Develop Reference

How to split a String by a comma, but from the second comma

I have a string as:
"model=iPhone12,3,os_version=13.6.1,os_update_exist=1,status=1"
How can I convert this into:
model=iPhone12,3
os_version=13.6.1
os_update_exist=1
status=1

Split the string from the first comma, then re-join the first two elements of the resulting string array.

I doubt there's a "clean" way to do this but this would work for your case:
String str = "model=iPhone12,3,os_version=13.6.1,os_update_exist=1,status=1";
String[] sp = str.split(",");
sp[0] += "," + sp[1];
sp[1] = sp[2];
sp[2] = sp[3];
sp[3] = sp[4];
sp[4] = "";

You can try this:
public String[] splitString(String source) {
// Split the source string based on a comma followed by letters and numbers.
// Basically "model=iPhone12,3,os_version=13.6.1,os_update_exist=1,status=1" will be split
// like this:
// model=iPhone12,3
// ,os_version=13.6.1
// ,os_update_exist=1
// ,status=1"
String[] result = source.split("(?=,[a-z]+\\d*)");
for (int i = 0; i < result.length; i++) {
// Removes the comma at the beginning of the string if present
if (result[i].matches(",.*")) {
result[i] = result[i].substring(1);
}
}
return result;
}

if you are parsing always the same kind of String a regex like this will be do the job
String str = "model=iPhone12,3,os_version=13.6.1,os_update_exist=1,status=1";
Matcher m = Pattern.compile("model=(.*),os_version=(.*),os_update_exist=(.*),status=(.*)").matcher(str);
if (m.find()) {
model = m.group(1)); // iPhone12,3
os = m.group(2)); // 13.6.1
update = m.group(3)); // 1
status = m.group(4)); // 1
}
If you really wants to use a split you can still use that kind of trick
String[] split = str.replaceAll(".*?=(.*?)(,[a-z]|$)", "$1#")
.split("#");
split[0] // iPhone12,3
split[1] // 13.6.1
split[2] // 1
split[3] // 1

How to split string at every nth occurrence of character in Java

I would like to split a string at every 4th occurrence of a comma ,.
How to do this? Below is an example:
String str = "1,,,,,2,3,,1,,3,,";
Expected output:
array[0]: 1,,,,
array[1]: ,2,3,,
array[2]: 1,,3,,
I tried using Google Guava like this:
Iterable<String> splitdata = Splitter.fixedLength(4).split(str);
output: [1,,,, ,,2,, 3,,1, ,,3,, ,]
I also tried this:
String [] splitdata = str.split("(?<=\\G.{" + 4 + "})");
output: [1,,,, ,,2,, 3,,1, ,,3,, ,]
Yet this is is not the output I want. I just want to split the string at every 4th occurrence of a comma.
Thanks.

Take two int variable. One is to count the no of ','. If ',' occurs then the count will move. And if the count is go to 4 then reset it to 0. The other int value will indicate that from where the string will be cut off. it will start from 0 and after the first string will be detected the the end point (char position in string) will be the first point of the next. Use the this start point and current end point (i+1 because after the occurrence happen the i value will be incremented). Finally add the string in the array list. This is a sample code. Hope this will help you. Sorry for my bad English.
String str = "1,,,,,2,3,,1,,3,,";
int k = 0;
int startPoint = 0;
ArrayList<String> arrayList = new ArrayList<>();
for (int i = 0; i < str.length(); i++)
{
if (str.charAt(i) == ',')
{
k++;
if (k == 4)
{
String ab = str.substring(startPoint, i+1);
System.out.println(ab);
arrayList.add(ab);
startPoint = i+1;
k = 0;
}
}
}

Here's a more flexible function, using an idea from this answer:
static List<String> splitAtNthOccurrence(String input, int n, String delimiter) {
List<String> pieces = new ArrayList<>();
// *? is the reluctant quantifier
String regex = Strings.repeat(".*?" + delimiter, n);
Matcher matcher = Pattern.compile(regex).matcher(input);
int lastEndOfMatch = -1;
while (matcher.find()) {
pieces.add(matcher.group());
lastEndOfMatch = matcher.end();
}
if (lastEndOfMatch != -1) {
pieces.add(input.substring(lastEndOfMatch));
}
return pieces;
}
This is how you call it using your example:
String input = "1,,,,,2,3,,1,,3,,";
List<String> pieces = splitAtNthOccurrence(input, 4, ",");
pieces.forEach(System.out::println);
// Output:
// 1,,,,
// ,2,3,,
// 1,,3,,
I use Strings.repeat from Guava.

try this also, if you want result in array
String str = "1,,,,,2,3,,1,,3,,";
System.out.println(str);
char c[] = str.toCharArray();
int ptnCnt = 0;
for (char d : c) {
if(d==',')
ptnCnt++;
}
String result[] = new String[ptnCnt/4];
int i=-1;
int beginIndex = 0;
int cnt=0,loopcount=0;
for (char ele : c) {
loopcount++;
if(ele==',')
cnt++;
if(cnt==4){
cnt=0;
result[++i]=str.substring(beginIndex,loopcount);
beginIndex=loopcount;
}
}
for (String string : result) {
System.out.println(string);
}

This work pefectly and tested in Java 8
public String[] split(String input,int at){
String[] out = new String[2];
String p = String.format("((?:[^/]*/){%s}[^/]*)/(.*)",at);
Pattern pat = Pattern.compile(p);
Matcher matcher = pat.matcher(input);
if (matcher.matches()) {
out[0] = matcher.group(1);// left
out[1] = matcher.group(2);// right
}
return out;
}
//Ex: D:/folder1/folder2/folder3/file1.txt
//if at = 2, group(1) = D:/folder1/folder2 and group(2) = folder3/file1.txt

The accepted solution above by Saqib Rezwan does not add the leftover string to the list, if it divides the string after every 4th comma and the length of the string is 9 then it will leave the 9th character, and return the wrong list.
A complete solution would be :
private static ArrayList<String> splitStringAtNthOccurrence(String str, int n) {
int k = 0;
int startPoint = 0;
ArrayList<String> list = new ArrayList();
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == ',') {
k++;
if (k == n) {
String ab = str.substring(startPoint, i + 1);
list.add(ab);
startPoint = i + 1;
k = 0;
}
}
// if there is no comma left and there are still some character in the string
// add them to list
else if (!str.substring(i).contains(",")) {
list.add(str.substring(startPoint));
break;
}
}
return list;
}
}

How to split the string in java?

String str = "AlwinX-road-9:00pm-kanchana travels-25365445421";
String[] names = str.split("-");
I want output like following:
AlwinX-road
9:00pm
kanchana travels
25365445421

Use pattern matching to match your requirement
String str = "AlwinX-road-9:00pm-kanchana travels-25365445421";
String regex = "(^[A-Z-a-z ]+)[-]+(\\d+:\\d+pm)[-]([a-z]+\\s+[a-z]+)[-](\\d+)";
Matcher matcher = Pattern.compile( regex ).matcher( str);
while (matcher.find( ))
{
String roadname = matcher.group(1);
String time = matcher.group(2);
String travels = matcher.group(3);
String digits= matcher.group(4);
System.out.println("time="+time);
System.out.println("travels="+travels);
System.out.println("digits="+digits);
}

Since you want to include the delimiter in your first output line, you can do the split, and merge the first two element with a -: -
String[] names = str.split("-");
System.out.println(names[0] + "-" + names[1])
for (int i = 2;i < names.length; i++) {
System.out.println(names[i])
}

The split() method can't distinguish the dash in AlwinX-road and the other dashes in the string, it treats all the dashes the same. You will need to do some sort of post processing on the resulting array. If you will always need the first two strings in the array joined you can just do that. If your strings are more complex you will need to add additional logic to join the strings in the array.

One way you could do it, assuming the first '-' is always part of a two part identifier.
String str = "AlwinX-road-9:00pm-kanchana travels-25365445421";
String[] tokens = str.split("-");
String[] output = new String[tokens.length - 1];
output[0] = tokens[0] + '-' + tokens[1];
System.out.println(output[0]);
for(int i = 1; i < output.length; i++){
output[i] = tokens[i+1];
System.out.println(output[i]);
}

Looks like you want to split (with removal of all dashes but the first one).
String str = "AlwinX-road-9:00pm-kanchana travels-25365445421";
String[] names = str.split("-");
for (String value : names)
{
System.out.println(value);
}
So its produces:
AlwinX
road
9:00pm
kanchana travels
25365445421
Notice that "AlwinX" and "road" we split as well since they had a dash in between. So you will need custom logic to handle this case. here is an example how to do it (I used StringTokenizer):
StringTokenizer tk = new StringTokenizer(str, "-", true);
String firstString = null;
String secondString = null;
while (tk.hasMoreTokens())
{
final String token = tk.nextToken();
if (firstString == null)
{
firstString = token;
continue;
}
if (secondString == null && firstString != null && !token.equals("-"))
{
secondString = token;
System.out.println(firstString + "-" + secondString);
continue;
}
if (!token.equals("-"))
{
System.out.println(token);
}
}
This will produce:
AlwinX-road
9:00pm
kanchana travels
25365445421

from your format, I think you want to split the first one just before the time part. You can do it this way:
String str =yourString;
String beforetime=str.split("-\\d+:\\d+[ap]m")[0]; //this is your first token,
//AlwinX-road in your example
String rest=str.substring(beforetime.length()+1);
String[] restNames=rest.split("-");
If you really need it all together in one array then see the code below:
String[] allTogether=new String[restNames.length+1];//the string with all your tokens
allTogether[0]=beforetime;
System.arraycopy(restNames, 0, allTogether, 1, restNames.length);

If you use "_" as a separator instead of "-": AlwinX-road_9:00pm_kanchana travels_25365445421
New code:
String str = new String("AlwinX-road_9:00pm_kanchana travels_25365445421");
String separator = new String("_");
String[] names = str.split(separator);
for(int i=0; i<names.length; i++){
System.out.println(names[i]);
}

How to flip two words in a string, Java

So say I have a string called x that = "Hello world". I want to somehow make it so that it will flip those two words and instead display "world Hello". I am not very good with loops or arrays and obviously am a beginner. Could I accomplish this somehow by splitting my string? If so, how? If not, how could I do this? Help would be appreciated, thanks!

1) split string into String array on space.
String myArray[] = x.split(" ");
2) Create new string with words in reverse order from array.
String newString = myArray[1] + " " + myArray[0];
Bonus points for using a StringBuilder instead of concatenation.

String abc = "Hello world";
String cba = abc.replace( "Hello world", "world Hello" );
abc = "This is a longer string. Hello world. My String";
cba = abc.replace( "Hello world", "world Hello" );
If you want, you can explode your string as well:
String[] pieces = abc.split(" ");
for( int i=0; i<pieces.length-1; ++i )
if( pieces[i]=="Hello" && pieces[i+1]=="world" ) swap(pieces[i], pieces[i+1]);
There are many other ways you can do it too. Be careful for capitalization. You can use .toUpperCase() in your if statements and then make your matching conditionals uppercase, but leave the results with their original capitalization, etc.

Here's the solution:
import java.util.*;
public class ReverseWords {
public String reverseWords(String phrase) {
List<String> wordList = Arrays.asList(phrase.split("[ ]"));
Collections.reverse(wordList);
StringBuilder sbReverseString = new StringBuilder();
for(String word: wordList) {
sbReverseString.append(word + " ");
}
return sbReverseString.substring(0, sbReverseString.length() - 1);
}
}
The above solution was coded by me, for Google Code Jam and is also blogged here: Reverse Words - GCJ 2010

Just use this method, call it and pass the string that you want to split out
static String reverseWords(String str) {
// Specifying the pattern to be searched
Pattern pattern = Pattern.compile("\\s");
// splitting String str with a pattern
// (i.e )splitting the string whenever their
// is whitespace and store in temp array.
String[] temp = pattern.split(str);
String result = "";
// Iterate over the temp array and store
// the string in reverse order.
for (int i = 0; i < temp.length; i++) {
if (i == temp.length - 1) {
result = temp[i] + result;
} else {
result = " " + temp[i] + result;
}
}
return result;
}

Depending on your exact requirements, you may want to split on other forms of whitespace (tabs, multiple spaces, etc.):
static Pattern p = Pattern.compile("(\\S+)(\\s+)(\\S+)");
public String flipWords(String in)
{
Matcher m = p.matcher(in);
if (m.matches()) {
// reverse the groups we found
return m.group(3) + m.group(2) + m.group(1);
} else {
return in;
}
}
If you want to get more complex see the docs for Pattern http://docs.oracle.com/javase/6/docs/api/java/util/regex/Pattern.html

Try something as follows:
String input = "how is this";
List<String> words = Arrays.asList(input.split(" "));
Collections.reverse(words);
String result = "";
for(String word : words) {
if(!result.isEmpty()) {
result += " ";
}
result += word;
}
System.out.println(result);
Output:
this is how

Too much?
private static final Pattern WORD = Pattern.compile("^(\\p{L}+)");
private static final Pattern NUMBER = Pattern.compile("^(\\p{N}+)");
private static final Pattern SPACE = Pattern.compile("^(\\p{Z}+)");
public static String reverseWords(final String text) {
final StringBuilder sb = new StringBuilder(text.length());
final Matcher wordMatcher = WORD.matcher(text);
final Matcher numberMatcher = NUMBER.matcher(text);
final Matcher spaceMatcher = SPACE.matcher(text);
int offset = 0;
while (offset < text.length()) {
wordMatcher.region(offset, text.length());
numberMatcher.region(offset, text.length());
spaceMatcher.region(offset, text.length());
if (wordMatcher.find()) {
final String word = wordMatcher.group();
sb.insert(0, reverseCamelCase(word));
offset = wordMatcher.end();
} else if (numberMatcher.find()) {
sb.insert(0, numberMatcher.group());
offset = numberMatcher.end();
} else if (spaceMatcher.find()) {
sb.insert(0, spaceMatcher.group(0));
offset = spaceMatcher.end();
} else {
sb.insert(0, text.charAt(offset++));
}
}
return sb.toString();
}
private static final Pattern CASE_REVERSAL = Pattern
.compile("(\\p{Lu})(\\p{Ll}*)(\\p{Ll})$");
private static String reverseCamelCase(final String word) {
final StringBuilder sb = new StringBuilder(word.length());
final Matcher caseReversalMatcher = CASE_REVERSAL.matcher(word);
int wordEndOffset = word.length();
while (wordEndOffset > 0 && caseReversalMatcher.find()) {
sb.insert(0, caseReversalMatcher.group(3).toUpperCase());
sb.insert(0, caseReversalMatcher.group(2));
sb.insert(0, caseReversalMatcher.group(1).toLowerCase());
wordEndOffset = caseReversalMatcher.start();
caseReversalMatcher.region(0, wordEndOffset);
}
sb.insert(0, word.substring(0, wordEndOffset));
return sb.toString();
}

How to get the right substring after split?

I have a string like this:[name ;24, name;23, name;22]. How can i split this string to obtain only the numbers after";"?

String s = "[name ;24, name;23, name;22]";
String couples[] = s.replace("]", "").split(",");
int ages[] = new int[couples.length];
for (int i=0; i< couples.length; i++)
ages[i] = Integer.parseInt(couples[i].split(";")[1]);

// Your input looks like this.
String s = "[name ;24, name;23, name;22]";
String[] numberStrings = s
// First get rid of the known prefix and suffix
.substring("[name ;".length(), s.length - "]".length())
// Then split on the repeated portion that occurs between numbers.
.split(", name;");

yet another method
String str = "name ;24, name;23, name;22";
int p = 0;
while ((p = str.indexOf(";", p + 1)) > -1) {
System.out.println(str.substring(p+1).split("[^0-9]")[0]);
}