tried to use the below regex but it didnt help me
%s/\\([^\"]\\)\\n/\\1/

Try this:
public static void main(String[] args) {
String input = "\"AHLR150\",\"CDS\",\"-1\",\"MDCPBusinessRelationshipID\","
+ ",,\"Investigating\",\"1600 Amphitheatre Pkwy\n"
+ "California\",,\"Mountain View\",,\"United\n"
+ "States\",,\"California\",,,\"94043-1351\",\"9958\"\n";
Matcher matcher = Pattern.compile("\"([^\"]*[\n\r].*?)\"").matcher(input);
Pattern patternRemoveLineBreak = Pattern.compile("[\n\r]");
String result = input;
while(matcher.find()) {
String quoteWithLineBreak = matcher.group(1);
String quoteNoLineBreaks = patternRemoveLineBreak.matcher(quoteWithLineBreak).replaceAll(" ");
result = result.replaceFirst(quoteWithLineBreak, quoteNoLineBreaks);
}
//Output
System.out.println(result);
}
Output:
"AHLR150","CDS","-1","MDCPBusinessRelationshipID",,,"Investigating","1600 Amphitheatre Pkwy California",,"Mountain View",,"United States",,"California",,,"94043-1351","9958"

Create a RegEx surrounding the text you want to keep by parentheses and that will create a group of matched characters. Then replace the string using the group index to compose as you wish.
String test = "\"AHLR150\",\"CDS\",\"-1\",\"MDCPBusinessRelationshipID\","
+ ",,\"Investigating\",\"1600 Amphitheatre Pkwy\n"
+ "California\",,\"Mountain View\",,\"United\n"
+ "States\",,\"California\",,,\"94043-1351\",\"9958\"\n";
System.out.println(test.replaceAll("(\"[^\"]*)\n([^\"]*\")", "$1$2"));
So when we replace the matching string ("United\nStates") by $1$2 we are removing the line break because it not belongs to any group:
$1 => the first group (\"[^\"]*) that will match "United
$2 => the second group ([^\"]*\")" that will match States"

Based on this you can try with:
/\r?\n|\r/
I checked it here and seems to be fine

Need help in regex matching

It may be very simple, but I am extremely new to regex and have a requirement where I need to do some regex matches in a string and extract the number in it. Below is my code with sample i/p and required o/p. I tried to construct the Pattern by referring to https://www.freeformatter.com/java-regex-tester.html, but my regex match itself is returning false.
Pattern pattern = Pattern.compile(".*/(a-b|c-d|e-f)/([0-9])+(#[0-9]?)");
String str = "foo/bar/Samsung-Galaxy/a-b/1"; // need to extract 1.
String str1 = "foo/bar/Samsung-Galaxy/c-d/1#P2";// need to extract 2.
String str2 = "foo.com/Samsung-Galaxy/9090/c-d/69"; // need to extract 69
System.out.println("result " + pattern.matcher(str).matches());
System.out.println("result " + pattern.matcher(str1).matches());
System.out.println("result " + pattern.matcher(str1).matches());
All of above SOPs are returning false. I am using java 8, is there is any way by which in a single statement I can match the pattern and then extract the digit from the string.
I would be great if somebody can point me on how to debug/develop the regex.Please feel free to let me know if something is not clear in my question.

You may use
Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");
See the regex demo
When used with matches(), the pattern above does not require explicit anchors, ^ and $.
Details
.* - any 0+ chars other than line break chars, as many as possible
/ - the rightmost / that is followed with the subsequent subpatterns
(?:a-b|c-d|e-f) - a non-capturing group matching any of the alternatives inside: a-b, c-d or e-f
/ - a / char
[^/]*? - any chars other than /, as few as possible
([0-9]+) - Group 1: one or more digits.
Java demo:
List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");
for (String s : strs) {
Matcher m = pattern.matcher(s);
if (m.matches()) {
System.out.println(s + ": \"" + m.group(1) + "\"");
}
}
A replacing approach using the same regex with anchors added:
List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
String pattern = "^.*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)$";
for (String s : strs) {
System.out.println(s + ": \"" + s.replaceFirst(pattern, "$1") + "\"");
}
See another Java demo.
Output:
foo/bar/Samsung-Galaxy/a-b/1: "1"
foo/bar/Samsung-Galaxy/c-d/1#P2: "2"
foo.com/Samsung-Galaxy/9090/c-d/69: "69"

Because you match always the last number in your regex, I would Like to just use replaceAll with this regex .*?(\d+)$ :
String regex = ".*?(\\d+)$";
String strResult1 = str.replaceAll(regex, "$1");
System.out.println(!strResult1.isEmpty() ? "result " + strResult1 : "no result");
String strResult2 = str1.replaceAll(regex, "$1");
System.out.println(!strResult2.isEmpty() ? "result " + strResult2 : "no result");
String strResult3 = str2.replaceAll(regex, "$1");
System.out.println(!strResult3.isEmpty() ? "result " + strResult3 : "no result");
If the result is empty then you don't have any number.
Outputs
result 1
result 2
result 69

Here is a one-liner using String#replaceAll:
public String getDigits(String input) {
String number = input.replaceAll(".*/(?:a-b|c-d|e-f)/[^/]*?(\\d+)$", "$1");
return number.matches("\\d+") ? number : "no match";
}
System.out.println(getDigits("foo.com/Samsung-Galaxy/9090/c-d/69"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/a-b/some other text/1"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/9090/a-b/69ace"));
69
no match
no match
This works on the sample inputs you provided. Note that I added logic which will display no match for the case where ending digits could not be matched fitting your pattern. In the case of a non-match, we would typically be left with the original input string, which would not be all digits.

Can't undertand why my regex in Java doesn't work

I'm trying to find pieces of text on the webpage I fetch that lay between 'align="left">\n" and '</form>\n</td>' substrings.
I wrote a regex:
(align=\"left\">\\n)(?<part>.*?)(<\/form>\\n<\/td>)
and tested it at https://www.freeformatter.com/java-regex-tester.html where it works just as I need.
But in the Java code it can't find anything.
My test code that I'm trying make working:
String frontPage = "<html>\n<head>\n<title>Hello</title>\n</head>\n" +
"<body>\n<table>\n<tr align=\"left\">\n" +
"<td>Hello \n<form>\n<input type=\"submit\" value=\"ok\">\n" +
"</form>\n</td>\n" +
"<td>World \n<form>\n<input type=\"submit\" value=\"ok\">\n" +
"</form>\n</td>\n" +
"</tr>\n</table>\n</body>\n</html>";
java.util.regex.Pattern p =
java.util.regex.Pattern.compile(
"(align=\"left\">\\n)(?<part>.*?)(<\\/form>\\n<\\/td>)");
java.util.regex.Matcher m = p.matcher(frontPage);
List<String> parts = new ArrayList<>();
while (m.find()) {
parts.add(m.group("part"));
}
if (parts.size() == 0)
System.out.println("No page parts found");
else {
System.out.println("Something matches at least");
}
It finds matches if only first two groups specified, but when I add at least simple (form) sequence to the last group, it stops matching anything, and I can't even guess why.

Add DOTALL to the compile. Like
java.util.regex.Pattern.compile(
"(align=\"left\">\\n)(?<part>.*?)(<\\/form>\\n<\\/td>)",
java.util.regex.Pattern.DOTALL
);
See it here at ideone.

take replaced variable from replaceAll

I have a big string and I wanna take links from that string. I can print link.
Pattern pattern = Pattern.compile(".*(?<=overlay-link\" href=\").*?(?=\">).*");
with that code. Example output:
<a title="TITLE" class="overlay-link" href="LINK HERE"></a>
when I try string.replaceAll, regex deleting link and printing another variables.
EX: <a title="TITLE" class="overlay-link" href=""></a>
I am new on regex. Can you help me?
Here is full code :
String content;
Pattern pattern = Pattern.compile(".*(?<=overlay-link\" href=\").*?(?=\">).*");
try {
Scanner scanner = new Scanner(new File("sourceCode.txt"));
while (scanner.hasNext()) {
content = scanner.nextLine();
if (pattern.matcher(content).matches()) {
System.out.println(content.replaceAll("(?<=overlay-link\" href=\").*?(?=\">)", ""));
}
}
} catch (IOException ex) {
Logger.getLogger(SourceCodeExample.class.getName()).log(Level.SEVERE, null, ex);
}

If I understand your question correctly you are looking to pull out just the link specified in the href tag.
To do this you should use a capture group in your regex itself instead of trying to replaceAll.
The replaceAll method is accurately finding the link and replacing it with an empty string and returning the full resulting string as per the docs which is not the desired result.
The regex you should use is as such: .*(?<=overlay-link\" href=\")(.*?)(?=\">).* Notice the capture group () around the link.
This will allow you to find the matches and access the capture group 1. I found a good example of how to do this in this other question. (important snippet pasted below)
String line = "This order was placed for QT3000! OK?"; //<a> tag string
Pattern pattern = Pattern.compile("(.*?)(\\d+)(.*)"); //insert regex provided above
Matcher matcher = pattern.matcher(line);
while (matcher.find()) {
System.out.println("group 1: " + matcher.group(1)); //This will be your link
System.out.println("group 2: " + matcher.group(2));
System.out.println("group 3: " + matcher.group(3));
}
Comments added by me
Note: index 0 represents the whole Pattern

Java Regex ReplaceAll with grouping

I want to surround all tokens in a text with tags in the following manner:
Input: " abc fg asd "
Output:" <token>abc</token> <token>fg</token> <token>asd</token> "
This is the code I tried so far:
String regex = "(\\s)([a-zA-Z]+)(\\s)";
String text = " abc fg asd ";
text = text.replaceAll(regex, "$1<token>$2</token>$3");
System.out.println(text);
Output:" <token>abc</token> fg <token>asd</token> "
Note: for simplicity we can assume that the input starts and ends with whitespaces

Use lookaround:
String regex = "(?<=\\s)([a-zA-Z]+)(?=\\s)";
...
text = text.replaceAll(regex, "<token>$1</token>");

If your tokens are only defined with a character class you don't need to describe what characters are around. So this should suffice since the regex engine walks from left to right and since the quantifier is greedy:
String regex = "[a-zA-Z]+";
text = text.replaceAll(regex, "<token>$0</token>");

// meaning not a space, 1+ times
String result = input.replaceAll("([^\\s]+)", "<token>$1</token>");
this matches everything that isn't a space. Prolly the best fit for what you need. Also it's greedy meaning it will never leave out a character that it shouldn't ( it will never find the string "as" in the string "asd" when there is another character with which it matches)

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