I need to be able to replace all occurrences of the word "and" ONLY when it occurs between single quotes. For example replacing "and" with "XXX" in the string:
This and that 'with you and me and others' and not 'her and him'
Results in:
This and that 'with you XXX me XXX others' and not 'her XXX him'
I have been able to come up with regular expressions which nearly gets every case, but I'm failing with the "and" between the two sets of quoted text.
My code:
String str = "This and that 'with you and me and others' and not 'her and him'";
String patternStr = ".*?\\'.*?(?i:and).*?\\'.*";
Pattern pattern= Pattern.compile(patternStr);
Matcher matcher = pattern.matcher(str);
System.out.println(matcher.matches());
while(matcher.matches()) {
System.out.println("in matcher");
str = str.replaceAll("(?:\\')(.*?)(?i:and)(.*?)(?:\\')", "'$1XXX$2'");
matcher = pattern.matcher(str);
}
System.out.println(str);
Try this code:
str = "This and that 'with you and me and others' and not 'her and him'";
Matcher matcher = Pattern.compile("('[^']*?')").matcher(str);
StringBuffer sb = new StringBuffer();
while (matcher.find()) {
matcher.appendReplacement(sb, matcher.group(1).replaceAll("and", "XXX"));
}
matcher.appendTail(sb);
System.out.println("Output: " + sb);
OUTPUT
Output: This and that 'with you XXX me XXX others' and not 'her XXX him'
String str = "This and that 'with you and me and others' and not 'her and him'";
Pattern p = Pattern.compile("(\\s+)and(\\s+)(?=[^']*'(?:[^']*+'[^']*+')*+[^']*+$)");
System.out.println(p.matcher(str).replaceAll("$1XXX$2"));
The idea is, each time you find the complete word and, you you scan from the current match position to the end of the string, looking for an odd number of single-quotes. If the lookahead succeeds, the matched word must be between a pair of quotes.
Of course, this assumes quotes always come in matched pairs, and that quotes can't be escaped. Quotes escaped with backslashes can be dealt with, but it makes the regex much longer.
I'm also assuming the target word never appears at the beginning or end of a quoted sequence, which seems reasonable for the word and. If you want to allow for target words that are not surrounded by whitespace, you could use something like "\\band\\b" instead, but be aware of Java's problems in the area of word characters vs word boundaries.
Related
I was trying to replace concatenation symbol '+' with '||' in given multi-line script, however it seems that java regex just replaces 1 occurrence, instead of all.
String ss="A+B+C+D";
Matcher mm=Pattern.compile("(?imc)(.+)\\s*\\+\\s*(.+)").matcher(ss);
while(mm.find())
{
System.out.println(mm.group(1));
System.out.println(mm.group(2));
ss=mm.replaceAll("$1 \\|\\| $2");
}
System.out.println(ss); // Output: A+B+C||D, Expected: A||B||C||D
The reason you only replace one element, is because you match the entire line. The regular expression you use "(?imc)(.+)\\s*\\+\\s*(.+)", matches anything (.+) until the end, then reverts, so it can match the rest \\s*\\+.... So basically your group 1 is .+ almost everything, but the last + and beyond. Therefore replaceAll can only match once, and will terminate after that one replacement.
What you need is a replacement that finds + optionally wrapped in spaces:
Pattern.compile("(?imc)\\s*\\+\\s*");
This should match all you want to match, and does not match the entire line, but only your replacement character.
You could just use:
ss = ss.replaceAll("\\+", "||")
as #ernest_k has pointed out. If you really want to continue using a matcher with iteration, then use Matcher#appendReplacement with a StringBuffer:
String ss = "A+B+C+D";
Matcher mm = Pattern.compile("\\+").matcher(ss);
StringBuffer sb = new StringBuffer();
while (mm.find()) {
mm.appendReplacement(sb, "||");
}
mm.appendTail(sb);
System.out.println(sb);
I thing maybe we would just need a simple string replace:
Demo
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\\+";
final String string = "A+B+C+D";
final String subst = "||";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
final String result = matcher.replaceAll(subst);
System.out.println(result);
This link on the right panel explains your original expression. The first capturing group does match between one and unlimited times, as many times as possible, thus it would not work here. If we would have changed them to (.+?), it would have partially worked, yet still unnecessary.
I have the following code to extract the string within double quotes using Regex.
String str ="\"Java\",\"programming\"";
final Pattern pattern = Pattern.compile("\"([^\"]*)\"");
final Matcher matcher = pattern.matcher(str);
while(matcher.find()){
System.out.println(matcher.group(1));
}
The output I get now is java programming.But from the String str I want the content in the second double quotes which is programming. Can any one tell me how to do that using Regex.
If you take your example, and change it slightly to:
String str ="\"Java\",\"programming\"";
final Pattern pattern = Pattern.compile("\"([^\"]*)\"");
final Matcher matcher = pattern.matcher(str);
int i = 0
while(matcher.find()){
System.out.println("match " + ++i + ": " + matcher.group(1) + "\n");
}
You should find that it prints:
match 1: Java
match 2: programming
This shows that you are able to loop over all of the matches. If you only want the last match, then you have a number of options:
Store the match in the loop, and when the loop is finished, you have the last match.
Change the regex to ignore everything until your pattern, with something like: Pattern.compile(".*\"([^\"]*)\"")
If you really want explicitly the second match, then the simplest solution is something like Pattern.compile("\"([^\"]*)\"[^\"]*\"([^\"]*)\""). This gives two matching groups.
If you want the last token inside double quotes, add an end-of-line archor ($):
final Pattern pattern = Pattern.compile("\"([^\"]*)\"$");
In this case, you can replace while with if if your input is a single line.
Great answer from Paul. Well,You can also try this pattern
final Pattern pattern = Pattern.compile(",\"(\\w+)\"");
Java program
String str ="\"Java\",\"programming\"";
final Pattern pattern = Pattern.compile(",\"(\\w+)\"");
final Matcher matcher = pattern.matcher(str);
while(matcher.find()){
System.out.println(matcher.group(1));
}
Explanation
,\": matches a comma, followed by a quotation mark "
(\\w+): matches one or more words
\": matches the last quotation mark "
Then the group(\\w+) is captured (group 1 precisely)
Output
programming
I ping a host. In result a standard output. Below a REGEXP but it do not work correct. Where I did a mistake?
String REGEXP ="time=(\\\\d+)ms";
Pattern pattern = Pattern.compile(REGEXP);
Matcher matcher = pattern.matcher(result);
if (matcher.find()) {
result = matcher.group(1);
}
You only need \\d+ in your regex because
Matcher looks for the pattern (using which it is created) and then tries to find every occurance of the pattern in the string being matched.
Use while(matcher.group(1) in case of multiple occurances.
each () represents a captured group.
You have too many backslashes. Assuming you want to get the number from a string like "time=32ms", then you need:
String REGEXP ="time=(\\d+)ms";
Pattern pattern = Pattern.compile(REGEXP);
Matcher matcher = pattern.matcher(result);
if (matcher.find()) {
result = matcher.group(1);
}
Explanation: The search pattern you are looking for is "\d", meaning a decimal number, the "+" means 1 or more occurrences.
To get the "\" to the matcher, it needs to be escaped, and the escape character is also "\".
The brackets define the matching group that you want to pick out.
With "\\\\d+", the matcher sees this as "\\d+", which would match a backslash followed by one or more "d"s. The first backslash protects the second backslash, and the third protects the fourth.
I have some punctuation [] punctuation = {'.', ',' , '!', '?'};. And I want create a regex that can match the word that was combined from those punctuations.
For example some string I want to find: "....???", "!!!!!......", "??.....!", so on.
Thanks for any advice.
Use String.matches() with the posix regex for "punctuation":
str.matches("\\p{Punct}+");
FYI according to the Pattern javadoc, \p{Punct} is one of
!"#$%&'()*+,-./:;<=>?#[\]^_`{|}~
Also, The ^ and $ aren't needed in the expression either, because matches() must matche the whole input to return true, so start and end are implied.
Try this, it should match and group all the symbols written between []:
([.,!?]+)
Tested it with
??..,..!fsdgsdfgsdfgsdfg
And output was
??..,..!
Also tested with this:
String s = "??.....!fsdgsdfgsdfgsdfg?.,!0000a";
Pattern p = Pattern.compile("([.,!?]+)");
Matcher m = p.matcher(s);
while(m.find()) {
System.out.println(m.group(1));
}
And output was
??.....!
?.,!
You can try with a Unicode category for punctuation and a while loop to match your input, as such:
String test = "!...abcd??...!!efgh....!!??abc!";
Pattern pattern = Pattern.compile("\\p{Punct}{2,}");
Matcher matcher = pattern.matcher(test);
while (matcher.find()) {
System.out.println(matcher.group());
}
Output:
!...
??...!!
....!!??
Note: this has the advantage of matching any punctuation character sequence larger than 1 character (hence, the last "!" is not matched by design). To decide the minimum length of the punctuation sequence, just play with the {2,} part of the Pattern.
I'm a Java user but I'm new to regular expressions.
I just want to have a tiny expression that, given a word (we assume that the string is only one word), answers with a boolean, telling if the word is valid or not.
An example... I want to catch all words that is plausible to be in a dictionary... So, i just want words with chars from a-z A-Z, an hyphen (for example: man-in-the-middle) and an apostrophe (like I'll or Tiffany's).
Valid words:
"food"
"RocKet"
"man-in-the-middle"
"kahsdkjhsakdhakjsd"
"JESUS", etc.
Non-valid words:
"gipsy76"
"www.google.com"
"me#gmail.com"
"745474"
"+-x/", etc.
I use this code, but it won't gave the correct answer:
Pattern p = Pattern.compile("[A-Za-z&-&']");
Matcher m = p.matcher(s);
System.out.println(m.matches());
What's wrong with my regex?
Add a + after the expression to say "one or more of those characters":
Escape the hyphen with \ (or put it last).
Remove those & characters:
Here's the code:
Pattern p = Pattern.compile("[A-Za-z'-]+");
Matcher m = p.matcher(s);
System.out.println(m.matches());
Complete test:
String[] ok = {"food","RocKet","man-in-the-middle","kahsdkjhsakdhakjsd","JESUS"};
String[] notOk = {"gipsy76", "www.google.com", "me#gmail.com", "745474","+-x/" };
Pattern p = Pattern.compile("[A-Za-z'-]+");
for (String shouldMatch : ok)
if (!p.matcher(shouldMatch).matches())
System.out.println("Error on: " + shouldMatch);
for (String shouldNotMatch : notOk)
if (p.matcher(shouldNotMatch).matches())
System.out.println("Error on: " + shouldNotMatch);
(Produces no output.)
This should work:
"[A-Za-z'-]+"
But "-word" and "word-" are not valid. So you can uses this pattern:
WORD_EXP = "^[A-Za-z]+(-[A-Za-z]+)*$"
Regex - /^([a-zA-Z]*('|-)?[a-zA-Z]+)*/
You can use above regex if you don't want successive "'" or "-".
It will give you accurate matching your text.
It accepts
man-in-the-middle
asd'asdasd'asd
It rejects following string
man--in--midle
asdasd''asd
Hi Aloob please check with this, Bit lengthy, might be having shorter version of this, Still...
[A-z]*||[[A-z]*[-]*]*||[[A-z]*[-]*[']*]*