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I Want to extract a certain number from string in java but not able to do so

I have a string like String a = "I have 102 string but 123453 is best"
So, In the above string i want to extract only 123453
This string keep getting changes but i always want to get that second number from this string.
What is the best possible way to do this?

Here is a complete example. First define your pattern:
Pattern p = Pattern.compile("[0-9]+");
Then define the string you want to test
String test = "I have 102 string but 123453 is best";
Create a matcher for that string
Matcher matcher = p.matcher(test);
Find once
matcher.find(); // find once
Find again, and if found, get the string
if (matcher.find()) { // find twice
System.out.println(matcher.group());
}
For a complete, minimal and repeatable example, consider defining a method and defining the pattern as a constant:
static final Pattern P = Pattern.compile("[0-9]+");
static String findSecondNumber(String test) {
Matcher matcher = P.matcher(test);
matcher.find(); // find once
if (matcher.find()) { // find twice
return matcher.group();
}
return null; // or alternatively return an empty string or a default value
}
Where matcher.group() returns the full match found; see also the documentation here.

It can be done in a single find() with capturing groups. Here is how:
String a = "I have 102 string but 123453 is best"
Pattern p = Pattern.compile("^.*\d+.*(\d*).*$");
Matcher m = p.matcher(a);
if (m.find()) {
System.out.println(m.group(1)); // this will print 123453
}

JAVA Get text from String

Hi I get this String from server :
id_not="autoincrement"; id_obj="-"; id_tr="-"; id_pgo="-"; typ_not=""; tresc="Nie wystawił"; datetime="-"; lon="-"; lat="-";
I need to create a new String e.x String word and send a value which I get from String tresc="Nie wystawił"

Like #Jan suggest in comment you can use regex for example :
String str = "id_not=\"autoincrement\"; id_obj=\"-\"; id_tr=\"-\"; id_pgo=\"-\"; typ_not=\"\"; tresc=\"Nie wystawił\"; datetime=\"-\"; lon=\"-\"; lat=\"-\";";
Pattern p = Pattern.compile("tresc(.*?);");
Matcher m = p.matcher(str);
if (m.find()) {
System.out.println(m.group());
}
Output
tresc="Nie wystawił";
If you want to get only the value of tresc you can use :
Pattern p = Pattern.compile("tresc=\"(.*?)\";");
Matcher m = p.matcher(str);
if (m.find()) {
System.out.println(m.group(1));
}
Output
Nie wystawił

Something along the lines of
Pattern p = Pattern.compile("tresc=\"([^\"]+)\");
Matcher m = p.matcher(stringFromServer);
if(m.find()) {
String whatYouWereLookingfor = m.group(1);
}
should to the trick. JSON parsing might be much better in the long run if you need additional values

Your question is unclear but i think you get a string from server and from that string you want the string/value for tresc. You can first search for tresc in the string you get. like:
serverString.substring(serverString.indexOf("tresc") + x , serverString.length());
Here replace x with 'how much further you want to pick characters.
Read on substring and delimiters
As values are separated by semicolon so annother solution could be:
int delimiter = serverstring.indexOf(";");
//in string thus giving you the index of where it is in the string
// Now delimiter can be -1, if lets say the string had no ";" at all in it i.e. no ";" is not found.
//check and account for it.
if (delimiter != -1)
String subString= serverstring.substring(5 , iend);
Here 5 means tresc is on number five in string, so it will five you tresc part.
You can then use it anyway you want.

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

Extracting Number from URL in Java via Regex

Take URL http://www.abc.com/alpha/beta/33445566778899/gamma/delta
i need to return the number 33445566778899 (with forward slashes removed, number is of variable length but between 10 & 20 digits)
Simple enough (or so i thought) except everything I've tried doesn't seem to work but why?
Pattern pattern = Pattern.compile("\\/([0-9])\\d{10,20}\\/");
Matcher matcher = pattern.matcher(fullUrl);
if (matcher.find()) {
return matcher.group(1);
}

Try this one-liner:
String number = url.replaceAll(".*/(\\d{10,20})/.*", "$1");

This regex works -
"\\/(\\d{10,20})\\/"
Testing it-
String fullUrl = "http://www.abc.com/alpha/beta/33445566778899/gamma/delta";
Pattern pattern = Pattern.compile("\\/(\\d{10,20})\\/");
Matcher matcher = pattern.matcher(fullUrl);
if (matcher.find()) {
System.out.println(matcher.group(1));
}
OUTPUT - 33445566778899

Try,
String url = "http://www.abc.com/alpha/beta/33445566778899/gamma/delta";
String digitStr = null;
for(String str : url.split("/")){
System.out.println(str);
if(str.matches("[0-9]{10,20}")){
digitStr = str;
break;
}
}
System.out.println(digitStr);
Output:
33445566778899

Instead of saying it "doesn't seem to work", you should have given use what it was returning. Testing it confirmed what I thought: your code would return 3 for this input.
This is simply because your regexp as written will capture a digit following a / and followed by 10 to 20 digits themselves followed by a /.
The regex you want is "/(\\d{10,20})/" (you don't need to escape the /). Below you'll find the code I tested this with.
public static void main(String[] args) {
String src = "http://www.abc.com/alpha/beta/33445566778899/gamma/delta";
Pattern pattern = Pattern.compile("/(\\d{10,20})/");
Matcher matcher = pattern.matcher(src);
if (matcher.find()) {
System.out.println(matcher.group(1));
}
}

rules for use reg exp java

i have pattern:
host=([a-z0-9./:]*)
it's find for me host address. And i have content
host=http//:sdf3452.domain.com/
And my code is:
Matcher m;
Pattern hostP = Pattern.compile("host=([a-z0-9./:]*)");
m=hostP.matcher(content);//string 1
String match = m.group();//string 2
Log.i("host", ""+hostP.matcher(content).find());
if i delete string 1 and 2 i see true in logcat. If left as is I got exception nothing found.
I've tried all kinds of pattern. Through debug looked m variable, finds no match. Please teach me use reg exp!

Before you group() a match, you need to invoke find().
Try it like this:
Pattern hostP = Pattern.compile("host=([a-z0-9./:]*)");
Matcher m = hostP.matcher(content);
if(m.find()) {
String match = m.group();
// ...
}
EDIT
and a little demo that shows what each match-group contains:
Pattern p = Pattern.compile("host=([a-z0-9./:]*)");
Matcher m = p.matcher("host=http://sdf3452.domain.com/");
if (m.find()) {
for(int i = 0; i <= m.groupCount(); i++) {
System.out.printf("m.group(%d) = '%s'\n", i, m.group(i));
}
}
which will print:
m.group(0) = 'host=http://sdf3452.domain.com/'
m.group(1) = 'http://sdf3452.domain.com/'
As you can see, group(0), which is the same as group(), contains what the entire pattern matches.
But realize that a URL can contain much more than what your defined in [a-z0-9./:]*!

String content = "host=http://sdf3452.domain.com/";
Matcher mm;
Pattern hostP = Pattern.compile("host=([a-z0-9./:]*)");
mm=hostP.matcher(content);
String match = "";
if (mm.find()){//use m.find() first
match = mm.group(1);//1 is order number of brackets
}