I have a requirement where a string needs to be matched and then extract further value from a that string
I will receive a header in request whose value will be a DN name from ssl certificate. Here need to match a specific string 1.2.3.47 in the header and extract remaining text.
Sample String passed to method:
O=ABC Bank Plc/1.2.3.47=ABC12-PQR-121878, CN=7ltM2wQ3bqlDJdBEURGAMq, L=INDIA, C=INDIA, E=xyz#gmail.com
Here is my code:
private String extractDN(String dnHeader) {
if(!ValidatorUtil.isEmpty(dnHeader)){
String tokens[]=dnHeader.split(",");
if(tokens[0].contains("1.2.3.47")){
int index=tokens[0].lastIndexOf("1.2.3.47");
String id=tokens[0].substring(index+9);
System.out.println(id);
}
}
return id;
}
Can a regex pattern be used here to match and extract value? Is there any better way to achieve this? Please help.
If you want to use a pattern and if you know that the value always starts with a forward slash and if followed by one or more digits separated by a dot and then an equals sign, you could use a capturing group:
/[0-9](?:\\.[0-9]+)+=([^,]+)
/ Match /
[0-9]+ Match 1+ digit 0-9
(?: Non capturing group
\\.[0-9]+ match . and 1+ digits 0-9
)+ Close non capturing group and repeat 1+ times
= Match =
([^,]+) Capture group 1, match 1+ times any char except a ,
Regex demo | Java demo
For example
final String regex = "/[0-9]+(?:\\.[0-9]+)+=([^,]+)";
final String string = "O=ABC Bank Plc/1.2.3.47=ABC12-PQR-121878, CN=7ltM2wQ3bqlDJdBEURGAMq, L=INDIA, C=INDIA, E=xyz#gmail.com";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
if (matcher.find()) {
System.out.println(matcher.group(1));
}
Output
ABC12-PQR-121878
If you want a more precise match, you could also specify the start of the pattern:
\\bO=\\w+(?:\\h+\\w+)*/[0-9]+(?:\\.[0-9]+)+=([^,]+)
Regex demo
Related
I want to replace a part of email address using regex. How to do it ?
For example : An email address test.email+alex#gmail.com is there and I want to replace the part of that email address from + to before # with '' so that final string will be test.email#gmail.com.
I tried with this given below :
str.replaceAll("[^+[a-z]]","");
You can try with that:
\+[^#]*
Explanation:
\+ matches + where \ is the escape character
[^#]* matches anything until it reaches #, where * means zero or more
The code is given below:
final String string = "test.email+alex#gmail.com";
final Pattern pattern = Pattern.compile("\\+[^#]*");
final Matcher matcher = pattern.matcher(string);
final String result = matcher.replaceAll("");
Regex Test Case
If you want to match either a dot or a plus sign till an #, you could use a positive lookahead to assert an # on the right for both cases and list each option using an alternation.
(?:\.|\+[^#]*)(?=.*#)
Explanation
(?: Non capture group
\. Match a dot
| Or
\+[^#]* Match + and 0+ times any char except a dot
) Close group
(?=.*#) Positive lookahead, assert an # to the right
Regex demo | Java demo
In Java
str.replaceAll("(?:\\.|\\+[^#]*)(?=.*#)","")
I am new to regular expression and i want to find a string between two characters,
I tried below but it always returns false. May i know whats wrong with this ?
public static void main(String[] args) {
String input = "myFunction(hello ,world, test)";
String patternString = "\\(([^]]+)\\)";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
System.out.println(matcher.group());
}
}
Input:
myFunction(hello,world,test) where myFunction can be any characters. before starting ( there can be any characters.
Output:
hello
world
test
You could match make use of the \G anchor which asserts the position at the end of the previous match and and capture your values in a group:
(?:\bmyFunction\(|\G(?!^))([^,]+)(?:\h*,\h*)?(?=[^)]*\))
In Java:
String regex = "(?:\\bmyFunction\\(|\\G(?!^))([^,]+)(?:\\h*,\\h*)?(?=[^)]*\\))";
Explanation
(?: Non capturing group
\bmyFunction\( Word boundary to prevent the match being part of a larger word, match myFunction and an opening parentheses (
| Or
\G(?!^) Assert position at the end of previous match, not at the start of the string
) Close non capturing group
([^,]+) Capture in a group matching 1+ times not a comma
(?:\h*,\h*)? Optionally match a comma surrounded by 0+ horizontal whitespace chars
(?=[^)]*\)) Positive lookahead, assert what is on the right is a closing parenthesis )
Regex demo | Java demo
For example:
String patternString = "(?:\\bmyFunction\\(|\\G(?!^))([^,]+)(?:\\h*,\\h*)?(?=[^)]*\\))";
String input = "myFunction(hello ,world, test)";
Pattern pattern = Pattern.compile(patternString);
Matcher matcher = pattern.matcher(input);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Result
hello
world
test
I'd suggest you to achieve this in a two-step process:
Step 1: Capture all the content between ( and )
Use the regex: ^\S+\((.*)\)$
Demo
The first and the only capturing group will contain the required text.
Step 2: Split the captured string above on ,, thus yielding all the comma-separated parameters independently.
See this you may get idea
([\w]+),([\w]+),([\w]+)
DEMO: https://rubular.com/r/9HDIwBTacxTy2O
I have string inside brackets like following format:
[space string space]
I want to extract the string if the string is in UUID format.
example : [ d6a413f4-059c-11e8-ba89-0ed5f89f718b ]
With java regular expression how can I get d6a413f4-059c-11e8-ba89-0ed5f89f718b ?
For your given example, you could use a lookaround to match what is between the [ and the ]:
(?<=\[ ).*?(?= \])
Explanation
(?= \]) positive lookbehind to assert that what is before is [
.*? match any character zero or more times non greedy
(?= \]) positive lookahead to assert that what follows is ]
For example:
String regex = "(?<=\\[ ).*?(?= \\])";
String string = "[ d6a413f4-059c-11e8-ba89-0ed5f89f718b ]";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
Java example output
Using regex
\[ ([a-f0-9]{8}-[a-f0-9]{4}-[a-f0-9]{4}-[a-f0-9]{4}-[a-f0-9]{12}) ]
Regex101
Why you don't want to do this
If you know that your string will definitely have the right format then you can just use substring to get the UUID
class Main {
public static void main(String... args) {
String s = "[ d6a413f4-059c-11e8-ba89-0ed5f89f718b ]";
System.out.println(s.substring(2, s.length()-2));
}
}
Try it online!
This will be faster than using the regex option.
Regex to check if given String contains valid UUID:
"\\[ ([a-f0-9]{8}\\-(?:[a-f0-9]{4}\\-){3}[a-f0-9]{12}) \\]"
So, what is going on in this regex:
\\[ - character ‘[‘ and whitespace after it
[a-f0-9]{8} – characters from ‘a’ to ‘f’ and from ‘0’ to ‘9’ exactly eight times (123e5670 part)
\\- - ‘-‘ character
(?:[a-f0-9]{4}\\-){3} – non-capturing group that you want to be present exactly three times (this non-capturing group should contain exactly 4 characters that are in the range from ‘a’ to ‘f’ or from ‘0’ to ‘9’. After these 4 characters there must be present ‘-‘ character) (a234-b234-c234- part)
[a-f0-9]{12} - characters from ‘a’ to ‘f’ and from ‘0’ to ‘9’ exactly twelve times (d23456789012 part)
\\] – whitespace and ‘]’ character
After searching String for match with find() method, you only print capturing group #1 with group(1) method ( capturing group #1 is contained in parenthesis () )
Your UUID is in capture group 1. Here is a simple example how you can get UUID from source String:
String source = "[ 123e5670-a234-b234-c234-d23456789012 ]";
Pattern p = Pattern.compile("\\[ ([a-f0-9]{8}\\-(?:[a-f0-9]{4}\\-){3}[a-f0-9]{12}) \\]");
Matcher m = p.matcher(source);
if(m.find()) {
System.out.println( m.group(1));
}
Scenario: I want to check whether string contains only numbers and 2 predefined special characters, a dash and a comma.
My string contains numbers (0 to 9) and 2 special characters: a dash (-) defines a range and a comma (,) defines a sequence.
Tried attempt :
Tried following regex [0-9+-,]+, but not working as expected.
Possible inputs :
1-5
1,5
1-5,6
1,3,5-10
1-5,6-10
1,3,5-7,8,10
The regex should not accept these types of strings:
-----
1--4
,1,5
5,6,
5,4,-
5,6-
-5,6
Please can any one help me to create regex for above scenario?
You may use
^\d+(?:-\d+)?(?:,\d+(?:-\d+)?)*$
See the regex demo
Regex details:
^ - start of string
\d+ - 1 or more digits
(?:-\d+)? - an optional sequence of - and 1+ digits
(?:,\d+(?:-\d+)?)* - zero or more seuqences of:
, - a comma
\d+(?:-\d+)? - same pattern as described above
$ - end of string.
Change your regex [0-9+-,]+ to [0-9,-]+
final String patternStr = "[0-9,-]+";
final Pattern p = Pattern.compile(patternStr);
String data = "1,3,5-7,8,10";
final Matcher m = p.matcher(data);
if (m.matches()) {
System.out.println("SUCCESS");
}else{
System.out.println("ERROR");
}
I have a String which contains " Dear user BAL= 1,234/ ".
I want to extract 1,234 from the String using the regular expression. It can be 1,23, 1,2345, 5,213 or 500
final Pattern p=Pattern.compile("((BAL)=*(\\s{1}\\w+))");
final Matcherm m = p.matcher(text);
if(m.find())
return m.group(3);
else
return "";
This returns 3.
What regular expression should I make? I am new to regular expressions.
You search in your regex for word characters \w+ but you should search for digits with \d+.
Additionally there is the comma, so you need to match that as well.
I'd use
/.BAL=\s([\d,]+(?=/)./
as pattern and get only the number in the resulting group.
Explanation:
.* match anything before
BAL= match the string "BAL="
\s match a whitespace
( start matching group
[\d,]+ matches every digit or comma one ore more times
(?=/) match the former only if followed by a slash
) end matching group
.* matches anything thereaft
This is untestet, but it should work like this:
final Pattern p=Pattern.compile(".*BAL=\\s([\\d,]+(?=/)).*");
final Matcherm m = p.matcher(text);
if(m.find())
return m.group(1);
else
return "";
According to an online tester, the pattern above matches the text:
BAL= 1,234/
If it didn't have to be extracted by the regular expression you could simply do:
// split on any whitespace into a 4-element array
String[] foo = text.split("\\s+");
return foo[3];