Develop Reference

regular expression to match a string in order

I have string as follows
"ValueFilter("val1") AND ColumnFilter("val2") AND ValueFilter("val3")"
I have stored the following regex in a array. Using for loop I tried to match the pattern
"ValueFilter\\((.*?)\\)","ColumnFilter\\((.*?)\\)"
what I will do is I will replace the value in the bracket and copy it to a new string.
When I run this above regex against the string in the first loop i have XFilter so it will match both occurrence. But I want to do this in order.
Here is the i thing i want to achieve
first i want to match ValueFilter first then ColumnFilter then again ValueFilter. How can I achieve this?
Edit : Added Code
String expr = "\"ValueFilter(\"val1\") AND ColumnFilter(\"val2\") AND ValueFilter(\"val3\")\"";
String patterns = {"ValueFilter\\((.*?)\\)", "ColumnFilter\\((.*?)\\)"}
for (String pattern : patterns) {
Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(expr);
while (m.find()) {
//do something
}
}
Expected Output
ValueFilter("val1")
ColumnFilter("val2")
ValueFilter("val3")

You can use this regex [XY]Filter\((.*?)\) with pattern and you have to loop throw the matches using :
String str = "\"XFilter(\"val1\") AND YFilter(\"val2\") AND XFilter(\"val3\")\"";
String regex = "[XY]Filter\\((.*?)\\)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group());
}
Note you can i use [XY] which mean to match both X or Y,
Output
XFilter("val1")
YFilter("val2")
XFilter("val3")
regex demo
If you want to get only the value you can get the group 1 like matcher.group(1) instead, the output should be :
"val1"
"val2"
"val3"
Edit
what if I have filtername as "ValueFilter" and "ColumnFilter" instead
of X and Y
In this case you can use (Value|Column) instead of [XY] which mean match ValueFilter or ColumnFilter, the regex should look like :
String str = "\"ValueFilter(\"val1\") AND ColumnFilter(\"val2\") AND ValueFilter(\"val3\")\"";
String regex = "(Value|Column)Filter\\((.*?)\\)";
Pattern pattern = Pattern.compile(regex);
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group());
}
Output
ValueFilter("val1")
ColumnFilter("val2")
ValueFilter("val3")
Check code demo

Java regex does not match as expected

I'm starting with regex in Java recently, and I cant wrap my head around this problem.
Pattern p = Pattern.compile("[^A-Z]+");
Matcher matcher = p.matcher("GETs");
if (matcher.matches()) {
System.out.println("Matched.");
} else {
System.out.println("Did not match.");
}
Result: Did not Match(Unexpected result) Explain this
I get the output "Did not match." This is strange to me, while reading https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html,
I'm using the X+, which matches "One, or more times".
I thought my code in words would go something like this:
"Check if there is one or more characters in the string "GETs" which does not belong in A to Z."
So I'm expecting the following result:
"Yes, there is one character that does not belong to A-Z in "GETs", the regex was a match."
However this is not the case, I'm confused to why this is.
I tried the following:
Pattern p = Pattern.compile("[A-Z]+");
Matcher matcher = p.matcher("GETs");
if (matcher.matches()) {
System.out.println("Matched.");
} else {
System.out.println("Did not match.");
}
Result: Did not match. (Expected result)
Pattern p = Pattern.compile("[A-Z]+");
Matcher matcher = p.matcher("GET");
if (matcher.matches()) {
System.out.println("Matched.");
} else {
System.out.println("Did not match.");
}
Result: Matched. (Expected result)
Please, explain why my first example did not work.

Matcher.matches returns true only if the ENTIRE region
matches the pattern.
For the output you are looking for, use Matches.find instead
Explanation of each case:
Pattern p = Pattern.compile("[^A-Z]+");
Matcher matcher = p.matcher("GETs");
if (matcher.matches()) {
Fails because the ENTIRE region 'GETs' isn't lowercase
Pattern p = Pattern.compile("[A-Z]+");
Matcher matcher = p.matcher("GETs");
if (matcher.matches()) {
This fails because the ENTIRE region 'GETs' isn't uppercase
Pattern p = Pattern.compile("[A-Z]+");
Matcher matcher = p.matcher("GET");
if (matcher.matches()) {
The ENTIRE region 'GET' is uppercase, the pattern matches.

You're very first regex asks to match any character that is not in an uppercase range of A-Z. The match is on the lowercase "s" in GETs.

if you want a regex to match either in UPPERCASE and lowercase, you can use this:
String test = "yes";
String test2= "YEs";
test.matches("(?i).*\\byes\\b.*");
test2.matches("(?i).*\\byes\\b.*");
will return true in the two cases

How to create a regEx to extract value from a string with square brackets in Java

I have a string "Something[Anything]".
First I want to check whether string contains Square brackets if yes then I want this string to be separated in "Something" & "[Anything]".
Need some help with regEx for this.
Thanks in Advance.

Try this:
String test = "Something[Anything]";
if (test.matches(".*\\[.*\\].*")) { // checks if in the string presents open and close square brackets
Pattern pat = Pattern.compile("(.*?)(\\[.*?\\])");
Matcher matcher = pat.matcher(test);
matcher.find();
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}
Outputs:
Something
[Anything]
Or, as suggested by #madatx, without first check:
Pattern pat = Pattern.compile("(.*?)(\\[.*?\\])");
Matcher matcher = pat.matcher(test);
if (matcher.find()) {
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}
Same output.

You can use regex withour checking the square brackets.
This could work:
String strPattern = "(.+)(\\[.+\\])";
Pattern p = Pattern.compile(strPattern);
Matcher m = p.matcher("<yout string>");
if (m.matches()){
<your code here>
}

Java regex not matching, regex looks OK

The following returns no matches:
String patternStr = "((19\\d{2}|20\\d{2})-([0-2]\\d{2}|3[0-5]\\d)-(([0-1]\\d|2[0-3])[0-5]\\d[0-5]\\d))";
String fullPath = aFile.getAbsolutePath();
// fullPath should expand to this: "/home/user1/2013-023-135159_abcd_001/File.txt"
Pattern p = Pattern.compile(patternStr);
Matcher m = p.matcher(fullPath);
if (m.matches())
{
System.out.println("Matches found");
}
It should match the date portion, 2013-023-135159. I tested it online and the regex looks OK.

You will need to use:
m.find()
instead of:
m.matches()
As your regex is matching the parts of the input string not fully as expected by m.matches()
RegEx Demo

Splitting a string composed of numbers and alphabets

I want to break a string like :
String s = "xyz213123kop234430099kpf4532";
into tokens where each token starts with an alphabet and ends with a number. So the above string can be broken down into 3 tokens :
xyz213123
kop234430099
kpf4532
This string s could be very big but the pattern will remain the same, i.e each token will start with 3 alphabets and end with a number.
How do I split them ?

Try this:
\w+?\d+
Java Matcher:
Pattern pattern = Pattern.compile("\\w+?\\d+"); //compiles the pattern we want to use
Matcher matcher = pattern.matcher("xyz213123kop234430099kpf4532"); //we create the matcher on certain string using our pattern
while(matcher.find()) //while the matcher can find the next match
{
System.out.println(matcher.group()); //print it
}
And then you could use Regex.Matches C#:
foreach(Match m in Regex.Matches("xyz213123kop234430099kpf4532", #"\w+?\d+"))
{
Console.WriteLine(m.Value);
}
And for the future this:
RegExr

Do it like this,
String s = "xyz213123kop234430099kpf4532";
Pattern p = Pattern.compile("\\w+?\\d+");
Matcher match = p.matcher(s);
while(match.find()){
System.out.println(match.group());
}
OUTPUT
xyz213123
kop234430099
kpf4532

You can start from such regexp: (\w+?\d+)
http://regexr.com?36utt