We Keep Coding

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);
}

Regex to get value between two colon excluding the colons

I have a string like this:
something:POST:/some/path
Now I want to take the POST alone from the string. I did this by using this regex
:([a-zA-Z]+):
But this gives me a value along with colons. ie I get this:
:POST:
but I need this
POST
My code to match the same and replace it is as follows:
String ss = "something:POST:/some/path/";
Pattern pattern = Pattern.compile(":([a-zA-Z]+):");
Matcher matcher = pattern.matcher(ss);
if (matcher.find()) {
System.out.println(matcher.group());
ss = ss.replaceFirst(":([a-zA-Z]+):", "*");
}
System.out.println(ss);
EDIT:
I've decided to use the lookahead/lookbehind regex since I did not want to use replace with colons such as :*:. This is my final solution.
String s = "something:POST:/some/path/";
String regex = "(?<=:)[a-zA-Z]+(?=:)";
Matcher matcher = Pattern.compile(regex).matcher(s);
if (matcher.find()) {
s = s.replaceFirst(matcher.group(), "*");
System.out.println("replaced: " + s);
}
else {
System.out.println("not replaced: " + s);
}

There are two approaches:
Keep your Java code, and use lookahead/lookbehind (?<=:)[a-zA-Z]+(?=:), or
Change your Java code to replace the result with ":*:"
Note: You may want to define a String constant for your regex, since you use it in different calls.

As pointed out, the reqex captured group can be used to replace.
The following code did it:
String ss = "something:POST:/some/path/";
Pattern pattern = Pattern.compile(":([a-zA-Z]+):");
Matcher matcher = pattern.matcher(ss);
if (matcher.find()) {
ss = ss.replaceFirst(matcher.group(1), "*");
}
System.out.println(ss);

UPDATE
Looking at your update, you just need ReplaceFirst only:
String result = s.replaceFirst(":[a-zA-Z]+:", ":*:");
See the Java demo
When you use (?<=:)[a-zA-Z]+(?=:), the regex engine checks each location inside the string for a * before it, and once found, tries to match 1+ ASCII letters and then assert that there is a : after them. With :[A-Za-z]+:, the checking only starts after a regex engine found : character. Then, after matching :POST:, the replacement pattern replaces the whole match. It is totlally OK to hardcode colons in the replacement pattern since they are hardcoded in the regex pattern.
Original answer
You just need to access Group 1:
if (matcher.find()) {
System.out.println(matcher.group(1));
}
See Java demo
Your :([a-zA-Z]+): regex contains a capturing group (see (....) subpattern). These groups are numbered automatically: the first one has an index of 1, the second has the index of 2, etc.
To replace it, use Matcher#appendReplacement():
String s = "something:POST:/some/path/";
StringBuffer result = new StringBuffer();
Matcher m = Pattern.compile(":([a-zA-Z]+):").matcher(s);
while (m.find()) {
m.appendReplacement(result, ":*:");
}
m.appendTail(result);
System.out.println(result.toString());
See another demo

This is your solution:
regex = (:)([a-zA-Z]+)(:)
And code is:
String ss = "something:POST:/some/path/";
ss = ss.replaceFirst("(:)([a-zA-Z]+)(:)", "$1*$3");
ss now contains:
something:*:/some/path/
Which I believe is what you are looking for...

Extract every complete word that contains a certain substring

I'm trying to write a function that extracts each word from a sentence that contains a certain substring e.g. Looking for 'Po' in 'Porky Pork Chop' will return Porky Pork.
I've tested my regex on regexpal but the Java code doesn't seem to work. What am I doing wrong?
private static String foo()
{
String searchTerm = "Pizza";
String text = "Cheese Pizza";
String sPattern = "(?i)\b("+searchTerm+"(.+?)?)\b";
Pattern pattern = Pattern.compile ( sPattern );
Matcher matcher = pattern.matcher ( text );
if(matcher.find ())
{
String result = "-";
for(int i=0;i < matcher.groupCount ();i++)
{
result+= matcher.group ( i ) + " ";
}
return result.trim ();
}else
{
System.out.println("No Luck");
}
}

In Java to pass \b word boundaries to regex engine you need to write it as \\b. \b represents backspace in String object.
Judging by your example you want to return all words that contains your substring. To do this don't use for(int i=0;i < matcher.groupCount ();i++) but while(matcher.find()) since group count will iterate over all groups in single match, not over all matches.
In case your string can contain some special characters you probably should use Pattern.quote(searchTerm)
In your code you are trying to find "Pizza" in "Cheese Pizza" so I assume that you also want to find strings that same as searched substring. Although your regex will work fine for it, you can change your last part (.+?)?) to \\w* and also add \\w* at start if substring should also be matched in the middle of word (not only at start).
So your code can look like
private static String foo() {
String searchTerm = "Pizza";
String text = "Cheese Pizza, Other Pizzas";
String sPattern = "(?i)\\b\\w*" + Pattern.quote(searchTerm) + "\\w*\\b";
StringBuilder result = new StringBuilder("-").append(searchTerm).append(": ");
Pattern pattern = Pattern.compile(sPattern);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
result.append(matcher.group()).append(' ');
}
return result.toString().trim();
}

While the regex approach is certainly a valid method, I find it easier to think through when you split the words up by whitespace. This can be done with String's split method.
public List<String> doIt(final String inputString, final String term) {
final List<String> output = new ArrayList<String>();
final String[] parts = input.split("\\s+");
for(final String part : parts) {
if(part.indexOf(term) > 0) {
output.add(part);
}
}
return output;
}
Of course it is worth nothing that doing this will effectively be doing two passes through your input String. The first pass to find the characters that are whitespace to split on, and the second pass looking through each split word for your substring.
If one pass is necessary though, the regex path is better.

I find nicholas.hauschild's answer to be the best.
However if you really wanted to use regex, you could do it as such:
String searchTerm = "Pizza";
String text = "Cheese Pizza";
Pattern pattern = Pattern.compile("\\b" + Pattern.quote(searchTerm)
+ "\\b", Pattern.CASE_INSENSITIVE);
Matcher matcher = pattern.matcher(text);
while (matcher.find()) {
System.out.println(matcher.group());
}
Output:
Pizza

The pattern should have been
String sPattern = "(?i)\\b("+searchTerm+"(?:.+?)?)\\b";
You want to capture the whole (pizza)string.?: ensures you don't capture a part of the string twice.

Try this pattern:
String searchTerm = "Po";
String text = "Porky Pork Chop oPod zzz llPo";
Pattern p = Pattern.compile("\\p{Alpha}+" + substring + "|\\p{Alpha}+" + substring + "\\p{Alpha}+|" + substring + "\\p{Alpha}+");
Matcher m = p.matcher(myString);
while(m.find()) {
System.out.println(">> " + m.group());
}

Ok, I give you a pattern in raw style (not java style, you must double escape yourself):
(?i)\b[a-z]*po[a-z]*\b
And that's all.

In Java how do you replace all instances of a character except the first one?

In Java trying to find a regular expression that will match all instances of a specific character (:) except the first instance, want to replace all instances except first with nothing.
I can do this,
Pattern p = Pattern.compile(":");
Matcher m = p.matcher(input);
String output = m.replaceAll("");
and there is also m.replaceFirst() but I want to replace everything but first.

Naive approach:
String[] parts = str.split(":", 2);
str = parts[0] + ":" + parts[1].replaceAll(":", "");

For regex replace use match pattern \G((?!^).*?|[^:]*:.*?): and as replacement use first group $1
See and test the regex code in Perl here.

public static void main(String[] args) {
String name ="1_2_3_4_5";
int index = name.indexOf("_");
String name1 = name.substring(index+1);
name1 = name1.replace("_", "#");
System.out.println(name.substring(0,index+1)+ name1);
}

You can use reg ex
String str1 = "A:B:C:D:E:F:G:H:I:J:K:L:M";
str1= str1.replaceAll("([:|_].*?):", "$1_");
str1= str1.replaceAll("([:|_].*?):", "$1_");
Here I cant modify the regex to have output in first replace itself. Actually first replaceAll do replace ':' with '_' in alternate positions.

if (matcher.find()) {
String start = originalString.substring(0, matcher.end());
matcher.reset(originalString.substring(matcher.end(), originalString.length()));
replacedString = start + matcher.replaceAll("");
}

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)