How to get ricCode:.ABC from following string.
My matcher is
Matcher matcher = Pattern.compile("ricCode:([A-Za-z]+),$").matcher(str);
String str = "{AMX:{ricCode:.ABC,indexDetailEnable:true,price:20,648.15,netChange:<spanclass="md-down">-41.09</span>,percentChange:<spanclass="md-down">-0.20％</span>,tradeDate:17/04/05,tradeTime:16:40,chartDate:17/04/05,chartTime:16:40}";
What is missing in the regex?
Change your regex from this:
ricCode:([A-Za-z]+),$
to this:
ricCode:([A-Za-z.]+)(?=,)
Your original regex would only allow alphabetic characters after ricCode:, but your example has a period . character. Also you were matching the , character, but this would also include the comma in your match, you dont want this - so I added a positive lookahead for the comma so it looks for it there but does not match it. Finally you had the $ character at the end of your regex which matches the end of the string, you dont want to look for the end of the string immediately after the comma, so I removed it.
It helps to use regexr.com to test out your expression.
Related
I have a string which needs to be split based on a delimiter(:). This delimiter can be escaped by a character (say '?'). Basically the delimiter can be preceded by any number of escape character. Consider below example string:
a:b?:c??:d???????:e
Here, after the split, it should give the below list of string:
a
b?:c??
d???????:e
Basically, if the delimiter (:) is preceded by even number of escape characters, it should split. If it is preceded by odd number of escape characters, it should not split. Is there a solution to this with regex?
Any help would be greatly appreciated.
Similar question has been asked earlier here, But the answers are not working for this use case.
Update:
The solution with the regex: (?:\?.|[^:?])* correctly split the string. However, this also gives few empty strings. If + is given instead of *, even the real empty matches also ignored. (Eg:- a::b gives only a,b)
Scenario 1: No empty matches
You may use
(?:\?.|[^:?])+
Or, following the pattern in the linked answer
(?:\?.|[^:?]++)+
See this regex demo
Details
(?: - start of a non-capturing group
\?. - a ? (the delimiter) followed with any char
| - or
[^:?] - any char but the : (your delimiter char) and ? (the escape char)
)+ - 1 or more repetitions.
In Java:
String regex = "(?:\\?.|[^:?]++)+";
In case the input contains line breaks, prepend the pattern with (?s) (like (?s)(?:\\?.|[^:?])+) or compile the pattern with Pattern.DOTALL flag.
Scenario 2: Empty matches included
You may add (?<=:)(?=:) alternative to the above pattern to match empty strings between : chars, see this regex demo:
String s = "::a:b?:c??::d???????:e::";
Pattern pattern = Pattern.compile("(?>\\?.|[^:?])+|(?<=:)(?=:)");
Matcher matcher = pattern.matcher(s);
while (matcher.find()){
System.out.println("'" + matcher.group() + "'");
}
Output of the Java demo:
''
'a'
'b?:c??'
''
'd???????:e'
''
Note that if you want to also match empty strings at the start/end of the string, use (?<![^:])(?![^:]) rather than (?<=:)(?=:).
I am searching for regex pattern that matches the following String. I am using this regex pattern as,
^;[A-za-z0-9,:]+
Above regex doesn't matches the following.
I am looking for all given string to be matched with regex pattern.
:a123,234,444:322 //String started with semicolon and values are separated with comma and colon
;123,A234:123;123,345,456:999,456 // Above case with repeated condition
;;123,345,C555:123 //String started with double semicolon
Can anyone provide regex pattern that matches above string.
This one
[;:]+[A-za-z0-9,;:]+
will work for all three you want, see online on regex101.
[;:]+: Started with one or more ; or : .
[A-za-z0-9,;:]+: You miss' a : here.
You can match the above with this regex
^;+[A-za-z0-9,;:]+
Modifications:
;+ will match 1 or more semicolons
colon : has been added in characters you want to match
I'm new with regex and just can't find what the regex is to prohibit a backslash.
Thanks for any advice.
EDIT:
I'm using the regex for JTextField, to avoid that the user writes an unvalid input. This regex currently doesnt allow the user to write a space character.
I'm doing this with
String regex = "\\S{1}";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
So how I change my regex to prohibit backslash as well?
Based on your example snippet, the following expression should work similarly, but also disallowing the backslash:
String regex = "[^\\s\\\\]{1}";
It is a bit strange that you are looking for a one-char non-space and non-backslash pattern, but I guess you are iterating through and checking for consecutive matches.
I would use the following regex though:
String regex = "[^\\s\\\\]+";
and check whether it matched the whole String (matcher.matches()).
The regex pattern with java is \\\\
String somestring;
somestring = somestring.replaceAll("\\\\", "");
Would remove them. Semantically it equates down to \\ at the regex level, which becomes a literal \ match.
You can also use a Pattern match if you want to just compare, or just use String#contains
String somestring;
if (somestring.contains("\\")) {...}
To test if a string has a backslash:
if (input.matches(".*\\\\.*"))
To remove backslashes from strings:
input = input.replace("\\", "");
Note the use of the non-regex based replace() method.
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
what is the pattern to validate the following regular expression in java.
string with only letters, spaces and apostrophes (')
I know that for letters is this ("^[a-zA-Z]+$")
For spaces is this ("\\s)
I don't know what apostrophe is.
But most of all i just want a single expression. Not 3 individual ones.
You can create your own class with the characters that you need to match, like this:
Pattern.compile("[a-zA-Z\\s']+");
there is no single class that could match something so specific, but you can build your own from the existing ones
(\p{Alpha}|\s|')*
matches any number of characters, spaces or apostrophes, in any order.
Take a look at http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html
Pattern p = Pattern.compile("^[a-zA-Z ]*$");
Matcher m = p.matcher("Tester String");
System.out.println(m.matches());// true
Matcher m2 = p.matcher("Tester String 123");
System.out.println(m2.matches());// false
This will accept only alphabets.