Develop Reference

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. ''

Need help in regex matching

It may be very simple, but I am extremely new to regex and have a requirement where I need to do some regex matches in a string and extract the number in it. Below is my code with sample i/p and required o/p. I tried to construct the Pattern by referring to https://www.freeformatter.com/java-regex-tester.html, but my regex match itself is returning false.
Pattern pattern = Pattern.compile(".*/(a-b|c-d|e-f)/([0-9])+(#[0-9]?)");
String str = "foo/bar/Samsung-Galaxy/a-b/1"; // need to extract 1.
String str1 = "foo/bar/Samsung-Galaxy/c-d/1#P2";// need to extract 2.
String str2 = "foo.com/Samsung-Galaxy/9090/c-d/69"; // need to extract 69
System.out.println("result " + pattern.matcher(str).matches());
System.out.println("result " + pattern.matcher(str1).matches());
System.out.println("result " + pattern.matcher(str1).matches());
All of above SOPs are returning false. I am using java 8, is there is any way by which in a single statement I can match the pattern and then extract the digit from the string.
I would be great if somebody can point me on how to debug/develop the regex.Please feel free to let me know if something is not clear in my question.

You may use
Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");
See the regex demo
When used with matches(), the pattern above does not require explicit anchors, ^ and $.
Details
.* - any 0+ chars other than line break chars, as many as possible
/ - the rightmost / that is followed with the subsequent subpatterns
(?:a-b|c-d|e-f) - a non-capturing group matching any of the alternatives inside: a-b, c-d or e-f
/ - a / char
[^/]*? - any chars other than /, as few as possible
([0-9]+) - Group 1: one or more digits.
Java demo:
List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
Pattern pattern = Pattern.compile(".*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)");
for (String s : strs) {
Matcher m = pattern.matcher(s);
if (m.matches()) {
System.out.println(s + ": \"" + m.group(1) + "\"");
}
}
A replacing approach using the same regex with anchors added:
List<String> strs = Arrays.asList("foo/bar/Samsung-Galaxy/a-b/1","foo/bar/Samsung-Galaxy/c-d/1#P2","foo.com/Samsung-Galaxy/9090/c-d/69");
String pattern = "^.*/(?:a-b|c-d|e-f)/[^/]*?([0-9]+)$";
for (String s : strs) {
System.out.println(s + ": \"" + s.replaceFirst(pattern, "$1") + "\"");
}
See another Java demo.
Output:
foo/bar/Samsung-Galaxy/a-b/1: "1"
foo/bar/Samsung-Galaxy/c-d/1#P2: "2"
foo.com/Samsung-Galaxy/9090/c-d/69: "69"

Because you match always the last number in your regex, I would Like to just use replaceAll with this regex .*?(\d+)$ :
String regex = ".*?(\\d+)$";
String strResult1 = str.replaceAll(regex, "$1");
System.out.println(!strResult1.isEmpty() ? "result " + strResult1 : "no result");
String strResult2 = str1.replaceAll(regex, "$1");
System.out.println(!strResult2.isEmpty() ? "result " + strResult2 : "no result");
String strResult3 = str2.replaceAll(regex, "$1");
System.out.println(!strResult3.isEmpty() ? "result " + strResult3 : "no result");
If the result is empty then you don't have any number.
Outputs
result 1
result 2
result 69

Here is a one-liner using String#replaceAll:
public String getDigits(String input) {
String number = input.replaceAll(".*/(?:a-b|c-d|e-f)/[^/]*?(\\d+)$", "$1");
return number.matches("\\d+") ? number : "no match";
}
System.out.println(getDigits("foo.com/Samsung-Galaxy/9090/c-d/69"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/a-b/some other text/1"));
System.out.println(getDigits("foo/bar/Samsung-Galaxy/9090/a-b/69ace"));
69
no match
no match
This works on the sample inputs you provided. Note that I added logic which will display no match for the case where ending digits could not be matched fitting your pattern. In the case of a non-match, we would typically be left with the original input string, which would not be all digits.

Java multiple regular expression search

I have a string some thing like this:
If message contains sensitive info like: {Password:123456, tmpPwd : tesgjadgj, TEMP_PASSWORD: kfnda}
My pattern should look for the particular words Password or tmpPwd or TEMP_PASSWORD.
How can I create a pattern for this kind of search?

I think you are looking for the values after these words. You need to set capturing groups to extract those values, e.g.
String content = "If message contains sensitive info like: {Password:123456, tmpPwd : tesgjadgj, TEMP_PASSWORD: kfnda} ";
Pattern p = Pattern.compile("\\{Password\\s*:\\s*([^,]+)\\s*,\\s*tmpPwd\\s*:\\s*([^,]+)\\s*,\\s*TEMP_PASSWORD:\\s*([^,]+)\\s*\\}");
Matcher m = p.matcher(content);
while (m.find()) {
System.out.println(m.group(1) + ", " + m.group(2) + ", " + m.group(3));
}
See IDEONE demo
This will output 123456, tesgjadgj, kfnda.
To just find out if there are any of the substrings, use contains method:
System.out.println(content.contains("Password") ||
content.contains("tmpPwd") ||
content.contains("TEMP_PASSWORD"));
See another demo
And if you want a regex-solution for the keywords, here it is:
String str = "If message contains sensitive info like: {Password:123456, tmpPwd : tesgjadgj, TEMP_PASSWORD: kfnda} ";
Pattern ptrn = Pattern.compile("Password|tmpPwd|TEMP_PASSWORD");
Matcher m = ptrn.matcher(str);
while (m.find()) {
System.out.println("Match found: " + m.group(0));
}
See Demo 3

Finally I am using it like as per my requirement .
private final static String censoredWords =
"(?i)PASSWORD|pwd";
The (?i) makes it case-insensitive

How to properly escape this regex patterns in java?

This is the input I want to process. I want to extract the value of the operation attribute:
<h:outputLink value="#" id="temp_solution">
<rich:componentContro
for="panel"
attachTo="temp_solution"
operation="show"
event="onclick"/>
</h:outputLink>
With the help of an online regex tester I came up with the following regular expression
(?<=operation=")(\w+)(?=")
To be a bit more dynamic, I replaced operation with %s so I can use this template for different situations. But I encountered a problem, while trying to test my "creation" with the help of a small test program:
public class Main {
private static final String INPUT = "<h:outputLink value=\"#\" id=\"temp_solution\">\n"
+ " <rich:componentControl \n"
+ " for=\"panel\" \n"
+ " attachTo=\"temp_solution\" \n"
+ " operation=\"show\""
+ " event=\"onclick\"/> \n"
+ "</h:outputLink>";
private static final String REGEX_TEMPLATE = "(?<=%s=\")(\\w+)(?=\")";
public static void main(String[] args) throws IOException {
final String actualRegex = String.format(REGEX_TEMPLATE, "operation");
final Pattern pattern = Pattern.compile(actualRegex);
final Matcher matcher = pattern.matcher(INPUT);
System.out.println("Regex: " + pattern);
System.out.println(matcher.matches() ? matcher.group(0) : "Nothing found");
}
}
Output:
Regex: (?<=operation=")(\w+)(?=")
Nothing found
Even double escaping the regex inside my code:
private static final String REGEX_TEMPLATE = "(?<=%s=\\\")(\\\\w+)(?=\\\")";
doesn't help:
Regex: (?<=operation=\")(\\w+)(?=\")
Nothing found
Please give me some advise on this.

There is nothing wrong with your regex. However, it doesn't match the entire input, so you can't use matches(). Change it to find(), which only tries to find a matching subsequence:
System.out.println(matcher.find() ? matcher.group(0) : "Nothing found");

Try regex like this :
(?<=operation=\")(\w+)
demo

extract values with java regex

I begin with regex and i want extract values from a String like this
String test="[ABC]Name:User:Date: Adresse ";
I want extract Name, User , Date and Adresse
I can do the trick with substring and split
String test = "String test="[ABC]Name:User:Date: Adresse ";
String test2= test.substring(5,test.length());
System.out.println(test2);
String[] chaine = test2.split(":");
for(String s :chaine)
{
System.out.println("Valeur " + s);
}
but i want try with regex , i did
pattern = Pattern.compile("^[(ABC)|:].");
but it doesn ' t work
Can you help me please ?
Thanks a lot

String#split is really the best way to accomplish what you are trying to do. Having said that, with regex, the following will give you the same output:
Pattern p = Pattern.compile("^(?:\\[ABC\\])([^:]+):([^:]+):([^:]+):([^:]+)$");
Matcher m = p.matcher(test);
while (m.find()) {
System.out.println("Valeur " + m.group(1)); // Name
System.out.println("Valeur " + m.group(2)); // User
System.out.println("Valeur " + m.group(3)); // Date
System.out.println("Valeur " + m.group(4)); // Address
}

You have to escape the [ and ] here is a working example.
^\[(.*)\](.*):(.*):(.*):(.*)$
Note that your code is probably more easily maintained than regular expressions in cases where the regular expression becomes complex.
Some people, when confronted with a problem, think "I know, I'll use
regular expressions." Now they have two problems. - Jamie Zawinski