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Get index of 3rd comma

For example I have this string params: Blabla,1,Yooooooo,Stackoverflow,foo,chinese
And I want to get the string testCaseParams until the 3rd comma: Blabla,1,Yooooooo
and then remove it and the comma from the original string so I get thisStackoverflow,foo,chinese
I'm trying this code but testCaseParams only shows the first two values (gets index of the 2nd comma, not 3rd...)
//Get how many parameters this test case has and group the parameters
int amountOfInputs = 3;
int index = params.indexOf(',', params.indexOf(',') + amountOfInputs);
String testCaseParams = params.substring(0,index);
params = params.replace(testCaseParams + ",","");

You can hold the index of the currently-found comma in a variable and iterate until the third comma is found:
int index = 0;
for (int i=0; i<3; i++) index = str.indexOf(',', index);
String left = str.substring(0, index);
String right = str.substring(index+1); // skip comma
Edit: to validate the string, simply check if index == -1. If so, there are not 3 commas in the string.

One option would be a clever use of String#split:
String input = "Blabla,1,Yooooooo,Stackoverflow,foo,chinese";
String[] parts = input.split("(?=,)");
String output = parts[0] + parts[1] + parts[2];
System.out.println(output);
Demo

One can use split with a limit of 4.
String input = "Blabla,1,Yooooooo,Stackoverflow,foo,chinese";
String[] parts = input.split(",", 4);
if (parts.length == 4) {
String first = parts[0] + "," + parts[1] + "," + parts[2];
String second = parts[3]; // "Stackoverflow,foo,chinese"
}

You can split with this regex to get the 2 pats:
String[] parts = input.split("(?<=\\G.*,.*,.*),");
It will result in parts equal to:
{ "Blabla,1,Yooooooo", "Stackoverflow,foo,chinese" }
\\G refers to the previous match or the start of the string.
(?<=) is positive look-behind.
So it means match a comma for splitting, if it is preceded by 2 other commas since the previous match or the start of the string.
This will keep empty strings between commas.

I offer this here just as a "fun" one line solution:
public static int nthIndexOf(String str, String c, int n) {
return str.length() - str.replace(c, "").length() < n ? -1 : n == 1 ? str.indexOf(c) : c.length() + str.indexOf(c) + nthIndexOf(str.substring(str.indexOf(c) + c.length()), c, n - 1);
}
//Usage
System.out.println(nthIndexOf("Blabla,1,Yooooooo,Stackoverflow,foo,chinese", ",", 3)); //17
(It's recursive of course, so will blow up on large strings, it's relatively slow, and certainly isn't a sensible way to do this in production.)
As a more sensbile one liner using a library, you can use Apache commons ordinalIndexOf(), which achieves the same thing in a more sensible way!

Java: Split string by number of characters but with guarantee that string will be split only after whitespace

I want to achieve something like this.
String str = "This is just a sample string";
List<String> strChunks = splitString(str,8);
and strChunks should should be like:
"This is ","just a ","sample ","string."
Please note that string like "sample " have only 7 characters as with 8 characters it will be "sample s" which will break down my next word "string".
Also we can go with the assumption that a word will never be larger than second argument of method (which is 8 in example) because in my use case second argument is always static with value 32000.
The obvious approach that I can think of is looping thru the given string, breaking the string after 8 chars and than searching the next white space from the end. And then repeating same thing again for remaining string.
Is there any more elegant way to achieve the same. Is there any utility method already available in some standard third libraries like Guava, Apache Commons.

Splitting on "(?<=\\G.{7,}\\s)" produces the result that you need (demo).
\\G means the end of previous match; .{7,} means seven or more of any characters; \\s means a space character.

Not a standard method, but this might suit your needs
See it on http://ideone.com/2RFIZd
public static List<String> splitString(String str, int chunksize) {
char[] chars = str.toCharArray();
ArrayList<String> list = new ArrayList<String>();
StringBuilder builder = new StringBuilder();
int count = 0;
for(char character : chars) {
if(count < chunksize - 1) {
builder.append(character);
count++;
}
else {
if(character == ' ') {
builder.append(character);
list.add(builder.toString());
count = 0;
builder.setLength(0);
}
else {
builder.append(character);
count++;
}
}
}
list.add(builder.toString());
builder.setLength(0);
return list;
}
Please note, I used the human notation for string length, because that's what your sample reflects( 8 = postion 7 in string). that's why the chunksize - 1 is there.
This method takes 3 milliseconds on a text the size of http://catdir.loc.gov/catdir/enhancements/fy0711/2006051179-s.html

Splitting String using method 1.
String text="This is just a sample string";
List<String> strings = new ArrayList<String>();
int index = 0;
while (index < text.length()) {
strings.add(text.substring(index, Math.min(index + 8,text.length())));
index += 8;
}
for(String s : strings){
System.out.println("["+s+"]");
}
Splitting String using Method 2
String[] s=text.split("(?<=\\G.{"+8+"})");
for (int i = 0; i < s.length; i++) {
System.out.println("["+s[i]+"]");
}

This uses a hacked reduction to get it done without much code:
String str = "This is just a sample string";
List<String> parts = new ArrayList<>();
parts.add(Arrays.stream(str.split("(?<= )"))
.reduce((a, b) -> {
if (a.length() + b.length() <= 8)
return a + b;
parts.add(a);
return b;
}).get());
See demo using edge case input (that breaks some other answers!)
This splits after each space, then either joins up parts or adds to the list depending on the length of the pair.

How to insert a space in a charArray in an exact position [Java]

My problem is that I'm getting a String and I need to check if there is a space in the 4th position but starting from the end. If in this position there is not a space, I should insert it.
For example:
I get this String: TW12EF, need to get it like this: TW1 2EF
First of all I get the 4 last characters in a char array because I also need to check if they are numbers or letters.
With this method I check if there is a space:
public static boolean isSpace(){
return String.valueOf(charArray[0]).matches("[ \\t\\n\\x0B\\f\\r]");
}
charArray contains the last 4 characters of the input String
If charArray[0] wouldn't be a space, I want to insert a space in the 2nd place (charArray[1])
If there is something that I can correct in the question to make it easier to understand, just let me know and I will try to make it better for next questions.

A simple and direct solution (most likely faster than using a regular expression) is to get the 4th to the last character (if it exists), and if it isn't a white-space, insert a space at that position.
public static void main(String[] args) {
String str = "TW12EF";
int insertPos = str.length() - 4;
if (insertPos >= 0) {
char ch = str.charAt(insertPos);
if (!Character.isWhitespace(ch)) {
str = new StringBuilder(str).insert(insertPos + 1, ' ').toString();
}
}
System.out.println(str);
}
A whitespace is determined by invoking isWhitespace, which returns true for space but also tabs or line feeds, like you did in your question. The character is inserted by leveraging the StringBuilder#insert method, which is more direct that taking 2 substrings and concatenating them.

A quick, dirty regex will help :
String p = "TW12EF";
System.out.println(p.replaceAll("(.)\\s*(\\S.{2})$", "$1 $2")); // Select a character followed by 0 or more spaces and followed by 3 non-space characters. And replace multiple spaces if they exist with a single space
O/P :
TW1 2EF
Also works if there are one or more spaces after the 3rd char (from the left)

As char is a primitive data type, the comparison can be done simply with
if (charArray[0] == ' ') {
char[] temp = new char[5];
temp[0] = ' ';
for (int i = 1; i <= 4; i++) {
temp[i] = charArray[i - 1];
}
charArray = temp;
}

You could use something like:
public static void main(String[] args) {
String str = "TW12EF";
processStr(str);
}
public static final int SPACE_POS = 4, OFFSET = 1;
public static String processStr(String str)
{
if(!Character.isWhitespace(str.charAt(str.length() - SPACE_POS)))
{
str = String.format("%s %s", str.substring(0, str.length() - SPACE_POS + OFFSET), str.substring(SPACE_POS - OFFSET));
}
return str;
}

Like this？
` String s="TW12EF";
String result="";
int length=s.length();
for(int i=length-1;i>-1;i--){
if(i==length-4&&s.charAt(i)!=' '){
result+=" ";
}
result+=s.charAt(length-i-1);
}
System.out.println(result);`

Copy the first N words in a string in java

I want to select the first N words of a text string.
I have tried split() and substring() to no avail.
What I want is to select the first 3 words of the following prayer and copy them to another variable.
For example if I have a string:
String greeting = "Hello this is just an example"
I want to get into the variable Z the first 3 words so that
Z = "Hello this is"

String myString = "Copying first N numbers of words to a string";
String [] arr = myString.split("\\s+");
//Splits words & assign to the arr[] ex : arr[0] -> Copying ,arr[1] -> first
int N=3; // NUMBER OF WORDS THAT YOU NEED
String nWords="";
// concatenating number of words that you required
for(int i=0; i<N ; i++){
nWords = nWords + " " + arr[i] ;
}
System.out.println(nWords);
NOTE : Here .split() function returns an array of strings computed by splitting a given string around matches of the given regular expression
so if i write the code like follows
String myString = "1234M567M98723651";
String[] arr = myString.split("M"); //idea : split the words if 'M' presents
then answers will be : 1234 and 567 where stored into an array.
This is doing by storing the split values into the given array. first split value store to arr[0], second goes to arr[1].
Later part of the code is for concatenating the required number of split words
Hope that you can get an idea from this!!!
Thank you!

public String getFirstNStrings(String str, int n) {
String[] sArr = str.split(" ");
String firstStrs = "";
for(int i = 0; i < n; i++)
firstStrs += sArr[i] + " ";
return firstStrs.trim();
}
Now getFirstNStrings("Hello this is just an example", 3); will output:
Hello this is

You could try something like:
String greeting = "Hello this is just an example";
int end = 0;
for (int i=0; i<3; i++) {
end = greeting.indexOf(' ', end) + 1;
}
String Z = greeting.substring(0, end - 1);
N.B. This assumes there are at least three space characters in your source string. Any less and this code will probably fail.

Add this in a utility class, such as Util.java
public static String getFirstNWords(String s, int n) {
if (s == null) return null;
String [] sArr = s.split("\\s+");
if (n >= sArr.length)
return s;
String firstN = "";
for (int i=0; i<n-1; i++) {
firstN += sArr[i] + " ";
}
firstN += sArr[n-1];
return firstN;
}
Usage:
Util.getFirstNWords("This will give you the first N words", 3);
---->
"This will give"

If you use Apache Commons Lang3, you can make it a little shorter like this:
public String firstNWords(String input, int numOfWords) {
String[] tokens = input.split(" ");
tokens = ArrayUtils.subarray(tokens, 0, numOfWords);
return StringUtils.join(tokens, ' ');
}

Most of the answers posted already use regular expressions which can become an overhead if we have to process a large number of strings. Even str.split(" ") uses regular expression operations internally. dave's answer is perhaps the mos efficient, but it does not handle correctly strings that have multiple spaces occurring together, beside assuming that regular space is the only word separator and that the input string has 3 or more words (an assumption he has already called out). If using Apache Commons in an option, then I would use the following code as it is not only concise and avoids using regular expression even internally but also handled gracefully input strings that have less than 3 words:
/* Splits by whitespace characters. All characters after the 3rd whitespace,
* if present in the input string, go into the 4th "word", which could really
* be a concanetation of multiple words. For the example in the question, the
* 4th "word" in the result array would be "just an example". Invoking the
* utility method with max-splits specified is slightly more efficient as it
* avoids the need to look for and split by space after the first 3 words have
* been extracted
*/
String[] words = StringUtils.split(greeting, null, 4);
String Z = StringUtils.join((String[]) ArrayUtils.subarray(words, 0, 3), ' ');

java replace substring in string specific index

How would I replace a string 10100 with 10010 using the algorithm "replace the last substring 10 with 01."
I tried
s=s.replace(s.substring(a,a+2), "01");
but this returns 01010, replacing both the first and the second substring of "10".
"a" represents s.lastindexOf("10");

Here's a simple and extensible function you can use. First its use/output and then its code.
String original = "10100";
String toFind = "10";
String toReplace = "01";
int ocurrence = 2;
String replaced = replaceNthOcurrence(original, toFind, toReplace, ocurrence);
System.out.println(replaced); // Output: "10010"
original = "This and This and This";
toFind = "This";
toReplace = "That";
ocurrence = 3;
replaced = replaceNthOcurrence(original, toFind, toReplace, ocurrence);
System.out.println(replaced); // Output: "This and This and That"
Function code:
public static String replaceNthOcurrence(String str, String toFind, String toReplace, int ocurrence) {
Pattern p = Pattern.compile(Pattern.quote(toFind));
Matcher m = p.matcher(str);
StringBuffer sb = new StringBuffer(str);
int i = 0;
while (m.find()) {
if (++i == ocurrence) { sb.replace(m.start(), m.end(), toReplace); break; }
}
return sb.toString();
}

If you want to access the last two indices of a string, then you can use: -
str.substring(str.length() - 2);
This gives you string from index str.length() - 2 to the last character, which is exactly the last two character.
Now, you can replace the last two indices with whatever string you want.
UPDATE: -
Of you want to access the last occurrence of a character or substring, you can use String#lastIndexOf method: -
str.lastIndexOf("10");
Ok, you can try this code: -
String str = "10100";
int fromIndex = str.lastIndexOf("10");
str = str.substring(0, fromIndex) + "01" + str.substring(fromIndex + 2);
System.out.println(str);

10100 with 10010
String result = "10100".substring(0, 2) + "10010".substring(2, 4) + "10100".substring(4, 5);

You can get the last index of a character or substring using string's lastIndexOf method. See the documentation link below for how to use it.
http://docs.oracle.com/javase/1.4.2/docs/api/java/lang/String.html#lastIndexOf(java.lang.String)
Once you know the index of your substring, you can get the substring of all characters before that index, and the substring of all characters after the last character in your search string, and concatenate.
This is a little drawn out, and I didn't actually run it (so I might have a syntax error), but it gives you the point of what I'm trying to convey at least. You could do this all in one line if you want, but it wouldn't illustrate the point as well.
string s = "10100";
string searchString = "10";
string replacementString = "01";
string charsBeforeSearchString = s.substring(0, s.lastIndexOf(searchString) - 1);
string charsAfterSearchString = s.substring(s.lastIndexIf(searchString) + 2);
s = charsBeforeSearchString + replacementString + charsAfterSearchString;

The easiest way:
String input = "10100";
String result = Pattern.compile("(10)(?!.*10.*)").matcher(input).replaceAll("01");
System.out.println(result);