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Is there a regex where if first expression is valid then check for next [duplicate]

I have several strings in the rough form:
[some text] [some number] [some more text]
I want to extract the text in [some number] using the Java Regex classes.
I know roughly what regular expression I want to use (though all suggestions are welcome). What I'm really interested in are the Java calls to take the regex string and use it on the source data to produce the value of [some number].
EDIT: I should add that I'm only interested in a single [some number] (basically, the first instance). The source strings are short and I'm not going to be looking for multiple occurrences of [some number].

Full example:
private static final Pattern p = Pattern.compile("^([a-zA-Z]+)([0-9]+)(.*)");
public static void main(String[] args) {
// create matcher for pattern p and given string
Matcher m = p.matcher("Testing123Testing");
// if an occurrence if a pattern was found in a given string...
if (m.find()) {
// ...then you can use group() methods.
System.out.println(m.group(0)); // whole matched expression
System.out.println(m.group(1)); // first expression from round brackets (Testing)
System.out.println(m.group(2)); // second one (123)
System.out.println(m.group(3)); // third one (Testing)
}
}
Since you're looking for the first number, you can use such regexp:
^\D+(\d+).*
and m.group(1) will return you the first number. Note that signed numbers can contain a minus sign:
^\D+(-?\d+).*

import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Regex1 {
public static void main(String[]args) {
Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher("hello1234goodboy789very2345");
while(m.find()) {
System.out.println(m.group());
}
}
}
Output:
1234
789
2345

Allain basically has the java code, so you can use that. However, his expression only matches if your numbers are only preceded by a stream of word characters.
"(\\d+)"
should be able to find the first string of digits. You don't need to specify what's before it, if you're sure that it's going to be the first string of digits. Likewise, there is no use to specify what's after it, unless you want that. If you just want the number, and are sure that it will be the first string of one or more digits then that's all you need.
If you expect it to be offset by spaces, it will make it even more distinct to specify
"\\s+(\\d+)\\s+"
might be better.
If you need all three parts, this will do:
"(\\D+)(\\d+)(.*)"
EDIT The Expressions given by Allain and Jack suggest that you need to specify some subset of non-digits in order to capture digits. If you tell the regex engine you're looking for \d then it's going to ignore everything before the digits. If J or A's expression fits your pattern, then the whole match equals the input string. And there's no reason to specify it. It probably slows a clean match down, if it isn't totally ignored.

In addition to Pattern, the Java String class also has several methods that can work with regular expressions, in your case the code will be:
"ab123abc".replaceFirst("\\D*(\\d*).*", "$1")
where \\D is a non-digit character.

In Java 1.4 and up:
String input = "...";
Matcher matcher = Pattern.compile("[^0-9]+([0-9]+)[^0-9]+").matcher(input);
if (matcher.find()) {
String someNumberStr = matcher.group(1);
// if you need this to be an int:
int someNumberInt = Integer.parseInt(someNumberStr);
}

This function collect all matching sequences from string. In this example it takes all email addresses from string.
static final String EMAIL_PATTERN = "[_A-Za-z0-9-\\+]+(\\.[_A-Za-z0-9-]+)*#"
+ "[A-Za-z0-9-]+(\\.[A-Za-z0-9]+)*(\\.[A-Za-z]{2,})";
public List<String> getAllEmails(String message) {
List<String> result = null;
Matcher matcher = Pattern.compile(EMAIL_PATTERN).matcher(message);
if (matcher.find()) {
result = new ArrayList<String>();
result.add(matcher.group());
while (matcher.find()) {
result.add(matcher.group());
}
}
return result;
}
For message = "adf#gmail.com, <another#osiem.osiem>>>> lalala#aaa.pl" it will create List of 3 elements.

Try doing something like this:
Pattern p = Pattern.compile("^.+(\\d+).+");
Matcher m = p.matcher("Testing123Testing");
if (m.find()) {
System.out.println(m.group(1));
}

Simple Solution
// Regexplanation:
// ^ beginning of line
// \\D+ 1+ non-digit characters
// (\\d+) 1+ digit characters in a capture group
// .* 0+ any character
String regexStr = "^\\D+(\\d+).*";
// Compile the regex String into a Pattern
Pattern p = Pattern.compile(regexStr);
// Create a matcher with the input String
Matcher m = p.matcher(inputStr);
// If we find a match
if (m.find()) {
// Get the String from the first capture group
String someDigits = m.group(1);
// ...do something with someDigits
}
Solution in a Util Class
public class MyUtil {
private static Pattern pattern = Pattern.compile("^\\D+(\\d+).*");
private static Matcher matcher = pattern.matcher("");
// Assumptions: inputStr is a non-null String
public static String extractFirstNumber(String inputStr){
// Reset the matcher with a new input String
matcher.reset(inputStr);
// Check if there's a match
if(matcher.find()){
// Return the number (in the first capture group)
return matcher.group(1);
}else{
// Return some default value, if there is no match
return null;
}
}
}
...
// Use the util function and print out the result
String firstNum = MyUtil.extractFirstNumber("Testing4234Things");
System.out.println(firstNum);

Look you can do it using StringTokenizer
String str = "as:"+123+"as:"+234+"as:"+345;
StringTokenizer st = new StringTokenizer(str,"as:");
while(st.hasMoreTokens())
{
String k = st.nextToken(); // you will get first numeric data i.e 123
int kk = Integer.parseInt(k);
System.out.println("k string token in integer " + kk);
String k1 = st.nextToken(); // you will get second numeric data i.e 234
int kk1 = Integer.parseInt(k1);
System.out.println("new string k1 token in integer :" + kk1);
String k2 = st.nextToken(); // you will get third numeric data i.e 345
int kk2 = Integer.parseInt(k2);
System.out.println("k2 string token is in integer : " + kk2);
}
Since we are taking these numeric data into three different variables we can use this data anywhere in the code (for further use)

How about [^\\d]*([0-9]+[\\s]*[.,]{0,1}[\\s]*[0-9]*).* I think it would take care of numbers with fractional part.
I included white spaces and included , as possible separator.
I'm trying to get the numbers out of a string including floats and taking into account that the user might make a mistake and include white spaces while typing the number.

Sometimes you can use simple .split("REGEXP") method available in java.lang.String. For example:
String input = "first,second,third";
//To retrieve 'first'
input.split(",")[0]
//second
input.split(",")[1]
//third
input.split(",")[2]

if you are reading from file then this can help you
try{
InputStream inputStream = (InputStream) mnpMainBean.getUploadedBulk().getInputStream();
BufferedReader br = new BufferedReader(new InputStreamReader(inputStream));
String line;
//Ref:03
while ((line = br.readLine()) != null) {
if (line.matches("[A-Z],\\d,(\\d*,){2}(\\s*\\d*\\|\\d*:)+")) {
String[] splitRecord = line.split(",");
//do something
}
else{
br.close();
//error
return;
}
}
br.close();
}
}
catch (IOException ioExpception){
logger.logDebug("Exception " + ioExpception.getStackTrace());
}

Pattern p = Pattern.compile("(\\D+)(\\d+)(.*)");
Matcher m = p.matcher("this is your number:1234 thank you");
if (m.find()) {
String someNumberStr = m.group(2);
int someNumberInt = Integer.parseInt(someNumberStr);
}

How to extract a number from a string in a particular format?

I have a String like this as shown below. From below string I need to extract number 123 and it can be at any position as shown below but there will be only one number in a string and it will always be in the same format _number_
text_data_123
text_data_123_abc_count
text_data_123_abc_pqr_count
text_tery_qwer_data_123
text_tery_qwer_data_123_count
text_tery_qwer_data_123_abc_pqr_count
Below is the code:
String value = "text_data_123_abc_count";
// this below code will not work as index 2 is not a number in some of the above example
int textId = Integer.parseInt(value.split("_")[2]);
What is the best way to do this?

With a little guava magic:
String value = "text_data_123_abc_count";
Integer id = Ints.tryParse(CharMatcher.inRange('0', '9').retainFrom(value)
see also CharMatcher doc

\\d+
this regex with find should do it for you.

Use Positive lookahead assertion.
Matcher m = Pattern.compile("(?<=_)\\d+(?=_)").matcher(s);
while(m.find())
{
System.out.println(m.group());
}

You can use replaceAll to remove all non-digits to leave only one number (since you say there will be only 1 number in the input string):
String s = "text_data_123_abc_count".replaceAll("[^0-9]", "");
See IDEONE demo
Instead of [^0-9] you can use \D (which also means non-digit):
String s = "text_data_123_abc_count".replaceAll("\\D", "");
Given current requirements and restrictions, the replaceAll solution seems the most convenient (no need to use Matcher directly).

u can get all parts from that string and compare with its UPPERCASE, if it is equal then u can parse it to a number and save:
public class Main {
public static void main(String[] args) {
String txt = "text_tery_qwer_data_123_abc_pqr_count";
String[] words = txt.split("_");
int num = 0;
for (String t : words) {
if(t == t.toUpperCase())
num = Integer.parseInt(t);
}
System.out.println(num);
}
}

Print out the last match of a regex

I have this code:
String responseData = "http://xxxxx-f.frehd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/.m3u8";
"http://xxxxx-f.frehd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/.m3u8";
String pattern = ^(https://.*\.54325)$;
Pattern pr = Pattern.compile(pattern);
Matcher math = pr.matcher(responseData);
if (math.find()) {
// print the url
}
else {
System.out.println("No Math");
}
I want to print out the last string that starts with http and ends with .m3u8. How do I do this? I'm stuck. All help is appreciated.
The problem I have now is that when I find a math and what to print out the string, I get everything from responseData.

In case you need to get some substring at the end that is preceded by similar substrings, you need to make sure the regex engine has already consumed as many characters before your required match as possible.
Also, you have a ^ in your pattern that means beginning of a string. Thus, it starts matching from the very beginning.
You can achieve what you want with just lastIndexOf and substring:
System.out.println(str.substring(str.lastIndexOf("http://")));
Or, if you need a regex, you'll need to use
String pattern = ".*(http://.*?\\.m3u8)$";
and use math.group(1) to print the value.
Sample code:
import java.util.regex.*;
public class HelloWorld{
public static void main(String []args){
String str = "http://xxxxx-f.akamaihd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/index_0_av.m3u8" +
"EXT-X-STREAM-INF:PROGRAM-ID=1,BANDWIDTH=2795000,RESOLUTION=1280x720,CODECS=avc1.64001f, mp4a.40.2" +
"http://xxxxx-f.akamaihd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/index_6_av.m3u8";
String rx = ".*(http://.*?\\.m3u8)$";
Pattern ptrn = Pattern.compile(rx);
Matcher m = ptrn.matcher(str);
while (m.find()) {
System.out.println(m.group(1));
}
}
}
Output:
http://xxxxx-f.akamaihd.net/i/world/open/20150426/1370235-005A/EPISOD-1370235-005A-016f1729028090bf_,892,144,252,360,540,1584,2700,.mp4.csmil/index_6_av.m3u8
Also tested on RegexPlanet

Replacing digits separated with commas using String.replace("","");

I have a string which looks like following:
Turns 13,000,000 years old
Now i want to convert the digits to words in English, I have a function ready for that however I am finding problems to detect the original numbers (13,000,000) in this case, because it is separated by commas.
Currently I am using the following regex to detect a number in a string:
stats = stats.replace((".*\\d.*"), (NumberToWords.start(Integer.valueOf(notification_data_greet))));
But the above seems not to work, any suggestions?

You need to extract the number using a RegEx wich allows for the commas. The most robust one I can think of right now is
\d{1,3}(,?\d{3})*
Wich matches any unsigned Integer both with correctly placed commas and without commas (and weird combinations thereof like 100,000000)
Then replace all , from the match by the empty String and you can parse as usual:
Pattern p = Pattern.compile("\\d{1,3}(,?\\d{3})*"); // You can store this as static final
Matcher m = p.matcher(input);
while (m.find()) { // Go through all matches
String num = m.group().replace(",", "");
int n = Integer.parseInt(num);
// Do stuff with the number n
}
Working example:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
public static void main(String[] args) throws InterruptedException {
String input = "1,300,000,000";
Pattern p = Pattern.compile("\\d{1,3}(,?\\d{3})*"); // You can store this as static final
Matcher m = p.matcher(input);
while (m.find()) { // Go through all matches
String num = m.group().replace(",", "");
System.out.println(num);
int n = Integer.parseInt(num);
System.out.println(n);
}
}
}
Gives output
1300000000
1300000000

Try this regex:
[0-9][0-9]?[0-9]?([0-9][0-9][0-9](,)?)*
This matches numbers that are seperated by a comma for each 1000. So it will match
10,000,000
but not
10,1,1,1

You can do it with the help of DecimalFormat instead of a regular expression
DecimalFormat format = (DecimalFormat) DecimalFormat.getInstance();
System.out.println(format.parse("10,000,000"));

Try the below regex to match the comma separted numbers,
\d{1,3}(,\d{3})+
Make the last part as optional to match also the numbers which aren't separated by commas,
\d{1,3}(,\d{3})*

Reusing the consumed characters in pattern matching in java?

Consider the following Pattern :-
aba
And the foll. source string :-
abababbbaba
01234567890 //Index Positions
Using Pattern and Matcher classes from java.util.regex package, finds this pattern only two times since regex does not consider already consumed characters.
What if I want to reuse a part of already consumed characters. That is, I want 3 matches here, one at position 0, one at 2 (which is ignored previously), and one at 8.
How do I do it??

I think you can use the indexOf() for something like that.
String str = "abababbbaba";
String substr = "aba";
int location = 0;
while ((location = str.indexOf(substr, location)) >= 0)
{
System.out.println(location);
location++;
}
Prints:
0, 2 and 8

You can use a look ahead for that. Now what you have is the first position in group(1) and the second match in group(2). Both making each String of length 3 in the sentence you are searching in.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Question8968432 {
public static void main(String args[]) {
final String needle = "aba";
final String sentence = "abababbbaba";
final Matcher m = Pattern.compile("(.)(?=(..))").matcher(sentence);
while (m.find()) {
final String match = m.group(1) + m.group(2);
final String hint = String.format("%s[%s]%s",
sentence.substring(0, m.start()), match,
sentence.substring(m.start() + match.length()));
if (match.equals(needle)) {
System.out.printf("Found %s starting at %d: %s\n",
match, m.start(), hint);
}
}
}
}
Output:
Found aba starting at 0: [aba]babbbaba
Found aba starting at 2: ab[aba]bbbaba
Found aba starting at 8: abababbb[aba]
You can skip the final String hint part, this is just to show you what it matches and where.

If you can change the regexp, then you can simply use something like:
a(?=ba)