I have a string which contains multiple url starting from http and https I need to fetch all those url and put into a list.
I have tried below code.
List<String> httpLinksList = new ArrayList<>();
String hyperlinkRegex = "((http:\/\/|https:\/\/)?(([a-zA-Z0-9-]){2,}\.){1,4}([a-zA-Z]){2,6}(\/([a-zA-Z-_\/\.0-9#:?=&;,]*)?)?)";
String synopsis = "This is http://stackoverflow.com/questions and https://test.com/method?param=wasd The code below catches all urls in text and returns urls in list";
Pattern pattern = Pattern.compile(hyperlinkRegex);
Matcher matcher = pattern.matcher(synopsis);
while(matcher.find()){
System.out.println(matcher.find()+" "+matcher.group(1)+" "+matcher.groupCount()+" "+matcher.group(2));
httpLinksList.add(matcher.group());
}
System.out.println(httpLinksList);
I need below result
[http://stackoverflow.com/questions,
https://test.com/method?param=wasd]
But getting below output
[https://test.com/method?param=wasd]
This regex will match all the valid urls, including FTP and other
String urlRegex = "((https?|ftp|gopher|telnet|file):((//)|(\\\\))+[\\w\\d:##%/;$()~_?\\+-=\\\\\\.&]*)";
import java.util.ArrayList;
import java.util.List;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class xmlValue {
public static void main(String[] args) {
String text = "This is http://stackoverflow.com/questions and https://test.com/method?param=wasd The code below catches all urls in text and returns urls in list";
System.out.println(extractUrls(text));
}
public static List<String> extractUrls(String text)
{
List<String> containedUrls = new ArrayList<String>();
String urlRegex = "((https?|ftp|gopher|telnet|file):((//)|(\\\\))+[\\w\\d:##%/;$()~_?\\+-=\\\\\\.&]*)";
Pattern pattern = Pattern.compile(urlRegex, Pattern.CASE_INSENSITIVE);
Matcher urlMatcher = pattern.matcher(text);
while (urlMatcher.find())
{
containedUrls.add(text.substring(urlMatcher.start(0),
urlMatcher.end(0)));
}
return containedUrls;
}
}
Output:
[http://stackoverflow.com/questions,
https://test.com/method?param=wasd]
credits #BullyWiiPlaza
So I know this is not exactly what you asked since you are specifically looking for regex, but I thought this would fun to try out with an indexOf variant. I will leave it here as an alternative to the regex someone comes up with:
public static void main(String[] args){
String synopsis = "This is http://stackoverflow.com/questions and https://test.com/method?param=wasd The code below catches all urls in text and returns urls in list";
ArrayList<String> list = splitUrl(synopsis);
for (String s : list) {
System.out.println(s);
}
}
public static ArrayList<String> splitUrl(String s)
{
ArrayList<String> list = new ArrayList<>();
int spaceIndex = 0;
while (true) {
int httpIndex = s.indexOf("http", spaceIndex);
if (httpIndex < 0) {
break;
}
spaceIndex = s.indexOf(" ", httpIndex);
if (spaceIndex < 0) {
list.add(s.substring(httpIndex));
break;
}
else {
list.add(s.substring(httpIndex, spaceIndex));
}
}
return list;
}
All the logic is contained in the splitUrl(String s) method, it takes in a String as a parameter and outputs the ArrayList<String> of all the split urls.
It first searches for the index of any http and then the first space that occurs after the url and substrings the difference. It then uses the space it found as the second parameter in indexOf(String, int) to start searching the String beginning after the http that was already found so it does not repeat the same ones.
Additionally a case had to be made when the http is the final part of the String as there is no space afterward. This is done when the indexOf the space returns negative, I use substring(int) instead of substring(int, int) which will take the current location and substring the rest of the String.
The loop ends when either indexOf returns with a negative, though if the space returns negative it does that final substring operation before the break.
Output:
http://stackoverflow.com/questions
https://test.com/method?param=wasd
Note: As someone mentioned in the comments too, this implementation will work with non-Latin characters such as Hiragana too, which could be an advantage over regex.
I will be getting the string as app1(down) and app2(up)
the words in the brackets indicate status of the app, they may be up or down depending,
now i need to use a regex to get the status of the apps like a comma seperated string
ex:ill get app1(UP) and app2(DOWN)
required result UP,DOWN
It's easy using RegEx like this:
\\((.*?)\\)
String x = "app1(UP) and app2(DOWN)";
Matcher m = Pattern.compile("\\((.*?)\\)").matcher(x);
String tmp = "";
while(m.find()) {
tmp+=(m.group(1))+",";
}
System.out.println(tmp);
Output:
UP,DOWN,
Java 8: using StringJoiner
String x = "app1(UP) and app2(DOWN)";
Matcher m = Pattern.compile("\\((.*?)\\)").matcher(x);
StringJoiner sj = new StringJoiner(",");
while(m.find()) {
sj.add((m.group(1)));
}
System.out.print(sj.toString());
Output:
UP,DOWN
(Last , is removed)
import java.util.ArrayList;
import java.util.List;
import java.util.regex.*;
public class ValidateDemo
{
public static void main(String[] args)
{
String input = "ill get app1(UP) and app2(DOWN)";
Pattern p = Pattern.compile("app[0-9]+\\(([A-Z]+)\\)");
Matcher m = p.matcher(input);
List<String> found = new ArrayList<String>();
while (m.find())
{
found.add(m.group(1));
}
System.out.println(found.toString());
}
}
my first java script, have mercy
Consider this code:
private static final Pattern RX_MATCH_APP_STATUS = Pattern.compile("\\s*(?<name>[^(\\s]+)\\((?<status>[^(\\s]+)\\)");
final String input = "app1(UP) or app2(down) let's have also app-3(DOWN)";
final Matcher m = RX_MATCH_APP_STATUS.matcher(input);
while (m.find()) {
final String name = m.group("name");
final String status = m.group("status");
System.out.printf("%s:%s\n", name, status);
}
This plucks from input line as many app status entries, as they really are there, and put each app name and its status into proper variable. It's then up to you, how you want to handle them (print or whatever).
Plus, this gives you advantage if there will come other states than UP and DOWN (like UNKNOWN) and this will still work.
Minus, if there are sentences in brackets prefixed with some name, that is actually not a name of an app and the content of the brackets is not an app state.
Use this as regex and test it on http://regexr.com/
[UP]|[DOWN]
I want to split the string "004-034556" into two strings by the delimiter "-":
part1 = "004";
part2 = "034556";
That means the first string will contain the characters before '-', and the second string will contain the characters after '-'.
I also want to check if the string has '-' in it.
Use the appropriately named method String#split().
String string = "004-034556";
String[] parts = string.split("-");
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556
Note that split's argument is assumed to be a regular expression, so remember to escape special characters if necessary.
there are 12 characters with special meanings: the backslash \, the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), and the opening square bracket [, the opening curly brace {, These special characters are often called "metacharacters".
For instance, to split on a period/dot . (which means "any character" in regex), use either backslash \ to escape the individual special character like so split("\\."), or use character class [] to represent literal character(s) like so split("[.]"), or use Pattern#quote() to escape the entire string like so split(Pattern.quote(".")).
String[] parts = string.split(Pattern.quote(".")); // Split on the exact string.
To test beforehand if the string contains certain character(s), just use String#contains().
if (string.contains("-")) {
// Split it.
} else {
throw new IllegalArgumentException("String " + string + " does not contain -");
}
Note, this does not take a regular expression. For that, use String#matches() instead.
If you'd like to retain the split character in the resulting parts, then make use of positive lookaround. In case you want to have the split character to end up in left hand side, use positive lookbehind by prefixing ?<= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?<=-)");
String part1 = parts[0]; // 004-
String part2 = parts[1]; // 034556
In case you want to have the split character to end up in right hand side, use positive lookahead by prefixing ?= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?=-)");
String part1 = parts[0]; // 004
String part2 = parts[1]; // -034556
If you'd like to limit the number of resulting parts, then you can supply the desired number as 2nd argument of split() method.
String string = "004-034556-42";
String[] parts = string.split("-", 2);
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556-42
An alternative to processing the string directly would be to use a regular expression with capturing groups. This has the advantage that it makes it straightforward to imply more sophisticated constraints on the input. For example, the following splits the string into two parts, and ensures that both consist only of digits:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class SplitExample
{
private static Pattern twopart = Pattern.compile("(\\d+)-(\\d+)");
public static void checkString(String s)
{
Matcher m = twopart.matcher(s);
if (m.matches()) {
System.out.println(s + " matches; first part is " + m.group(1) +
", second part is " + m.group(2) + ".");
} else {
System.out.println(s + " does not match.");
}
}
public static void main(String[] args) {
checkString("123-4567");
checkString("foo-bar");
checkString("123-");
checkString("-4567");
checkString("123-4567-890");
}
}
As the pattern is fixed in this instance, it can be compiled in advance and stored as a static member (initialised at class load time in the example). The regular expression is:
(\d+)-(\d+)
The parentheses denote the capturing groups; the string that matched that part of the regexp can be accessed by the Match.group() method, as shown. The \d matches and single decimal digit, and the + means "match one or more of the previous expression). The - has no special meaning, so just matches that character in the input. Note that you need to double-escape the backslashes when writing this as a Java string. Some other examples:
([A-Z]+)-([A-Z]+) // Each part consists of only capital letters
([^-]+)-([^-]+) // Each part consists of characters other than -
([A-Z]{2})-(\d+) // The first part is exactly two capital letters,
// the second consists of digits
Use:
String[] result = yourString.split("-");
if (result.length != 2)
throw new IllegalArgumentException("String not in correct format");
This will split your string into two parts. The first element in the array will be the part containing the stuff before the -, and the second element in the array will contain the part of your string after the -.
If the array length is not 2, then the string was not in the format: string-string.
Check out the split() method in the String class.
This:
String[] out = string.split("-");
should do the thing you want. The string class has many method to operate with a string.
// This leaves the regexes issue out of question
// But we must remember that each character in the Delimiter String is treated
// like a single delimiter
public static String[] SplitUsingTokenizer(String subject, String delimiters) {
StringTokenizer strTkn = new StringTokenizer(subject, delimiters);
ArrayList<String> arrLis = new ArrayList<String>(subject.length());
while(strTkn.hasMoreTokens())
arrLis.add(strTkn.nextToken());
return arrLis.toArray(new String[0]);
}
With Java 8:
List<String> stringList = Pattern.compile("-")
.splitAsStream("004-034556")
.collect(Collectors.toList());
stringList.forEach(s -> System.out.println(s));
Use org.apache.commons.lang.StringUtils' split method which can split strings based on the character or string you want to split.
Method signature:
public static String[] split(String str, char separatorChar);
In your case, you want to split a string when there is a "-".
You can simply do as follows:
String str = "004-034556";
String split[] = StringUtils.split(str,"-");
Output:
004
034556
Assume that if - does not exists in your string, it returns the given string, and you will not get any exception.
The requirements left room for interpretation. I recommend writing a method,
public final static String[] mySplit(final String s)
which encapsulate this function. Of course you can use String.split(..) as mentioned in the other answers for the implementation.
You should write some unit-tests for input strings and the desired results and behaviour.
Good test candidates should include:
- "0022-3333"
- "-"
- "5555-"
- "-333"
- "3344-"
- "--"
- ""
- "553535"
- "333-333-33"
- "222--222"
- "222--"
- "--4555"
With defining the according test results, you can specify the behaviour.
For example, if "-333" should return in [,333] or if it is an error.
Can "333-333-33" be separated in [333,333-33] or [333-333,33] or is it an error? And so on.
To summarize: there are at least five ways to split a string in Java:
String.split():
String[] parts ="10,20".split(",");
Pattern.compile(regexp).splitAsStream(input):
List<String> strings = Pattern.compile("\\|")
.splitAsStream("010|020202")
.collect(Collectors.toList());
StringTokenizer (legacy class):
StringTokenizer strings = new StringTokenizer("Welcome to EXPLAINJAVA.COM!", ".");
while(strings.hasMoreTokens()){
String substring = strings.nextToken();
System.out.println(substring);
}
Google Guava Splitter:
Iterable<String> result = Splitter.on(",").split("1,2,3,4");
Apache Commons StringUtils:
String[] strings = StringUtils.split("1,2,3,4", ",");
So you can choose the best option for you depending on what you need, e.g. return type (array, list, or iterable).
Here is a big overview of these methods and the most common examples (how to split by dot, slash, question mark, etc.)
You can try like this also
String concatenated_String="hi^Hello";
String split_string_array[]=concatenated_String.split("\\^");
Assuming, that
you don't really need regular expressions for your split
you happen to already use apache commons lang in your app
The easiest way is to use StringUtils#split(java.lang.String, char). That's more convenient than the one provided by Java out of the box if you don't need regular expressions. Like its manual says, it works like this:
A null input String returns null.
StringUtils.split(null, *) = null
StringUtils.split("", *) = []
StringUtils.split("a.b.c", '.') = ["a", "b", "c"]
StringUtils.split("a..b.c", '.') = ["a", "b", "c"]
StringUtils.split("a:b:c", '.') = ["a:b:c"]
StringUtils.split("a b c", ' ') = ["a", "b", "c"]
I would recommend using commong-lang, since usually it contains a lot of stuff that's usable. However, if you don't need it for anything else than doing a split, then implementing yourself or escaping the regex is a better option.
For simple use cases String.split() should do the job. If you use guava, there is also a Splitter class which allows chaining of different string operations and supports CharMatcher:
Splitter.on('-')
.trimResults()
.omitEmptyStrings()
.split(string);
The fastest way, which also consumes the least resource could be:
String s = "abc-def";
int p = s.indexOf('-');
if (p >= 0) {
String left = s.substring(0, p);
String right = s.substring(p + 1);
} else {
// s does not contain '-'
}
String Split with multiple characters using Regex
public class StringSplitTest {
public static void main(String args[]) {
String s = " ;String; String; String; String, String; String;;String;String; String; String; ;String;String;String;String";
//String[] strs = s.split("[,\\s\\;]");
String[] strs = s.split("[,\\;]");
System.out.println("Substrings length:"+strs.length);
for (int i=0; i < strs.length; i++) {
System.out.println("Str["+i+"]:"+strs[i]);
}
}
}
Output:
Substrings length:17
Str[0]:
Str[1]:String
Str[2]: String
Str[3]: String
Str[4]: String
Str[5]: String
Str[6]: String
Str[7]:
Str[8]:String
Str[9]:String
Str[10]: String
Str[11]: String
Str[12]:
Str[13]:String
Str[14]:String
Str[15]:String
Str[16]:String
But do not expect the same output across all JDK versions. I have seen one bug which exists in some JDK versions where the first null string has been ignored. This bug is not present in the latest JDK version, but it exists in some versions between JDK 1.7 late versions and 1.8 early versions.
There are only two methods you really need to consider.
Use String.split for a one-character delimiter or you don't care about performance
If performance is not an issue, or if the delimiter is a single character that is not a regular expression special character (i.e., not one of .$|()[{^?*+\) then you can use String.split.
String[] results = input.split(",");
The split method has an optimization to avoid using a regular expression if the delimeter is a single character and not in the above list. Otherwise, it has to compile a regular expression, and this is not ideal.
Use Pattern.split and precompile the pattern if using a complex delimiter and you care about performance.
If performance is an issue, and your delimiter is not one of the above, you should pre-compile a regular expression pattern which you can then reuse.
// Save this somewhere
Pattern pattern = Pattern.compile("[,;:]");
/// ... later
String[] results = pattern.split(input);
This last option still creates a new Matcher object. You can also cache this object and reset it for each input for maximum performance, but that is somewhat more complicated and not thread-safe.
You can split a string by a line break by using the following statement:
String textStr[] = yourString.split("\\r?\\n");
You can split a string by a hyphen/character by using the following statement:
String textStr[] = yourString.split("-");
public class SplitTest {
public static String[] split(String text, String delimiter) {
java.util.List<String> parts = new java.util.ArrayList<String>();
text += delimiter;
for (int i = text.indexOf(delimiter), j=0; i != -1;) {
String temp = text.substring(j,i);
if(temp.trim().length() != 0) {
parts.add(temp);
}
j = i + delimiter.length();
i = text.indexOf(delimiter,j);
}
return parts.toArray(new String[0]);
}
public static void main(String[] args) {
String str = "004-034556";
String delimiter = "-";
String result[] = split(str, delimiter);
for(String s:result)
System.out.println(s);
}
}
Please don't use StringTokenizer class as it is a legacy class that is retained for compatibility reasons, and its use is discouraged in new code. And we can make use of the split method as suggested by others as well.
String[] sampleTokens = "004-034556".split("-");
System.out.println(Arrays.toString(sampleTokens));
And as expected it will print:
[004, 034556]
In this answer I also want to point out one change that has taken place for split method in Java 8. The String#split() method makes use of Pattern.split, and now it will remove empty strings at the start of the result array. Notice this change in documentation for Java 8:
When there is a positive-width match at the beginning of the input
sequence then an empty leading substring is included at the beginning
of the resulting array. A zero-width match at the beginning however
never produces such empty leading substring.
It means for the following example:
String[] sampleTokensAgain = "004".split("");
System.out.println(Arrays.toString(sampleTokensAgain));
we will get three strings: [0, 0, 4] and not four as was the case in Java 7 and before. Also check this similar question.
One way to do this is to run through the String in a for-each loop and use the required split character.
public class StringSplitTest {
public static void main(String[] arg){
String str = "004-034556";
String split[] = str.split("-");
System.out.println("The split parts of the String are");
for(String s:split)
System.out.println(s);
}
}
Output:
The split parts of the String are:
004
034556
import java.io.*;
public class BreakString {
public static void main(String args[]) {
String string = "004-034556-1234-2341";
String[] parts = string.split("-");
for(int i=0;i<parts.length;i++) ｛
System.out.println(parts[i]);
}
}
}
You can use Split():
import java.io.*;
public class Splitting
{
public static void main(String args[])
{
String Str = new String("004-034556");
String[] SplittoArray = Str.split("-");
String string1 = SplittoArray[0];
String string2 = SplittoArray[1];
}
}
Else, you can use StringTokenizer:
import java.util.*;
public class Splitting
{
public static void main(String[] args)
{
StringTokenizer Str = new StringTokenizer("004-034556");
String string1 = Str.nextToken("-");
String string2 = Str.nextToken("-");
}
}
Here are two ways two achieve it.
WAY 1: As you have to split two numbers by a special character you can use regex
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class TrialClass
{
public static void main(String[] args)
{
Pattern p = Pattern.compile("[0-9]+");
Matcher m = p.matcher("004-034556");
while(m.find())
{
System.out.println(m.group());
}
}
}
WAY 2: Using the string split method
public class TrialClass
{
public static void main(String[] args)
{
String temp = "004-034556";
String [] arrString = temp.split("-");
for(String splitString:arrString)
{
System.out.println(splitString);
}
}
}
You can simply use StringTokenizer to split a string in two or more parts whether there are any type of delimiters:
StringTokenizer st = new StringTokenizer("004-034556", "-");
while(st.hasMoreTokens())
{
System.out.println(st.nextToken());
}
Check out the split() method in the String class on javadoc.
https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
String data = "004-034556-1212-232-232";
int cnt = 1;
for (String item : data.split("-")) {
System.out.println("string "+cnt+" = "+item);
cnt++;
}
Here many examples for split string but I little code optimized.
String str="004-034556"
String[] sTemp=str.split("-");// '-' is a delimiter
string1=004 // sTemp[0];
string2=034556//sTemp[1];
I just wanted to write an algorithm instead of using Java built-in functions:
public static List<String> split(String str, char c){
List<String> list = new ArrayList<>();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++){
if(str.charAt(i) != c){
sb.append(str.charAt(i));
}
else{
if(sb.length() > 0){
list.add(sb.toString());
sb = new StringBuilder();
}
}
}
if(sb.length() >0){
list.add(sb.toString());
}
return list;
}
You can use the method split:
public class Demo {
public static void main(String args[]) {
String str = "004-034556";
if ((str.contains("-"))) {
String[] temp = str.split("-");
for (String part:temp) {
System.out.println(part);
}
}
else {
System.out.println(str + " does not contain \"-\".");
}
}
}
To split a string, uses String.split(regex). Review the following examples:
String data = "004-034556";
String[] output = data.split("-");
System.out.println(output[0]);
System.out.println(output[1]);
Output
004
034556
Note:
This split (regex) takes a regex as an argument. Remember to escape the regex special characters, like period/dot.
String s = "TnGeneral|DOMESTIC";
String a[]=s.split("\\|");
System.out.println(a.toString());
System.out.println(a[0]);
System.out.println(a[1]);
Output:
TnGeneral
DOMESTIC
String s="004-034556";
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)=='-')
{
System.out.println(s.substring(0,i));
System.out.println(s.substring(i+1));
}
}
As mentioned by everyone, split() is the best option which may be used in your case. An alternative method can be using substring().
Imagine you have a List<String> collection, which can contain tens of thousands of Strings.
If some of them are in the format of:
"This is ${0}, he likes ${1},${2} ... ${n}"
What would be the most efficient way ( performance-wise ) to transform a string like the one above to:
"This is %1, he likes %2,%3 ... %n"
Note that the % way starts from 1. Here's my solution:
import java.util.regex.*;
...
String str = "I am ${0}. He is ${1}";
Pattern pat = Pattern.compile("\\\$\\{(\\d+)\\}");
Matcher mat = pat.matcher(str)
while(mat.find()) {
str = mat.replaceFirst("%"+(Integer.parseInt(mat.group(1))+1))
mat = pat.matcher(str);
}
System.out.println(str);
I hope it's valid Java code, I just wrote it now in a GroovyConsole.
I'm interested in more efficient solutions, since I'm thinking that applying so many regex substitutions on so many strings might be too slow. The end code will run as Java code not Groovy code, I just used Groovy for quick prototyping :)
Here's how I would do it:
import java.util.regex.*;
public class Test
{
static final Pattern PH_Pattern = Pattern.compile("\\$\\{(\\d++)\\}");
static String changePlaceholders(String orig)
{
Matcher m = PH_Pattern.matcher(orig);
if (m.find())
{
StringBuffer sb = new StringBuffer(orig.length());
do {
m.appendReplacement(sb, "");
sb.append("%").append(Integer.parseInt(m.group(1)) + 1);
} while (m.find());
m.appendTail(sb);
return sb.toString();
}
return orig;
}
public static void main (String[] args) throws Exception
{
String s = "I am ${0}. He is ${1}";
System.out.printf("before: %s%nafter: %s%n", s, changePlaceholders(s));
}
}
test it at ideone.com
appendReplacement() performs two major functions: it appends whatever text lay between the previous match and the current one; and it parses the replacement string for group references and inserts the captured text in their place. We don't need the second function, so we bypass it by feeding it an empty replacement string. Then we call StringBuffer's append() method ourselves with the generated replacement text.
In Java 7, this API will be opened up a bit more, making further optimizations possible. The appendReplacement() functionality will be broken out into separate methods, and we'll be able to use StringBuilders instead of StringBuffers (StringBuilder didn't exist yet when Pattern/Matcher were introduced in JDK 1.4).
But probably the most effective optimization is compiling the Pattern once and saving it in a static final variable.
You should begin your match from the last checked index of the string instead of the first index at each iterative step. As btilly alludes in a comment, your solution is O(n^2) where it should be O(n). To avoid unnecessary string copying, use a StringBuilder instead:
StringBuilder str = new StringBuilder("I am ${0}. He is ${1}");
Pattern pat = Pattern.compile("\\\$\\{(\\d+)\\}");
Matcher mat = pat.matcher(str);
int lastIdx = 0;
while (mat.find(lastIdx)) {
String group = mat.group(1);
str.replace(mat.start(1), mat.end(1), "%"+(Integer.parseInt(group)+1));
lastIdx = mat.start(1);
}
System.out.println(str);
Code is untested so there might be some off-by-one errors.
I think it would be more efficient to use appendReplacement since then you aren't making a ton of new String objects and the search doesn't resume from the beginning each time.
String str = "I am ${0}. He is ${1}";
Pattern pat = Pattern.compile("\\$\\{(\\d+)\\}");
Matcher mat = pat.matcher(str);
StringBuffer sb = new StringBuffer(str.length());
while (mat.find()) {
mat.appendReplacement(sb, "" + Integer.parseInt(mat.group(1)));
}
mat.appendTail(sb);
System.out.println(sb.toString());
Prints:
I am 0. He is 1
Try this:
String str = "I am ${0}. He is ${1}";
Pattern pat = Pattern.compile("\\$\\{(\\d+)\\}");
Matcher mat = pat.matcher(string);
StringBuffer output = new StringBuilder(string.length());
while(mat.find()) {
m.appendReplacement(output, "%"+(Integer.parseInt(mat.group(1))+1));
}
mat.appendTail(output);
System.out.println(output);
(Copied mainly from the Javadoc, with the added transformation from the question.)
I think this is really O(n).
Users may want to delimit numbers as they want.
What is the most efficient (or a simple standard function) to extract all the (natural) numbers from a string?
You could use a regular expression. I modified this example from Sun's regex matcher tutorial:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
public class Test {
private static final String REGEX = "\\d+";
private static final String INPUT = "dog dog 1342 dog doggie 2321 dogg";
public static void main(String[] args) {
Pattern p = Pattern.compile(REGEX);
Matcher m = p.matcher(INPUT); // get a matcher object
while(m.find()) {
System.out.println("start(): "+m.start());
System.out.println("end(): "+m.end());
}
}
}
It finds the start and end indexes of each number. Numbers starting with 0 are allowed with the regular expression \d+, but you could easily change that if you want to.
I'm not sure I understand your question exactly. But if all you want is to pull out all non-negative integers then this should work pretty nicely:
String foo = "12,34,56.0567 junk 6745 some - stuff tab tab 789";
String[] nums = foo.split("\\D+");
// nums = ["12", "34", "56", "0567", "6745", "789"]
and then parse out the strings as ints (if needed).
If you know the delimiter, then:
String X = "12,34,56";
String[] y = X.split(","); // d=delimiter
int[] z = new int[y.length];
for (int i = 0; i < y.length; i++ )
{
z[i] = java.lang.Integer.valueOf(y[i]).intValue();
}
If you don't, you probably need to pre-process - you could do x.replace("[A-Za-z]", " "); and replace all characters with spaces and use space as the delimiter.
Hope that helps - I don't think there is a built-in function.