I have a CSS style that I need to extract the color from using a Java regex.
eg
color:#000;
I need to extract the thing after : to ;. Can anyone give an example?
I'm not sure how to apply it to Java, but one regex to do this would be:
^color:\s*(#[0-9a-f]+);?$
To just extract from : up to ; do something like:
Pattern pattern = Pattern.compile("[^:]*:(.*);");
Matcher matcher = pattern.matcher(text);
if (matcher.matches()) {
String value = matcher.group(1);
System.out.println("'" + value+ "'"); // do something with value
}
[^:]* - any number of chars that are not ':'
: - one ':'
(...) - a capturing group
.*- any number of any character
;- the terminating ';'
use color:(.*); for only accepting values for 'color'.
/(?<=:).+(?=;)/
That will do it for you
Not sure how you implement regex in Java though.
www.regexr.com to help you text out your regex in real time.
The expression
":(#.+);"
should do it
Related
I am trying to capture following word, number:
stxt:usa,city:14
I can capture usa and 14 using:
stxt:(.*?),city:(\d.*)$
However, when text is;
stxt:usa
The regex did not work. I tried to apply or condition using | but it did not work.
stxt:(.*?),|city:(\d.*)$
You may use
(stxt|city):([^,]+)
See the regex demo (note the \n added only for the sake of the demo, you do not need it in real life).
Pattern details:
(stxt|city) - either a stxt or city substrings (you may add \b before the ( to only match a whole word) (Group 1)
: - a colon
([^,]+) - 1 or more characters other than a comma (Group 2).
Java demo:
String s = "stxt:usa,city:14";
Pattern pattern = Pattern.compile("(stxt|city):([^,]+)");
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}
Looking at your string, you could also find the word/digits after the colon.
:(\w+)
I want to create a regular expresion where I want match in case my number are separated by a coma.
For example:
1 OK
1,2,3 OK
1\n2,3 OK
1,\n Not OK
1,,2 Not OK
1,\n2 Not Ok
So far I create this expresion
\d+(([,.|\n])+\d+)*
If I change the last * to be at least 1 with +
\d+(([,.|\n])+\d+)+
Then all previous scenarios works but not this one
1 Not OK//And should be ok
I´m using matcher.find()
Matcher matcher = Pattern.compile(pattern).matcher(number);
if (matcher.find()) {
System.out.println("total number:" + matcher.group(0));;
}
Any idea what I´m doing wrong in my regex?
You can use this regex:
^\d+(?:(?:,|\n)\d+)*$
Java regex:
Pattern p = Pattern.compile("^\\d+(?:(?:,|\\n)\\d+)*$");
RegEx Demo
PS: To match literal \n you will need:
^\d+(?:(?:,|\\n)\d+)*$
Given the string
Content ID [9283745997] Content ID [9283005997] There can be text in between Content ID [9283745953] Content ID [9283741197] Content ID [928374500] There can be valid text here which should not be removed.
I want to remove the text starting Content ID followed by [9283745997] any numbers can be present between square brackets. Eventually I want the result string to be
There can be text in between There can be valid text here which should not be removed.
Could anyone please provide a valid regex to capture this recurring text but the numerals within square brackets are unique?
I appreciate your help!
My soulution to this was :
Pattern p = Pattern.compile("(Content ID \\[\\d*\\] )");
Matcher m = p.matcher(str);
StringBuffer sb = new StringBuffer();
while(m.find()) {
m.appendReplacement(sb, "");
}
m.appendTail(sb);
System.out.println(sb);
So basically you are trying to remove each of Content ID [one or more digits].
To do this you can use replaceAll("regex","replacement") method of String class. As replacement you can use empty String "".
Only problem that stays is what regex should you use.
to match Content ID just write it normally as "Content ID "
to match [ or ] you will have to add \ before each of them because they are regex metacharacters and you need to escape them (in Java you will need to write \ as "\\")
to represent one digit (character from range 0-9) regex uses \d (again in Java you will need to write \ as "\\" which will result in "\\d")
to say "one or more of previously described element" just add + after definition of such element. For example if you want to match one or more letters a you can write it as a+.
Now you should be able to create correct regex. If you will have some questions feel free to ask them in comments.
Try this one:
(Content ID \[[0-9]+\])
You can test it here: http://regexpal.com/
I would use the regex
Content ID \[\d+\] ?
Implement it like this:
str.replaceAll("Content ID \\[\\d+\\] ?", "");
You can find an explanation and demonstration here: http://regex101.com/r/qD5rJ6
The following code blocks on my system. Why?
System.out.println( Pattern.compile(
"^((?:[^'\"][^'\"]*|\"[^\"]*\"|'[^']*')*)/\\*.*?\\*/(.*)$",
Pattern.MULTILINE | Pattern.DOTALL ).matcher(
"\n\n\n\n\n\nUPDATE \"$SCHEMA\" SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';"
).matches() );
The pattern (designed to detect comments of the form /*...*/ but not within ' or ") should be fast, as it is deterministic...
Why does it take soooo long?
You're running into catastrophic backtracking.
Looking at your regex, it's easy to see how .*? and (.*) can match the same content since both also can match the intervening \*/ part (dot matches all, remember). Plus (and even more problematic), they can also match the same stuff that ((?:[^'"][^'"]*|"[^"]*"|'[^']*')*) matches.
The regex engine gets bogged down in trying all the permutations, especially if the string you're testing against is long.
I've just checked your regex against your string in RegexBuddy. It aborts the match attempt after 1.000.000 steps of the regex engine. Java will keep churning on until it gets through all permutations or until a Stack Overflow occurs...
You can greatly improve the performance of your regex by prohibiting backtracking into stuff that has already been matched. You can use atomic groups for this, changing your regex into
^((?>[^'"]+|"[^"]*"|'[^']*')*)(?>/\*.*?\*/)(.*)$
or, as a Java string:
"^((?>[^'\"]+|\"[^\"]*\"|'[^']*')*)(?>/\\*.*?\\*/)(.*)$"
This reduces the number of steps the regex engine has to go through from > 1 million to 58.
Be advised though that this will only find the first occurrence of a comment, so you'll have to apply the regex repeatedly until it fails.
Edit: I just added two slashes that were important for the expressions to work. Yet I had to change more than 6 characters.... :(
I recommend that you read Regular Expression Matching Can Be Simple And Fast (but is slow in Java, Perl, PHP, Python, Ruby, ...).
I think it's because of this bit:
(?:[^'\"][^'\"]*|\"[^\"]*\"|'[^']*')*
Removing the second and third alternatives gives you:
(?:[^'\"][^'\"]*)*
or:
(?:[^'\"]+)*
Repeated repeats can take a long time.
For comment /* and */ detection I would suggest having a code like this:
String str = "\n\n\n\n\n\nUPDATE \"$SCHEMA\" /*a comment\n\n*/ SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';";
Pattern pt = Pattern.compile("\"[^\"]*\"|'[^']*'|(/\\*.*?\\*/)",
Pattern.MULTILINE | Pattern.DOTALL);
Matcher matcher = pt.matcher(str);
boolean found = false;
while (matcher.find()) {
if (matcher.group(1) != null) {
found = true;
break;
}
}
if (found)
System.out.println("Found Comment: [" + matcher.group(1) + ']');
else
System.out.println("Didn't find Comment");
For above string it prints:
Found Comment: [/*a comment
*/]
But if I change input string to:
String str = "\n\n\n\n\n\nUPDATE \"$SCHEMA\" '/*a comment\n\n*/' SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';";
OR
String str = "\n\n\n\n\n\nUPDATE \"$SCHEMA\" \"/*a comment\n\n*/\" SET \"VERSION\" = 12 WHERE NAME = 'SOMENAMEVALUE';";
Output is:
Didn't find Comment
I want to split the string
String fields = "name[Employee Name], employeeno[Employee No], dob[Date of Birth], joindate[Date of Joining]";
to
name
employeeno
dob
joindate
I wrote the following java code for this but it is printing only name other matches are not printing.
String fields = "name[Employee Name], employeeno[Employee No], dob[Date of Birth], joindate[Date of Joining]";
Pattern pattern = Pattern.compile("\\[.+\\]+?,?\\s*" );
String[] split = pattern.split(fields);
for (String string : split) {
System.out.println(string);
}
What am I doing wrong here?
Thank you
This part:
\\[.+\\]
matches the first [, the .+ then gobbles up the entire string (if no line breaks are in the string) and then the \\] will match the last ].
You need to make the .+ reluctant by placing a ? after it:
Pattern pattern = Pattern.compile("\\[.+?\\]+?,?\\s*");
And shouldn't \\]+? just be \\] ?
The error is that you are matching greedily. You can change it to a non-greedy match:
Pattern.compile("\\[.+?\\],?\\s*")
^
There's an online regular expression tester at http://gskinner.com/RegExr/?2sa45 that will help you a lot when you try to understand regular expressions and how they are applied to a given input.
WOuld it be better to use Negated Character Classes to match the square brackets? \[(\w+\s)+\w+[^\]]\]
You could also see a good example how does using a negated character class work internally (without backtracking)?