This question already has answers here:
Regex whitespace word boundary
(3 answers)
Closed 2 years ago.
How do I build a regex pattern that searches over a text T and tries to find a search string S.
There are 2 requirements:
S could be made of any character.
S could be anywhere in the string but can't be part of a word.
I know that in order to escape special regex characters I put the search string between \Q and \E as such:
\EMySearch_String\Q
How do I prevent finding partial matching of S in T?
You can do like this if
can't be part of a word
is interpreted as
preceded by start-of-string or space and followed by end-of-string or space:
String s = "3894$75\\/^()";
String text = "fdsfsd3894$75\\/^()dasdasd 22348 3894$75\\/^()";
Matcher m = Pattern.compile("(?<=^|\\s)\\Q" + s + "\\E(?=\\s|$)").matcher(text);
while (m.find()) {
System.out.println("Found match! :'" + m.group() + "'");
}
This prints only one
Found match! :'3894$75/^()'
I think what you're trying to find can be easily solved with lookaheads and lookbehinds. Take a look at this for a good explanation.
Then there's a bit of flip-flopping booleans, but you're looking ahead and behind for NOT Non-Space characters (\S). You don't want to look for space characters only because S might be at the start or end of the string. Like so:
(?<!\S)S(?!\S)
Related
This question already has answers here:
How to use java regex to match a line
(2 answers)
Closed 4 years ago.
i'm looking for a way to check whether a multiline string (from a pdf) contains a certain letter combination which must not start with a specific prefix. Specifically, i'm trying to find Strings that contain ARC but don't contain NON-ARC.
I found this great example Regular expression for a string that does not start with a sequence but it seems it does not work with my problem. With my pattern ^(?!NON\\-)ARC.* i get the expected result in a single line test, with real input the negative look ahead assertion has a false positive. Here is what i did:
#Test
public void testRegexLookAhead() {
String strTestSimplePos = "ARC 0.1-1";
String strTestSimpleNeg = "NON-ARC 3.4-1";
String strTestRealPos = "HEADLINE\r\n" + "Subheader Author\r\n" + "ARC 0.1-1\r\n" + "20190211";
String strTestRealNeg = "HEADLINE\r\n" + "Subheader Author\r\n" + "NON-ARC 0.1-1\r\n" + "20190211";
//based on https://stackoverflow.com/questions/899422/regular-expression-for-a-string-that-does-not-start-with-a-sequence
String regexNoNON = "^(?!NON\\-)ARC.*";
Pattern noNONPatter = Pattern.compile(regexNoNON);
System.out.println(noNONPatter.matcher(strTestSimplePos).find()); //true OK
System.out.println(noNONPatter.matcher(strTestSimpleNeg).find()); //false OK
System.out.println(noNONPatter.matcher(strTestRealPos).find()); //false but should be true -> does not work as intended
System.out.println(noNONPatter.matcher(strTestRealNeg).find()); //false OK
Would be glad if anyone can point out what went wrong...
Edit: This was marked as a duplicate of How to use java regex to match a line - however i didn't try to use a regex to match a line at all. Just needed a way to find a specific sequence (with negative look-ahead) for a multiline text input. One approach to solve the other question is also the solution to this one (compile pattern with java.util.regex.Pattern.MULTILINE) - but the questions are at best related.
Your input strings have multiple lines and you're using the caret, you need to add the multi-line flag:
Pattern.compile(regexNoNON, java.util.regex.Pattern.MULTILINE);
About MULTILINE:
Enables multiline mode.
In multiline mode the expressions ^ and $ match just after or just before, respectively, a line terminator or the end of the input sequence. By default these expressions only match at the beginning and the end of the entire input sequence.
Try this Regex:
HEADLINE(?:(?!HEADLINE)[\s\S])*(?<!NON-)ARC(?:(?!HEADLINE)[\s\S])*
Click for Demo
JAVA Code
Explanation:
HEADLINE - matches the word HEADLINE
(?:(?!HEADLINE)[\s\S])* - matches 0+ occurrences of any character that does not start with the word HEADLINE
(?<!NON-)ARC - matches the word ARC if it is not immediately preceded by NON-
(?:(?!HEADLINE)[\s\S])* - matches 0+ occurrences of any character that does not start with the word HEADLINE
This question already has answers here:
Whitespace Matching Regex - Java
(11 answers)
Regexp Java for password validation
(17 answers)
Closed 5 years ago.
I have see numerous suggestions for regex to find whitespace in a string none of which have worked so far. Yes the concept of looping through the string with a for next loop will work. I would really like to learn how to do this with regex and Pattern Matcher ! My question is what and where do I need to add to my regex string so it will return FALSE? code below I have added numerous incarnations of (\\s) to no avail. I do not want to remove the whitespace.
I tested the code suggested as a duplicate and it does not work see the link suggested in the comments
String tstr = "^(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[$#$!%*?&])[A-Za-z\\d$#$!%*?&]";
String astr = etPW.getText().toString().trim();
Pattern regex = Pattern.compile(tstr);
Matcher regexMatcher = regex.matcher(astr);
boolean foundMatch = regexMatcher.find();
if(foundMatch == false){
Toast.makeText( MainActivity.this, "Password must have one Numeric Value\n"
+ "\nOne Upper & Lower Case Letters\n"
+ "\nOne Special Character $ # ! % * ? &", Toast.LENGTH_LONG ).show();
//etPW.setText("");
//etCPW.setText("");
// Two lines of code above are optional
// Also by design these fields can be set to input type Password in the XML file
etPW.requestFocus();
return ;
}
You can use negative lookahead to check for spaces:
^(?!.* )
^ - Start matching at the beginning of the string.
(?! - Begin a negative lookahead group (the pattern inside the parentheses must not come next.
.* - Any non-newline character any number of times followed by a space.
) - Close the negative lookahead group.
Combined with the full regex pattern (also cleaned up a bit to remove redundancy):
^(?!.* )(?=.*[a-z])(?=.*[A-Z])(?=.*\\d)(?=.*[!#$%&*?])[A-Za-z\\d!#$%&*?]+
This question already has answers here:
Search for a word in a String
(9 answers)
Closed 7 years ago.
In my application I'm giving dictionary word suggestions and replacing the selected word with the suggested word using .replaceAll(). However that is replacing every sub string in the entire string
for example in this String,
String sentence = "od and not odds as a sample sam. but not odinary";
If I suggest the first word as odd .replaceAll() will replace every occurrence of od with odd hence affecting the fourth word to oddds and changing the sentence to
sentence.replaceAll("od", "odd");
//sentence String becomes
sentence ="odd and not oddds as a sample sam. but not oddinary"
Replacing the od to odd has affected all the other words which have the od characters in them.
Can any one help me with a better aproach?
Use regex. For you example "\bod\b" will just match od as a whole word. \b is a word boundary, meaning either the start or the end of a word (whether it ends with a dot or a whitespace or anything else).
The replaceAll method can already take in a regex, but if you need more power you can look at the Matcher class.
String REPLACE_WORD = "od"
sentence.replaceAll("\\b" + REPLACE_WORD + "\\b", "odd");
will give you the correct answer. The \ tells java that you want to write \ instead of \b (it first parses the string, and than parses that string as regex).
As mentioned, you can use a Matcher from Java.util.regex.* which has a lot of useful functionality.
String text = "I detect quite an od odour.";
String searchTerm = "\\bod\\b";
Pattern pattern = Pattern.compile(searchTerm);
Matcher matcher = pattern.matcher(text);
text = matcher.replaceAll("odd");
System.out.println(text);
The output would be:
I detect quite an odd odour.
Use the regular expression in the replaceAll() method:
\bod\b
This will filter out occurrences of the od inside any other word.
Of course when you use it in Java method, you need to escape the \
So
replaceAll("\\bod\\b", "odd");
should do it.
This question already has an answer here:
Escaping special characters in java regex (not quoting)
(1 answer)
Closed 8 years ago.
I'm trying to define a regex pattern that searches for a caret character, but since ^ is used for negation, I'm not sure how to define the pattern. I'm trying to make the program find a string that is a letter then a caret then a number (as you may have guessed, this is a mathematical term), such as "x^23". This is the line I tried:
String caseFour = "[a-zA-Z]" + "^" + "\\d+";
It's not working. Can anyone help me out?
You need to escape that character also since it is a character of special meaning.
String regex = "[a-zA-Z]\\^\\d+";
This question already has answers here:
Regex to match the longest repeating substring
(5 answers)
Closed 9 years ago.
I need to write a regex, that would identify a word that have a repeating character set at the end. According to the following code fragment, the repeating character set is An. I need to write a regex so this will be spotted and displayed.
According to the following code, \\w will match any word character (including digit, letter, or special character). But i only want to identify english characters.
String stringToMatch = "IranAnAn";
Pattern p = Pattern.compile("(\\w)\\1+");
Matcher m = p.matcher(stringToMatch);
if (m.find())
{
System.out.println("Word contains duplicate characters " + m.group(1));
}
UPDATE
Word contains duplicate characters a
Word contains duplicate characters a
Word contains duplicate characters An
You want to catch as many characters in your set as possible, so instead of (\\w) you should use (\\w+) and you want the sequence to be at the end, so you need to add $ (and I have removed the + after \\1 which is not useful to detect repetition: only one repetition is needed):
Pattern p = Pattern.compile("(\\w+)\\1$");
Your program then outputs An as expected.
Finally, if you only want to capture ascii characters, you can use [a-zA-Z] instead of \\w:
Pattern p = Pattern.compile("([a-zA-Z]+)\\1$");
And if you want the character set to be at least 2 characters:
Pattern p = Pattern.compile("([a-zA-Z]{2,})\\1$");
If by "only English characters" you mean A-Z and a-z, the follow regex will work:
".*([A-Za-z]{2,})\\1$"