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Regex to capture the staring with specific word or character and ending with either one of the word

Want to capture the string after the last slash and before either a (; sid=) word or a (?) character.
sample data:
sessionId=30a793b1-ed7e-464a-a630; Url=https://www.example.com/mybook/order/newbooking/itemSummary; sid=KJ4dgQGdhg7dDn1h0TLsqhsdfhsfhjhsdjfhjshdjfhjsfddscg139bjXZQdkbHpzf9l6wy1GdK5XZp; targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=122;
sessionId=sfdsdfsd-ba57-4e21-a39f-34; Url=https://www.example.com/mybook/order/newbooking/itemList?id=76734&para=jhjdfhj&type=new&ordertype=kjkf&memberid=273647632&iSearch=true; sid=Q4hWgR1GpQb8xWTLpQB2yyyzmYRgXgFlJLGTc0QJyZbW targetUrl=https://www.example.com/ mybook/order/newbooking/page1?id=123;
sessionId=0e1acab1-45b8-sdf3454fds-afc1-sdf435sdfds; Url=https://www.example.com/mybook/order/newbooking/; sid=hkm2gRSL2t5ScKSJKSJn3vg2sfdsfdsfdsfdsfdfdsfdsfdsfvJZkDD3ng0kYTjhNQw8mFZMn; targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=343;
Expecting the below output:
1. itemSummary
2. itemList
3. ''(empty string)
Have build the below regex to capture it but its 100% accurate. It is capturing some additional part.
Regex
Url=.*\/(.*)(; sid|\?)
Could you please help me to improve the regex to get desired output?
Thanks in advance!

You may use this regex in Java with a greedy match after Url=:
\bUrl=\S+/([^?;/]+)(?=; sid|\?)
RegEx Demo
RegEx Demo:
\b: Word boundary
Url=: Match text Url=
\S+/: Match 1+ non-whitespace characters followed by a /
([^?;/]+): Match 1+ of a character that not ? and ; and /
(?=; sid|\?): Lookahead to assert that we have ; sid or ? ahead

Alternative solution:
Used regex:
"^Url=.*/(\\w+|)$"
Regex in test bench and context:
public static void main(String[] args) {
String input1 = "sessionId=30a793b1-ed7e-464a-a630; "
+ "Url=https://www.example.com/mybook/order/newbooking/itemSummary; "
+ "sid=KJ4dgQGdhg7dDn1h0TLsqhsdfhsfhjhsdjfhjshdjfhjsfddscg139bjXZQdkbHpzf9l6wy1GdK5XZp; "
+ "targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=122;";
String input2 = "sessionId=sfdsdfsd-ba57-4e21-a39f-34; "
+ "Url=https://www.example.com/mybook/order/newbooking/itemList?id=76734&para=jhjdfhj&type=new&ordertype=kjkf&memberid=273647632&iSearch=true; "
+ "sid=Q4hWgR1GpQb8xWTLpQB2yyyzmYRgXgFlJLGTc0QJyZbW "
+ "targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=123;";
String input3 = "sessionId=0e1acab1-45b8-sdf3454fds-afc1-sdf435sdfds; "
+ "Url=https://www.example.com/mybook/order/newbooking/; "
+ "sid=hkm2gRSL2t5ScKSJKSJn3vg2sfdsfdsfdsfdsfdfdsfdsfdsfvJZkDD3ng0kYTjhNQw8mFZMn; "
+ "targetUrl=https://www.example.com/mybook/order/newbooking/page1?id=343;";
List<String> inputList = Arrays.asList(input1, input2, input3);
// Pre-compiled Patterns should not be in loops - that is why they are placed outside the loops
Pattern replaceWithNewLinePattern = Pattern.compile(";?\\s|\\?");
Pattern extractWordFromUrlPattern = Pattern.compile("^Url=.*/(\\w+|)$", Pattern.MULTILINE);
int count = 0;
for(String input : inputList) {
String inputWithNewLines = replaceWithNewLinePattern.matcher(input).replaceAll("\n");
// System.out.println(inputWithNewLines); // Check the change...
Matcher matcher = extractWordFromUrlPattern.matcher(inputWithNewLines);
while (matcher.find()) {
System.out.printf( "%d. '%s'%n", ++count, matcher.group(1));
}
}
}
Output:
1. 'itemSummary'
2. 'itemList'
3. ''

Get text in the URL with dynamic date - Regex Java

I need to get the text between the URL which has a date in Java
Input 1:
/test1/raw/2019-06-11/testcustomer/usr/pqr/DATA/mn/export/
Output: testcustomer
Only /raw/ remains, date will change and testcustomer will change
Input 2:
/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/
Output: newcustomer
String url = "/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/";
String customer = getCustomer(url);
public String getCustomer (String _url){
String source = "default";
String regex = basePath + "/raw/\\d{4}-\\d{2}-\\d{2}/usr*";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(_url);
if (m.find()) {
source = m.group(1);
} else {
logger.error("Cant get customer with regex " + regex);
}
return source;
}
It's returning 'default' :(

Your regex /raw/\\d{4}-\\d{2}-\\d{2}/usr* is missing the part for the value you want, you need a regex that find the date, and keep what's next :
/\w*/raw/[0-9-]+/(\w+)/.* or (?<=raw\/\d{4}-\d{2}-\d{2}\/)(\w+) will be good
Pattern p = Pattern.compile("/\\w*/raw/[0-9-]+/(\\w+)/.*");
Matcher m = p.matcher(str);
if (m.find()) {
String value = m.group(1);
System.out.println(value);
}
Or if it's always the 4th part, use split()
String value = str.split("/")[4];
System.out.println(value);
And here a >> code demo

Here, we can likely use raw followed by the date as a left boundary, then we would collect our desired output in a capturing group, we would add an slash and consume the rest of our string, with an expression similar to:
.+raw\/[0-9]{4}-[0-9]{2}-[0-9]{2}\/(.+?)\/.+
Demo
Test
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = ".+raw\\/[0-9]{4}-[0-9]{2}-[0-9]{2}\\/(.+?)\\/.+";
final String string = "/test1/raw/2019-06-11/testcustomer/usr/pqr/DATA/mn/export/\n"
+ "/test3/raw/2018-09-01/newcustomer/usr/pqr/DATA/mn/export/";
final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println("Full match: " + matcher.group(0));
for (int i = 1; i <= matcher.groupCount(); i++) {
System.out.println("Group " + i + ": " + matcher.group(i));
}
}
RegEx
If this expression wasn't desired or you wish to modify it, please visit regex101.com.
RegEx Circuit
jex.im visualizes regular expressions:

Using Regular Expression in Java to extract information from a String

I have one input String like this:
"I am Duc/N Ta/N Van/N"
String "/N" present it is the Name of one person.
The expected output is:
Name: Duc Ta Van
How can I do it by using regular expression?

You can use Pattern and Matcher like this :
String input = "I am Duc/N Ta/N Van/N";
Pattern pattern = Pattern.compile("([^\\s]+)/N");
Matcher matcher = pattern.matcher(input);
String result = "";
while (matcher.find()) {
result+= matcher.group(1) + " ";
}
System.out.println("Name: " + result.trim());
Output
Name: Duc Ta Van
Another Solution using Java 9+
From Java9+ you can use Matcher::results like this :
String input = "I am Duc/N Ta/N Van/N";
String regex = "([^\\s]+)/N";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(input);
String result = matcher.results().map(s -> s.group(1)).collect(Collectors.joining(" "));
System.out.println("Name: " + result); // Name: Duc Ta Van

Here is the regex to use to capture every "name" preceded by a /N
(\w+)\/N
Validate with Regex101
Now, you just need to loop on every match in that String and concatenate the to get the result :
String pattern = "(\\w+)\\/N";
String test = "I am Duc/N Ta/N Van/N";
Matcher m = Pattern.compile(pattern).matcher(test);
StringBuilder sbNames = new StringBuilder();
while(m.find()){
sbNames.append(m.group(1)).append(" ");
}
System.out.println(sbNames.toString());
Duc Ta Van
It is giving you the hardest part. I let you adapt this to match your need.
Note :
In java, it is not required to escape a forward slash, but to use the same regex in the entire answer, I will keep "(\\w+)\\/N", but "(\\w+)/N" will work as well.

I've used "[/N]+" as the regular expression.
Regex101
[] = Matches characters inside the set
\/ = Matches the character / literally (case sensitive)
+ = Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)

Regex not matching words delimited by whitespace

I have an input string that will follow the pattern /user/<id>?name=<name>, where <id> is alphanumeric but must start with a letter, and <name> is a letter-only string that can have multiple spaces. Some examples of matches would be:
/user/ad?name=a a
/user/one111?name=one ONE oNe
/user/hello?name=world
I came up with the following regex:
String regex = "/user/[a-zA-Z]+\\w*\\?name=[a-zA-Z\\s]+";
All of the above examples match the regex, but it only looks at the first word in <name>. Shouldn't the sequence \s allow me to have white spaces?
The code that I made to test what it is doing is:
String regex = "/user/[a-zA-Z]+\\w*\\?name=[a-zA-Z\\s]+";
// Check to see that input matches pattern
if(Pattern.matches(regex, str) == true){
str = str.replaceFirst("/user/", "");
str = str.replaceFirst("name=", "");
String[] tokens = str.split("\\?");
System.out.println("size = " + tokens.length);
System.out.println("tokens[0] = " + tokens[0]);
System.out.println("tokens[1] = " + tokens[1]);
} else
System.out.println("Didn't match.");
So for example, one test might look like:
/user/myID123?name=firstName LastName
size = 2
tokens[0] = myID123
tokens[1] = firstName
whereas the desired output would be
tokens[1] = firstName LastName
How can I change my regex to do this?

Not sure what you think is the problem in your code. tokens[1] will indeed contain firstName LastName in your example.
Here's an ideone.com demo showing this.
However, have you considered using capturing groups for the id and the name.
If you write it like
String regex = "/user/(\\w+)\\?name=([a-zA-Z\\s]+)";
Matcher m = Pattern.compile(regex).matcher(input);
you can get hold of myID123 and firstName LastName through m.group(1) and m.group(2)

I don't find any fault in your code but you may capture group like this:
String str = "/user/myID123?name=firstName LastName ";
String regex = "/user/([a-zA-Z]+\\w*)\\?name=([a-zA-Z\\s]+)";
Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(str);
if(m.find()) {
System.out.println(m.group(1) + ", " + m.group(2));
}

The problem is that * is greedy by default (it matches the whole string), so you need to modify your regex by adding a ? (making it reluctant):
List<String> str = Arrays.asList("/user/ad?name=a a", "/user/one111?name=one ONE oNe", "/user/hello?name=world");
String regex = "/user/([a-zA-Z]+\\w*?)\\?name=([a-zA-Z\\s]+)";
for (String s : str) {
Matcher matcher = Pattern.compile(regex).matcher(s);
if (matcher.matches()) {
System.out.println("user: " + matcher.group(1));
System.out.println("name: " + matcher.group(2));
}
}
Output:
user: ad
name: a a
user: one111
name: one ONE oNe
user: hello
name: world

Pattern for pulling strings out a string

I'm not new to Java, but have not dealt with Regex and Patterns before. What I'm looking to do is take a string like
"Class: " + data1 + "\nFrom: " + data2 + " To: " + data3 + "\nOccures: " + data4 + " In: " + data5 + " " + data6;
and pull out only data_1 to data_n.
I appreciate any help.

Use this regex:
Pattern pattern = Pattern.compile("Class: (.+?)\nFrom: (.+?) To: (.+?)\nOccures: (.+?) In: (.+?) (.+?)");
Matcher matcher = pattern.matcher(yourInputString);
if (matcher.find())
{
String data1 = matcher.group(1);
String data2 = matcher.group(2);
String data3 = matcher.group(3);
String data4 = matcher.group(4);
String data5 = matcher.group(5);
String data6 = matcher.group(6);
} else
{
// String didn't match the specified format
}
Explanation:
.+? will match any character for undefined times, but non-greedy.
(), using brackets will create a group. A group is given an index starting by 1 (since group 0 is the entire match)
So, (.+?) will creates groups of any character.
And what the matcher does, is searching for the whole pattern somewhere in the input string. But since you specified the format, we know exactly how your entire string is going to look like. The only thing you have to do is copy the format and replace the data you want to extract with "something" (.+?), which you give an index by creating a group of it.
Afterwards, the matcher will try to find the pattern (done by matcher.find()) and you ask them what the content is of the groups 1 up to 6.

how about using split() with ":", then from the splitted String[] get string[2i+1] ? (i from 0)