I need to help in writing regex pattern to remove only part of the matcher from original string.
Original String: 2017-02-15T12:00:00.268+00:00
Expected String: 2017-02-15T12:00:00+00:00
Expected String removes everything in milliseconds.
My regex pattern looks like this: (:[0-5][0-9])\.[0-9]{1,3}
i need this regex to make sure i am removing only the milliseconds from some time field, not everything that comes after dot. But using above regex, I am also removing the minute part. Please suggest and help.
You have defined a capturing group with (...) in your pattern, and you want to have that part of string to be present after the replacement is performed. All you need is to use a backreference to the value stored in this capture. It can be done with $1:
String s = "2017-02-15T12:00:00.268+00:00";
String res = s.replaceFirst("(:[0-5][0-9])\\.[0-9]{1,3}", "$1");
System.out.println(res); // => 2017-02-15T12:00:00+00:00
See the Java demo and a regex demo.
The $1 in the replacement pattern tells the regex engine it should look up the captured group with ID 1 in the match object data. Since you only have one pair of unescaped parentheses (1 capturing group) the ID of the group is 1.
Change your pattern to (?::[0-5][0-9])(\.[0-9]{1,3}), run the find in the matcher and remove all it finds in the group(1).
The backslash will force the match with the '.' char, instead of any char, which is what the dot represents in a regex.
The (?: defines a non-capturing group, so it will not be considered in the group(...) on the matcher.
And adding a parenthesis around what you want will make it show up as group in the matcher, and in this case, the first group.
A good reference is the Pattern javadoc: http://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html
Use $1 and $2 variable for replace
string.replaceAll("(.*)\\.\\d{1,3}(.*)","$1$2");
Related
Forgive me. I am not familiarized much with Regex patterns.
I have created a regex pattern as below.
String regex = Pattern.quote(value) + ", [NnoneOoff0-9\\-\\+\\/]+|[NnoneOoff0-9\\-\\+\\/]+, "
+ Pattern.quote(value);
This regex pattern is failing with 2 different set of strings.
value = "207e/160";
Use Case 1 -
When channelStr = "207e/160, 149/80"
Then channelStr.matches(regex), returns "true".
Use Case 2 -
When channelStr = "207e/160, 149/80, 11"
Then channelStr.matches(regex), returns "false".
Not able to figure out why? As far I can understand it may be because of the multiple spaces involved when more than 2 strings are present with separated by comma.
Not sure what should be correct pattern I should write for more than 2 strings.
Any help will be appreciated.
If you print your pattern, it is:
\Q207e/160\E, [NnoneOoff0-9\-\+\/]+|[NnoneOoff0-9\-\+\/]+, \Q207e/160\E
It consists of an alternation | matching a mandatory comma as well on the left as on the right side.
Using matches(), should match the whole string and that is the case for 207e/160, 149/80 so that is a match.
Only for this string 207e/160, 149/80, 11 there are 2 comma's, so you do get a partial match for the first part of the string, but you don't match the whole string so matches() returns false.
See the matches in this regex demo.
To match all the values, you can use a repeating pattern:
^[NnoeOf0-9+/-]+(?:,\h*[NnoeOf0-90+/-]+)*$
^ Start of string
[NnoeOf0-9\\+/-]+
(?: Non capture group
,\h* Match a comma and optional horizontal whitespace chars
[NnoeOf0-90-9\\+/-]+ Match 1+ any of the listed in the character class
)* Close the non capture group and optionally repeat it (if there should be at least 1 comma, then the quantifier can be + instead of *)
$ End of string
Regex demo
Example using matches():
String channelStr1 = "207e/160, 149/80";
String channelStr2 = "207e/160, 149/80, 11";
String regex = "^[NnoeOf0-9+/-]+(?:,\\h*[NnoeOf0-90+/-]+)*$";
System.out.println(channelStr1.matches(regex));
System.out.println(channelStr2.matches(regex));
Output
true
true
Note that in the character class you can put - at the end not having to escape it, and the + and / also does not have to be escaped.
You can use regex101 to test your RegEx. it has a description of everything that's going on to help with debugging. They have a quick reference section bottom right that you can use to figure out what you can do with examples and stuff.
A few things, you can add literals with \, so \" for a literal double quote.
If you want the pattern to be one or more of something, you would use +. These are called quantifiers and can be applied to groups, tokens, etc. The token for a whitespace character is \s. So, one or more whitespace characters would be \s+.
It's difficult to tell exactly what you're trying to do, but hopefully pointing you to regex101 will help. If you want to provide examples of the current RegEx you have, what you want to match and then the strings you're using to test it I'll be happy to provide you with an example.
^(?:[NnoneOoff0-9\\-\\+\\/]+ *(?:, *(?!$)|$))+$
^ Start
(?: ... ) Non-capturing group that defines an item and its separator. After each item, except the last, the separator (,) must appear. Spaces (one, several, or none) can appear before and after the comma, which is specified with *. This group can appear one or more times to the end of the string, as specified by the + quantifier after the group's closing parenthesis.
Regex101 Test
I am struggling with the following issue: say there's a regex 1 and there's regex 2 which should match everything the regex 1 does not.
Let's have the regex 1:
/\$\d+/ (i.e. the dollar sign followed by any amount of digits.
Having a string like foo$12___bar___$34wilma buzz it detects $12 and $34.
How does the regex 2 should look in order to match the remained parts of the aforementioned string, i.e. foo, ___bar___ and wilma buzz? In other words it should pick up all the "remained" chunks of the source string.
You may use String#split to split on given regex and get remaining substrings in an array:
String[] arr = str.split( "\\$\\d+" );
//=> ["foo", "___bar___", "wilma buzz"]
RegEx Demo
It was tricky to get this working, but this regex will match everything besides \$\d+ for you. EDIT: no longer erroneously matches $44$444 or similar.
(?!\$\d+)(.+?)\$\d+|\$\d+|(?!\$\d+)(.+)
Breakdown
(?!\$\d+)(.+?)\$\d+
(?! ) negative lookahead: assert the following string does not match
\$\d+ your pattern - can be replaced with another pattern
(.+?) match at least one symbol, as few as possible
\$\d+ non-capturing match your pattern
OR
\$\d+ non-capturing group: matches one instance of your pattern
OR
(?!\$\d+)(.+)
(?!\$\d+) negative lookahead to not match your pattern
(.+) match at least one symbol, as few as possible
GENERIC FORM
(?!<pattern>)(.+?)<pattern>|<pattern>|(?!<pattern>)(.+)
By replacing <pattern>, you can match anything that doesn't match your pattern. Here's one that matches your pattern, and here's an example of arbitrary pattern (un)matching.
Good luck!
Try this one
[a-zA-Z_]+
Or even better
[^\$\d]+ -> With the ^symbol you can negotiate the search like ! in the java -> not equal
I have been trying to match a regex pattern within the following data:
String:
TestData to 1colon delimiter list has 1 rows.Set...value is: 1 Save Error: 267467374736437-TestInfo send Error
Words to match:
TestData
267467374736437-TestInfo
Regex pattern i m using:
(.+?\s)?.*(\s\d+-.*?\s)?
Scenario here is that 2nd match (267467374736437-TestInfo) can be absent in the string to be matched. So, i want it to be a match if it exists otherwise proceed with other matches. Due to this i added zero or one match quantifier ? to the group pattern above. But then it ignores the 2nd group all together.
If i use the below pattern:
`(.+?\s)?.*(\s\d+-.*?\s)`
It matches just fine but fails if string "267467374736437-TestInfo" from the matching string as it's not having the "?" quantifier.
Please help me understand where is it going wrong.
I would rather not use a complex regex, which will be ugly and a maintenance nightmare. Instead, one simple way would be to just split the string and grab the first term, and then use a smart regex to pinpoint the second term.
String input = "TestData to 1colon delimiter list has 1 rows.Set...value is: 1 Save Error: 267467374736437-TestInfo send Error";
String first = input.split(" ")[0];
String second = input.replaceAll(".*Save Error:\\s(.*)?\\s", "$1");
Explore the regex:
Regex101
The optional pattern at the end will almost never not be matched if a more generic pattern occurs. In your case, the greedy dot .* grabs the whole rest of the line up to the end, and since the last pattern is optional, the regex engine calls it a day and does not try to accommodate any text for it.
If you had a lazy dot .*?, the only position where it would work is right after the preceding subpattern, which is rarely the case.
Thus, you can only rely on a tempered greedy token:
^(\S+)(?:(?!\d+-\S).)*(\d+-\S+)?
See the regex demo.
Or an unrolled version:
^(\S+)\D*(?:\d(?!\d*-\S)\D*)*(\d+-\S+)?
I know it's a simple problem but i'm blocked on it : i want to retrieve all strings written in this form :
$F{ETIQX}
Where X is a number. i wrote this regular expression but i'm getting errors :
if (textField.getText().matches("$F{ETIQ\d}")){
System.out.println("matches!!");
}
Any help will be appreciated.
i want to retrieve all strings
Then you shouldn't be using .matches() in the first place. but a Matcher and .find(). .matches() is a misnomer. It will succeed only if the whole input matches the regex (in contradiction with the definiton of regex matching which can occur anywhere in the input).
Also, your regex should be:
"\\$F\\{ETIQ\\d\\}"
(you need to escape backslashes in a Java string)
$, { and } are regex metacharacters; the first is an anchor matching the end of input, the two latter are bounds for a repetition quantifier.
Your code should read:
private static final Pattern PATTERN = Pattern.compile("\\$F\\{ETIQ\\d\\}");
// ...
final Matcher m = PATTERN.matcher(textField.getText());
while (m.find())
// work with m.group()
\$F\{ETIQ\d\}
escape character which have meaning in regex.
$ means end of string
{ means start of a quantifier
} means end of a quantifier
for matching these you must escape them to match them literally.
here is a demo http://regex101.com/r/xT4mR6
In java \ has no meaning and will throuw an error , so we need to escape \ with \.
I have a String like this : String x = "return function ('ABC','DEF')";
I am using this:
Pattern pattern = Pattern.compile("'(.*?)'");
Matcher matcher = pattern.matcher(formula);
while (matcher.find()) {
System.out.println("------> " + matcher.group();
}
to retrieve strings between single quotes.
My question is: how can i adapt this regex so that it will check for strings between single quotes AND strings like " ,'DEF' " (meaning which start with ,' and end with ')?
You can use this pattern:
'[^']+'|"[^"]+"
Just to match with empty quoted string change '+' to '*'.
See test.
This pattern should do what you want:
"(?:,\s*)?'[^']*'"
The ? means the first group will match zero or one times.
I used (?:...) because this is a non-capturing group. It is better to use when you don't need to capture that portion of the match.
Also, I replaced .*? with [^']*, meaning the single-quoted string contains anything that is not a single quote. This is more efficient and less likely to lead to mistakes in your regex than .*?.
(Note: this regex allows there to be space between the comma and the start of the string. At first looking at your example, I thought that was true of your example. But now I see that it is not. Still, that might be useful depending on what your data looks like).
You could use the regex pattern:
Pattern.compile(",?'(.*?)'");
,? means 0 or 1 commas. The ? is greedy, so if there is a comma, it will be included in the match.
So: This will match:
A comma, followed by a string enclosed in single quotes
OR.. only a string enclosed in single quotes