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Get substring between "first two" occurrences of a character

I have a String:
String thestra = "/aaa/bbb/ccc/ddd/eee";
Every time, in my situation, for this Sting, a minimum of two slashes will be present without fail.
And I am getting the /aaa/ like below, which is the subString between "FIRST TWO occurrences" of the char / in the String.
System.out.println("/" + thestra.split("\\/")[1] + "/");
It solves my purpose but I am wondering if there is any other elegant and cleaner alternative to this?
Please notice that I need both slashes (leading and trailing) around aaa. i.e. /aaa/

You can use indexOf, which accepts a second argument for an index to start searching from:
int start = thestra.indexOf("/");
int end = thestra.indexOf("/", start + 1) + 1;
System.out.println(thestra.substring(start, end));
Whether or not it's more elegant is a matter of opinion, but at least it doesn't find every / in the string or create an unnecessary array.

Scanner::findInLine returning the first match of the pattern may be used:
String thestra = "/aaa/bbb/ccc/ddd/eee";
System.out.println(new Scanner(thestra).findInLine("/[^/]*/"));
Output:
/aaa/

Use Pattern and Matcher from java.util.regex.
Pattern pattern = Pattern.compile("/.*?/");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
String match = matcher.group(0); // output
}

Pattern.compile("/.*?/")
.matcher(thestra)
.results()
.map(MatchResult::group)
.findFirst().ifPresent(System.out::println);
You can test this variant :)
With best regards, Fr0z3Nn

Every time, in my situation, for this Sting, minimum two slashes will be present
if that is guaranteed, split at each / keeping those delimeters and take the first three substrings.
String str = String.format("%s%s%s",(thestra.split("((?<=\\/)|(?=\\/))")));

You could also match the leading forward slash, then use a negated character class [^/]* to optionally match any character except / and then match the trailing forward slash.
String thestra = "/aaa/bbb/ccc/ddd/eee";
Pattern pattern = Pattern.compile("/[^/]*/");
Matcher matcher = pattern.matcher(thestra);
if (matcher.find()) {
System.out.println(matcher.group());
}
Output
/aaa/

One of the many ways can be replacing the string with group#1 of the regex, [^/]*(/[^/].*?/).* as shown below:
public class Main {
public static void main(String[] args) {
String thestra = "/aaa/bbb/ccc/ddd/eee";
String result = thestra.replaceAll("[^/]*(/[^/].*?/).*", "$1");
System.out.println(result);
}
}
Output:
/aaa/
Explanation of the regex:
[^/]* : Not the character, /, any number of times
( : Start of group#1
/ : The character, /
[^/]: Not the character, /
.*?: Any character any number of times (lazy match)
/ : The character, /
) : End of group#1
.* : Any character any number of times
Updated the answer as per the following valuable suggestion from Holger:
Note that to the Java regex engine, the / has no special meaning, so there is no need for escaping here. Further, since you’re only expecting a single match (the .* at the end ensures this), replaceFirst would be more idiomatic. And since there was no statement about the first / being always at the beginning of the string, prepending the pattern with either , .*? or [^/]*, would be a good idea.

I am surprised nobody mentioned using Path as of Java 7.
String thestra = "/aaa/bbb/ccc/ddd/eee";
String path = Paths.get(thestra).getName(0).toString();
System.out.println("/" + path + "/");
/aaa/

String thestra = "/aaa/bbb/ccc/ddd/eee";
System.out.println(thestra.substring(0, thestra.indexOf("/", 2) + 1));

How to crop multiple text from String in Java?

I want to crop a portion of String:
" this is Test [ABC:123456] Sting with multiple properties [ABC:98765] ..."
So in result i want to crop String between "[ ]". {Here ABC:12345 and ABC:98765}
Note There can be n number of property.
what is the Best way to get result.

public static void main(String[] args) {
String input = "test bla [ABC56465:asd] asdasdqwd [DEF:345]";
Pattern pattern = Pattern.compile("\\[(.*?)\\]");
Matcher match = pattern.matcher(input);
while(match.find()){
System.out.println(match.group());
}
}
Follow the Tutorials from Niels. This could be solution.
To get the output without the "[ ]" just replace:
System.out.println(match.group());
With:
System.out.println(match.group(1));
as mentioned in the comments.

You need to define you pattern and do a regex matching, extrating the group you need.
See http://tutorials.jenkov.com/java-regex/matcher.html
for a tutorial on regex. (Especially the find() start() and end() section)
In your case the pattern should be very simple.

In Java, you can do this:
String resultString = subjectString.replaceAll("\\[[^\\]]*\\]", "");
Explanation
\[ matches the opening bracket
The negated character class [^\]]* matches any character that is not a closing bracket
\] matches the closing bracket
We replace with the empty string

regex pattern - extract a string only if separated by a hyphen

I've looked at other questions, but they didn't lead me to an answer.
I've got this code:
Pattern p = Pattern.compile("exp_(\\d{1}-\\d)-(\\d+)");
The string I want to be matched is: exp_5-22-718
I would like to extract 5-22 and 718. I'm not too sure why it's not working What am I missing? Many thanks

Try this one:
Pattern p = Pattern.compile("exp_(\\d-\\d+)-(\\d+)");
In your original pattern you specified that second number should contain exactly one digit, so I put \d+ to match as more digits as we can.
Also I removed {1} from the first number definition as it does not add value to regexp.

If the string is always prefixed with exp_ I wouldn't use a regular expression.
I would:
replaceFirst() exp_
split() the resulting string on -
Note: This answer is based on the assumptions. I offer it as a more robust if you have multiple hyphens. However, if you need to validate the format of the digits then a regular expression may be better.

In your regexp you missed required quantifier for second digit \\d. This quantifier is + or {2}.
String yourString = "exp_5-22-718";
Matcher matcher = Pattern.compile("exp_(\\d-\\d+)-(\\d+)").matcher(yourString);
if (matcher.find()) {
System.out.println(matcher.group(1)); //prints 5-22
System.out.println(matcher.group(2)); //prints 718
}

You can use the string.split methods to do this. Check the following code.
I assume that your strings starts with "exp_".
String str = "exp_5-22-718";
if (str.contains("-")){
String newStr = str.substring(4, str.length());
String[] strings = newStr.split("-");
for (String string : strings) {
System.out.println(string);
}
}

java regular expression

Can anyone please help me do the following in a java regular expression?
I need to read 3 characters from the 5th position from a given String ignoring whatever is found before and after.
Example : testXXXtest
Expected result : XXX

You don't need regex at all.
Just use substring: yourString.substring(4,7)
Since you do need to use regex, you can do it like this:
Pattern pattern = Pattern.compile(".{4}(.{3}).*");
Matcher matcher = pattern.matcher("testXXXtest");
matcher.matches();
String whatYouNeed = matcher.group(1);
What does it mean, step by step:
.{4} - any four characters
( - start capturing group, i.e. what you need
.{3} - any three characters
) - end capturing group, you got it now
.* followed by 0 or more arbitrary characters.
matcher.group(1) - get the 1st (only) capturing group.

You should be able to use the substring() method to accomplish this:
string example = "testXXXtest";
string result = example.substring(4,7);

This might help: Groups and capturing in java.util.regex.Pattern.
Here is an example:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Example {
public static void main(String[] args) {
String text = "This is a testWithSomeDataInBetweentest.";
Pattern p = Pattern.compile("test([A-Za-z0-9]*)test");
Matcher m = p.matcher(text);
if (m.find()) {
System.out.println("Matched: " + m.group(1));
} else {
System.out.println("No match.");
}
}
}
This prints:
Matched: WithSomeDataInBetween
If you don't want to match the entire pattern rather to the input string (rather than to seek a substring that would match), you can use matches() instead of find(). You can continue searching for more matching substrings with subsequent calls with find().
Also, your question did not specify what are admissible characters and length of the string between two "test" strings. I assumed any length is OK including zero and that we seek a substring composed of small and capital letters as well as digits.

You can use substring for this, you don't need a regex.
yourString.substring(4,7);
I'm sure you could use a regex too, but why if you don't need it. Of course you should protect this code against null and strings that are too short.

Use the String.replaceAll() Class Method
If you don't need to be performance optimized, you can try the String.replaceAll() class method for a cleaner option:
String sDataLine = "testXXXtest";
String sWhatYouNeed = sDataLine.replaceAll( ".{4}(.{3}).*", "$1" );
References
https://docs.oracle.com/javase/1.5.0/docs/api/java/lang/String.html
http://www.vogella.com/tutorials/JavaRegularExpressions/article.html#using-regular-expressions-with-string-methods

Regular Expression problem in Java

I am trying to create a regular expression for the replaceAll method in Java. The test string is abXYabcXYZ and the pattern is abc. I want to replace any symbol except the pattern with +. For example the string abXYabcXYZ and pattern [^(abc)] should return ++++abc+++, but in my case it returns ab++abc+++.
public static String plusOut(String str, String pattern) {
pattern= "[^("+pattern+")]" + "".toLowerCase();
return str.toLowerCase().replaceAll(pattern, "+");
}
public static void main(String[] args) {
String text = "abXYabcXYZ";
String pattern = "abc";
System.out.println(plusOut(text, pattern));
}
When I try to replace the pattern with + there is no problem - abXYabcXYZ with pattern (abc) returns abxy+xyz. Pattern (^(abc)) returns the string without replacement.
Is there any other way to write NOT(regex) or group symbols as a word?

What you are trying to achieve is pretty tough with regular expressions, since there is no way to express “replace strings not matching a pattern”. You will have to use a “positive” pattern, telling what to match instead of what not to match.
Furthermore, you want to replace every character with a replacement character, so you have to make sure that your pattern matches exactly one character. Otherwise, you will replace whole strings with a single character, returning a shorter string.
For your toy example, you can use negative lookaheads and lookbehinds to achieve the task, but this may be more difficult for real-world examples with longer or more complex strings, since you will have to consider each character of your string separately, along with its context.
Here is the pattern for “not ‘abc’”:
[^abc]|a(?!bc)|(?<!a)b|b(?!c)|(?<!ab)c
It consists of five sub-patterns, connected with “or” (|), each matching exactly one character:
[^abc] matches every character except a, b or c
a(?!bc) matches a if it is not followed by bc
(?<!a)b matches b if it is not preceded with a
b(?!c) matches b if it is not followed by c
(?<!ab)c matches c if it is not preceded with ab
The idea is to match every character that is not in your target word abc, plus every word character that, according to the context, is not part of your word. The context can be examined using negative lookaheads (?!...) and lookbehinds (?<!...).
You can imagine that this technique will fail once you have a target word containing one character more than once, like example. It is pretty hard to express “match e if it is not followed by x and not preceded by l”.
Especially for dynamic patterns, it is by far easier to do a positive search and then replace every character that did not match in a second pass, as others have suggested here.

[^ ... ] will match one character that is not any of ...
So your pattern "[^(abc)]" is saying "match one character that is not a, b, c or the left or right bracket"; and indeed that is what happens in your test.
It is hard to say "replace all characters that are not part of the string 'abc'" in a single trivial regular expression. What you might do instead to achieve what you want could be some nasty thing like
while the input string still contains "abc"
find the next occurrence of "abc"
append to the output a string containing as many "+"s as there are characters before the "abc"
append "abc" to the output string
skip, in the input string, to a position just after the "abc" found
append to the output a string containing as many "+"s as there are characters left in the input
or possibly if the input alphabet is restricted you could use regular expressions to do something like
replace all occurrences of "abc" with a single character that does not occur anywhere in the existing string
replace all other characters with "+"
replace all occurrences of the target character with "abc"
which will be more readable but may not perform as well

Negating regexps is usually troublesome. I think you might want to use negative lookahead. Something like this might work:
String pattern = "(?<!ab).(?!abc)";
I didn't test it, so it may not really work for degenerate cases. And the performance might be horrible too. It is probably better to use a multistep algorithm.
Edit: No I think this won't work for every case. You will probably spend more time debugging a regexp like this than doing it algorithmically with some extra code.

Try to solve it without regular expressions:
String out = "";
int i;
for(i=0; i<text.length() - pattern.length() + 1; ) {
if (text.substring(i, i + pattern.length()).equals(pattern)) {
out += pattern;
i += pattern.length();
}
else {
out += "+";
i++;
}
}
for(; i<text.length(); i++) {
out += "+";
}

Rather than a single replaceAll, you could always try something like:
#Test
public void testString() {
final String in = "abXYabcXYabcHIH";
final String expected = "xxxxabcxxabcxxx";
String result = replaceUnwanted(in);
assertEquals(expected, result);
}
private String replaceUnwanted(final String in) {
final Pattern p = Pattern.compile("(.*?)(abc)([^a]*)");
final Matcher m = p.matcher(in);
final StringBuilder out = new StringBuilder();
while (m.find()) {
out.append(m.group(1).replaceAll(".", "x"));
out.append(m.group(2));
out.append(m.group(3).replaceAll(".", "x"));
}
return out.toString();
}

Instead of using replaceAll(...), I'd go for a Pattern/Matcher approach:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static String plusOut(String str, String pattern) {
StringBuilder builder = new StringBuilder();
String regex = String.format("((?:(?!%s).)++)|%s", pattern, pattern);
Matcher m = Pattern.compile(regex).matcher(str.toLowerCase());
while(m.find()) {
builder.append(m.group(1) == null ? pattern : m.group().replaceAll(".", "+"));
}
return builder.toString();
}
public static void main(String[] args) {
String text = "abXYabcXYZ";
String pattern = "abc";
System.out.println(plusOut(text, pattern));
}
}
Note that you'll need to use Pattern.quote(...) if your String pattern contains regex meta-characters.
Edit: I didn't see a Pattern/Matcher approach was already suggested by toolkit (although slightly different)...