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Regex for extracting digits in version format

I am going to extract numbers from a string. Numbers represents a version.
It means, I am going to match numbers which are between:
_ and /
/ and /
I have prepared the following regex, but it doesn't work as expected:
.*[\/_](\d{1,2}[.]\d{1,2}[.]\d{1,2})\/.*
For the following example, the regex should match twice:
Input: name_1.1.1/9.10.0/abc. Expected result: 1.1.1 and 9.10.0
, but my regex returns only 9.10.0, 1.1.1 is omitted. Do you have any idea what is wrong?

You could just split the string on _ or /, and then retain components which appear to be versions:
List<String> versions = new ArrayList<>();
String input = "name_1.1.1/9.10.0/abc";
String[] parts = input.split("[_/]");
for (String part : parts) {
if (part.matches("\\d+(?:\\.\\d+)*")) {
versions.add(part);
}
}
System.out.println(versions); // [1.1.1, 9.10.0]

You can assert the / at the end instead of matching it, and omit the .*
Note that you don't have to escape the /
[/_](\d{1,2}[.]\d{1,2}[.]\d{1,2})(?=/)
Regex demo | Java demo
Example code
String regex = "[/_](\\d{1,2}[.]\\d{1,2}[.]\\d{1,2})(?=/)";
String string = "name_1.1.1/9.10.0/abc";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
Output
1.1.1
9.10.0
Another option could be using a positive lookbehind to assert either a / or _ to the left, and get a match only.
(?<=[/_])\d{1,2}[.]\d{1,2}[.]\d{1,2}(?=/)
regex demo

Code Demo
String regex = "(\\d+.\\d+.\\d+)";
String string = "name_1.1.1/9.10.0/abc";
String string2 = "randomversion4.5.6/09.7.8_9.88.9";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(string);
Matcher matcher2 = pattern.matcher(string2);
while (matcher.find()) {
System.out.println(matcher.group(1));
}
while (matcher2.find()) {
System.out.println(matcher2.group(1));
}
Out:
1.1.1
9.10.0
4.5.6
09.7.8
9.88.9
Just write regex for what you want to match. In this case just the version number.
Regex can be used to match whole strings or to find if there is a substring that exists in a string.
When using regex to find a substring, you cannot always match all filenames or any string. Hence only match on what you want to find.
This way you can find the versions no matter what string it is in.

Android Java regexp pattern

I ping a host. In result a standard output. Below a REGEXP but it do not work correct. Where I did a mistake?
String REGEXP ="time=(\\\\d+)ms";
Pattern pattern = Pattern.compile(REGEXP);
Matcher matcher = pattern.matcher(result);
if (matcher.find()) {
result = matcher.group(1);
}

You only need \\d+ in your regex because
Matcher looks for the pattern (using which it is created) and then tries to find every occurance of the pattern in the string being matched.
Use while(matcher.group(1) in case of multiple occurances.
each () represents a captured group.

You have too many backslashes. Assuming you want to get the number from a string like "time=32ms", then you need:
String REGEXP ="time=(\\d+)ms";
Pattern pattern = Pattern.compile(REGEXP);
Matcher matcher = pattern.matcher(result);
if (matcher.find()) {
result = matcher.group(1);
}
Explanation: The search pattern you are looking for is "\d", meaning a decimal number, the "+" means 1 or more occurrences.
To get the "\" to the matcher, it needs to be escaped, and the escape character is also "\".
The brackets define the matching group that you want to pick out.
With "\\\\d+", the matcher sees this as "\\d+", which would match a backslash followed by one or more "d"s. The first backslash protects the second backslash, and the third protects the fourth.

Find string after last underscore before dot extension

I need to find 20140809T0000Z in this string:
PREVIMER_F2-MARS3D-MENOR1200_20140809T0000Z.nc
I tried the following to keep the string before the .nc:
(?<=_)(.*)(?=.nc)
I have the following to start from the last underscore:
/_[^_]*$/
How can I find string after last underscore before dot extension, using a regex?

RegEx is not always the best solution... :)
String pattern="PREVIMER_F2-MARS3D-MENOR1200_20140809T0000Z.nc";
int start=pattern.lastIndexOf("_") + 1;
int end=pattern.lastIndexOf(".");
if(start != 0 && end != -1 && end > start) {
System.out.println(pattern.substring(start,end);
}

You just need lookahead for this requirement.
You can use:
[^._]+(?=[^_]*$)
// matches and returns 20140809T0000Z
RegEx Demo

You could use the below regex,
(?<=_)[^_]*(?=\.nc)
In your pattern just replace .* with [^_]* so that it would match the inner string.
DEMO
String s = "PREVIMER_F2-MARS3D-MENOR1200_20140809T0000Z.nc";
Pattern regex = Pattern.compile("(?<=_)[^_]*(?=\\.nc)");
Matcher regexMatcher = regex.matcher(s);
if (regexMatcher.find()) {
String ResultString = regexMatcher.group();
System.out.println(ResultString);
} //=> 20140809T0000Z

You could use a simpler pattern with a capturing group
.*_(.*)\.nc
By default the first .* will be "greedy" and consume as many characters as possible before the _, leaving just the desired string inside the (.*).
Demo: http://regex101.com/r/aI2xQ9/1
Java code:
String input = "PREVIMER_F2-MARS3D-MENOR1200_20140809T0000Z.nc";
Pattern pattern = Pattern.compile(".*_(.*)\\.nc");
Matcher matcher = pattern.matcher(input);
if (matcher.find()) {
String group = matcher.group(1);
// ...
}

So, you need a sequence of non-underscore characters that immediately precede the period character.
Try [^_.]+(?=\.)
Demo: https://regex101.com/r/sLAnVs/2
Thanks to Cary Swoveland for pointing out that "no need to escape a period in a character class".

java Pattern Matching issue

I have an issue to write proper regex to match URL.
String input = "AAAhttp://www.gmail.comBBBBabc#gmail.com"
String regex = "www.*.com" // To match www.gmail.com URL
Pattern p = Pattern.compile(regex)
Matcher m = p.matcher(input)
while(m.find()){
}
Here I want to remove the Url www.gmail.com. However it matches till end of string to match email address also which ends with gmail.com.
Can someone help me to get proper regex to match only the URL?

.* does a greedy match. You have to add ? after * to does an reluctant match.
"www\\..*?\\.com"
Your code would be,
String s = "AAAhttp://www.gmail.comBBBBabc#gmail.com";
Pattern p = Pattern.compile("www\\..*?\\.com");
Matcher m = p.matcher(s);
while (m.find()) {
System.out.println(m.group(0));
}
IDEONE

String regex = "www\\..*?\\.com"
Non-greedy repetition of the wildcard '.' and escape dot when literally

A negated character class is faster than .*?
Use this regex:
www\.[^.]+\.com
[^.]+ means any character that is not a dot.
In Java we need to escape some characters:
// for instance
Pattern regex = Pattern.compile("www\\.[^.]+\\.com");
// etc

Java regex validating special chars

This seems like a well known title, but I am really facing a problem in this.
Here is what I have and what I've done so far.
I have validate input string, these chars are not allowed :
&%$###!~
So I coded it like this:
String REGEX = "^[&%$###!~]";
String username= "jhgjhgjh.#";
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(username);
if (matcher.matches()) {
System.out.println("matched");
}

Change your first line of code like this
String REGEX = "[^&%$##!~]*";
And it should work fine. ^ outside the character class denotes start of line. ^ inside a character class [] means a negation of the characters inside the character class. And, if you don't want to match empty usernames, then use this regex
String REGEX = "[^&%$##!~]+";

i think you want this:
[^&%$###!~]*

To match a valid input:
String REGEX = "[^&%$##!~]*";
To match an invalid input:
String REGEX = ".*[&%$##!~]+.*";