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How can i replace this?

How can I replace this
String str = "KMMH12DE1433";
String pattern = "^[a-z]{2}([0-9]{2})[a-z]{1,2}([0-9]{4})$";
String str2 = str.replaceAll(pattern, "repl");
Log.e("Founded_words2",str2);
What I got: KMMH12DE1433
What I want: MH12DE1433

Try it like this using a proper java.util.regex.Pattern and a java.util.regex.Matcher:
String str = "KMMH12DE1433";
//Make the pattern, case-insensitive using (?i)
Pattern pattern = Pattern.compile("(?i)[a-z]{2}([0-9]{2})[a-z]{1,2}([0-9]{4})");
//Create the Matcher
Matcher m = pattern.matcher(str);
//Check if we find anything
if(m.find()) {
//Use what you found - with proper capturing groups you
//gain access to parts of your pattern as needed
System.out.println("Found this: " + m.group());
}

If you just want to remove the first two characters and if the first two characters will always be uppercase letters:
String str = "KMMH12DE1433";
String pattern = "^[A-Z]{2}";
String str2 = str.replaceAll(pattern, "");
Log.e("Output string: ", str2);

try this :
String a = "KMMH12DE1433";
String pattern = "^[A-Z]{2}";
String rs = a.replaceAll(pattern,"");

Please change like this
String ans=str.substring(0);

java regex take variable between two tag

I am very new in regex and need your help. I wanna take numbers and letters between two span.
<span>454.000 $</span>
I wanna take 454.000 $. There are 12 space before . Please help me.

This Should Work.
Regexp:
\s+<.+>(.+)<.+>
Input:
<span>454.000 $</span>
Output:
454.000 $
JAVA CODE:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
final String regex = "\\s+<.+>(.+)<.+>";
final String string = " <span>454.000 $</span>";
final Pattern pattern = Pattern.compile(regex);
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));
}
}
See: https://regex101.com/r/2zg5Ws/1

Capturing group using pattern matching is something like below
String x = " <span>454.000 $</span> ";
Pattern p = Pattern.compile("<span>(.*?)</span>");
Matcher m = p.matcher(x);
if (m.find()) {
System.out.println(">> "+ m.group(1)); // output 454.000 $
}
But for such cases I always prefer to use the replaceAll() as it is shorter version of code:
String num = x.replaceAll(".*<span>(.*?)</span>.*", "$1");
// num has 454.000 $
For the replace it is actually capturing the group from the text and replacing the whole text with that group ($1). This solution depends upon how your input string is.

Java Pattern regex whole word match

I am trying to match a keyword with following string
"abc,pqr(1),xyz"
It will be succesfull match if the whole one word matched for e.g. "par" or "abc" or "xyz"
Can anyone please help me in creating regex for this match ?
String text = "hello, hellos(1),bye";
String keyword = "account";
String patternString = "["+ keyword + "]";
Pattern pattern = Pattern.compile(patternString, Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(text);
boolean matches = matcher.matches();
System.out.println("matches = " + matches);

This Should Work.
([a-zA-Z]+)
Input:
"abc,pqr(1),xyz"
Output:
abc
pqr
xyz
See: https://regex101.com/r/Us6G3X/2

Splitting a string java

I have a string in format:
<+923451234567>: Hi here is the text.
Now I want to get the mobile number(without any non-alphanumeric characters) ie 923451234567 in the start of the string in-between < > symbols, and also the text ie Hi here is the text.
Now I can place a hardcoded logic, which I am currently doing.
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
String[] splitted = cpaMessage.getText().split(">: ", 2);
String mobileNumber=MyUtils.removeNonDigitCharacters(splitted[0]);
String text=splitted[1];
How can I neatly get the required strings from the string with regular expression? So that I don't have to change the code whenever the format of the string changes.

String stringReceivedInSms="<+923451234567>: Hi here is the text.";
Pattern pattern = Pattern.compile("<\\+?([0-9]+)>: (.*)");
Matcher matcher = pattern.matcher(stringReceivedInSms);
if(matcher.matches()) {
String phoneNumber = matcher.group(1);
String messageText = matcher.group(2);
}

Use a regex that matches the pattern - <\\+?(\\d+)>: (.*)
Use the Pattern and Matcher java classes to match the input string.
Pattern p = Pattern.compile("<\\+?(\\d+)>: (.*)");
Matcher m = p.matcher("<+923451234567>: Hi here is the text.");
if(m.matches())
{
System.out.println(m.group(1));
System.out.println(m.group(2));
}

You need to use regex, the following pattern will work:
^<\\+?(\\d++)>:\\s*+(.++)$
Here is how you would use it -
public static void main(String[] args) throws IOException {
final String s = "<+923451234567>: Hi here is the text.";
final Pattern pattern = Pattern.compile(""
+ "#start of line anchor\n"
+ "^\n"
+ "#literal <\n"
+ "<\n"
+ "#an optional +\n"
+ "\\+?\n"
+ "#match and grab at least one digit\n"
+ "(\\d++)\n"
+ "#literal >:\n"
+ ">:\n"
+ "#any amount of whitespace\n"
+ "\\s*+\n"
+ "#match and grap the rest of the string\n"
+ "(.++)\n"
+ "#end anchor\n"
+ "$", Pattern.COMMENTS);
final Matcher matcher = pattern.matcher(s);
if (matcher.matches()) {
System.out.println(matcher.group(1));
System.out.println(matcher.group(2));
}
}
I have added the Pattern.COMMENTS flag so the code will work with the comments embedded for future reference.
Output:
923451234567
Hi here is the text.

You can get your phone number by just doing :
stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">"))
So try this snippet:
public static void main(String[] args){
String stringReceivedInSms="<+923451234567>: Hi here is the text.";
System.out.println(stringReceivedInSms.substring(stringReceivedInSms.indexOf("<+") + 2, stringReceivedInSms.indexOf(">")));
}
You don't need to split your String.

Java how to setup regex for this string

So I'm trying to pull two strings via a matcher object from one string that is stored in my online databases.
Each string appears after s:64: and is in quotations
Example s:64:"stringhere"
I'm currently trying to get them as so but any regex that I've tried has failed,
Pattern p = Pattern.compile("I don't know what to put as the regex");
Matcher m = p.matcher(data);
So with that said, all I need is the regex that will return the two strings in the matcher so that m.group(1) is my first string and m.group(2) is my second string.

Try this regex:-
s:64:\"(.*?)\"
Code:
Pattern pattern = Pattern.compile("s:64:\"(.*?)\"");
Matcher matcher = pattern.matcher(YourStringVar);
// Check all occurance
int count = 0;
while (matcher.find() && count++ < 2) {
System.out.println("Group : " + matcher.group(1));
}
Here group(1) returns the each match.
OUTPUT:
Group : First Match
Group : Second Match
Refer LIVE DEMO

String data = "s:64:\"first string\" random stuff here s:64:\"second string\"";
Pattern p = Pattern.compile("s:64:\"([^\"]*)\".*s:64:\"([^\"]*)\"");
Matcher m = p.matcher(data);
if (m.find()) {
System.out.println("First string: '" + m.group(1) + "'");
System.out.println("Second string: '" + m.group(2) + "'");
}
prints:
First string: 'first string'
Second string: 'second string'

Regex you need should be compile("s:64:\"(.*?)\".*s:64:\"(.*?)\"")