Develop Reference

How to insert char at specific location throughout the string? [duplicate]

I'm getting in an int with a 6 digit value. I want to display it as a String with a decimal point (.) at 2 digits from the end of int. I wanted to use a float but was suggested to use String for a better display output (instead of 1234.5 will be 1234.50). Therefore, I need a function that will take an int as parameter and return the properly formatted String with a decimal point 2 digits from the end.
Say:
int j= 123456
Integer.toString(j);
//processing...
//output : 1234.56

As mentioned in comments, a StringBuilder is probably a faster implementation than using a StringBuffer. As mentioned in the Java docs:
This class provides an API compatible with StringBuffer, but with no guarantee of synchronization. This class is designed for use as a drop-in replacement for StringBuffer in places where the string buffer was being used by a single thread (as is generally the case). Where possible, it is recommended that this class be used in preference to StringBuffer as it will be faster under most implementations.
Usage :
String str = Integer.toString(j);
str = new StringBuilder(str).insert(str.length()-2, ".").toString();
Or if you need synchronization use the StringBuffer with similar usage :
String str = Integer.toString(j);
str = new StringBuffer(str).insert(str.length()-2, ".").toString();

int j = 123456;
String x = Integer.toString(j);
x = x.substring(0, 4) + "." + x.substring(4, x.length());

int yourInteger = 123450;
String s = String.format("%6.2f", yourInteger / 100.0);
System.out.println(s);

Using ApacheCommons3 StringUtils, you could also do
int j = 123456;
String s = Integer.toString(j);
int pos = s.length()-2;
s = StringUtils.overlay(s,".", pos, pos);
it's basically substring concatenation but shorter if you don't mind using libraries, or already depending on StringUtils

In most use-cases, using a StringBuilder (as already answered) is a good way to do this. However, if performance matters, this may be a good alternative.
/**
* Insert the 'insert' String at the index 'position' into the 'target' String.
*
* ````
* insertAt("AC", 0, "") -> "AC"
* insertAt("AC", 1, "xxx") -> "AxxxC"
* insertAt("AB", 2, "C") -> "ABC
* ````
*/
public static String insertAt(final String target, final int position, final String insert) {
final int targetLen = target.length();
if (position < 0 || position > targetLen) {
throw new IllegalArgumentException("position=" + position);
}
if (insert.isEmpty()) {
return target;
}
if (position == 0) {
return insert.concat(target);
} else if (position == targetLen) {
return target.concat(insert);
}
final int insertLen = insert.length();
final char[] buffer = new char[targetLen + insertLen];
target.getChars(0, position, buffer, 0);
insert.getChars(0, insertLen, buffer, position);
target.getChars(position, targetLen, buffer, position + insertLen);
return new String(buffer);
}

For Kotlin dudes ;) from the accepted answer (#MikeThomsen's)
fun String.insert(insertAt: Int, string: String): String {
return this.substring(0, insertAt) + string + this.substring(insertAt, this.length)
}
Test ✅
"ThisTest".insert(insertAt = 4, string = "Is").should.equal("ThisIsTest")

String.format("%0d.%02d", d / 100, d % 100);

You could use
System.out.printf("%4.2f%n", ((float)12345)/100));
As per the comments, 12345/100.0 would be better, as would the use of double instead of float.

If you are using a system where float is expensive (e.g. no FPU) or not allowed (e.g. in accounting) you could use something like this:
for (int i = 1; i < 100000; i *= 2) {
String s = "00" + i;
System.out.println(s.substring(Math.min(2, s.length() - 2), s.length() - 2) + "." + s.substring(s.length() - 2));
}
Otherwise the DecimalFormat is the better solution. (the StringBuilder variant above won't work with small numbers (<100)

I think a simpler and more elegant solution to insert a String in a certain position would be this one-liner:
target.replaceAll("^(.{" + position + "})", "$1" + insert);
For example, to insert a missing : into a time String:
"-0300".replaceAll("^(.{3})", "$1:");
What it does is, matches position characters from the beginning of the string, groups that, and replaces the group with itself ($1) followed by the insert string. Mind the replaceAll, even though there's always one occurrence, because the first parameter must be a regex.
Of course it does not have the same performance as the StringBuilder solution, but I believe the succinctness and elegance as a simple and easier to read one-liner (compared to a huge method) is sufficient for making it the preferred solution in most non performance-critical use-cases.
Note I'm solving the generic problem in the title for documentation reasons, of course if you are dealing with decimal numbers you should use the domain-specific solutions already proposed.

There are good answers here, but with Kotlin extensions addition we can do it even more simply:
val indexWhereInsertIsIntended = 2
val inputString = "2408"
val resultingString = inputString.toCharArray().toMutableList()
.also {
it.add(indexWhereInsertIsIntended, '/')
}.joinToString("")
Result = 24/08
This example shows a card expiry date, and slash (/) is intended at 2nd Index. So the resulting index in this case will have / at 2nd index.
If you want to replace and not add:
val indexWhereInsertIsIntended = 2
val inputString = "2408"
val resultingString = inputString.toCharArray()
.also {
it[indexWhereInsertIsIntended] = '/'
}.joinToString("")
Result = 24/0

public static void main(String[] args) {
char ch='m';
String str="Hello",k=String.valueOf(ch),b,c;
System.out.println(str);
int index=3;
b=str.substring(0,index-1 );
c=str.substring(index-1,str.length());
str=b+k+c;
}

// Create given String and make with size 30
String str = "Hello How Are You";
// Creating StringBuffer Object for right padding
StringBuffer stringBufferRightPad = new StringBuffer(str);
while (stringBufferRightPad.length() < 30) {
stringBufferRightPad.insert(stringBufferRightPad.length(), "*");
}
System.out.println("after Left padding : " + stringBufferRightPad);
System.out.println("after Left padding : " + stringBufferRightPad.toString());
// Creating StringBuffer Object for right padding
StringBuffer stringBufferLeftPad = new StringBuffer(str);
while (stringBufferLeftPad.length() < 30) {
stringBufferLeftPad.insert(0, "*");
}
System.out.println("after Left padding : " + stringBufferLeftPad);
System.out.println("after Left padding : " + stringBufferLeftPad.toString());

Try this :
public String ConvertMessage(String content_sendout){
//use unicode (004E00650077) need to change to hex (&#x004E&#x;0065&#x;0077;) first ;
String resultcontent_sendout = "";
int i = 4;
int lengthwelcomemsg = content_sendout.length()/i;
for(int nadd=0;nadd<lengthwelcomemsg;nadd++){
if(nadd == 0){
resultcontent_sendout = "&#x"+content_sendout.substring(nadd*i, (nadd*i)+i) + ";&#x";
}else if(nadd == lengthwelcomemsg-1){
resultcontent_sendout += content_sendout.substring(nadd*i, (nadd*i)+i) + ";";
}else{
resultcontent_sendout += content_sendout.substring(nadd*i, (nadd*i)+i) + ";&#x";
}
}
return resultcontent_sendout;
}

Java - Reverse String null error

The input is meant to appear like this, example.
\n
Kazan R
\n
6789
\n
Nzk462
\n
However the output I receive looks like this
kzn462nullnzk
Why is this? and How can i solve it?
private void btnGenerateActionPerformed(java.awt.event.ActionEvent evt) {
secondname = JOptionPane.showInputDialog("Enter your surname:");
firstname = JOptionPane.showInputDialog("Enter your firstname:");
idno = JOptionPane.showInputDialog("Enter your idno:");
nametag = firstname.substring(0, 1);
initials = secondname + " " + nametag;
int randnum;
do {
randnum = (int) (Math.random() * 900) + 100;
} while (randnum % 2 != 0);
code = secondname.replaceAll("[aeiou || AEIOU](?!\\b)", "")+randnum ;
txaDisplay.append(initials + '\n' + idno.substring(6,10) + '\n' + code);
int length = secondname.length();
for (int i = length - 1; i >= 0; i--) {
reverse = reverse + secondname.charAt(i);
}
String end = reverse + code;
txaDisplay.append( reverse);

Why don't you use
new StringBuilder(secondname).reverse().toString()
to reverse your String? It's better, simple and more maintanable.

Get the character array from your source string
Create a new char array of same length
Start iterating from 0 to (sourceStringLength-1)
In each iteration, get the last character
from the end in your source array and populate in your new array
Create a new string from this new array
String source = "abcdefg";
char[] chars = source.toCharArray();
char[] reverseChars = new char[source.length()];
int len = source.length();
for(int i= 0; i < len; i++){
reverseChars[i] = chars[len-1-i];
}
String reverse = new String(reverseChars);
System.out.println(reverse);

Since You don't want to use StringBuilder/StringBuffer.
Try this
String reversedString="";
for(int i=inputString.length-1;i>=0;){
reversedString+=inputString.charAt(i--);
}

I think the problem is your definition of reverse, maybe you have something like:
String reverse;
Then you don't initialize your "reverse" so when your program makes the first concatenation in your loop, it looks like this:
reverse = null + secondname.charAt(i);
The null value is converted to a string so it can be visible in the output.
I hope this information helps you.
Good Luck.

How to start replacing characters at some point?

It was very hard to form the question and I am sure it is still not clear.
I have a CSV file e.g.: Firstname;Lastname;Adress;product1;product2;product3;product4;
I would like to start replacing ";" with "::". The problem is, I want to start replacing after third semicolon.
I know it can be done in while loop where I check every character, when semicolon occurs I will count +1 and if counter is 3, I will start replacing. But isn't there a way how to do it without a loop?

You can use indexOf(char,fromIndex) method.
Your third semicolon position search can be inlined :
csvLine.indexOf(';', csvLine.indexOf(';', csvLine.indexOf(';') + 1) + 1)
We assume that our csvLine has a least 3 semi-colons...
String csvLine = "Firstname;Lastname;Adress;product1;product2;product3;product4";
//Index of "fromIndex" param is inclusive, that's why we need to add 1
int pos = csvLine.indexOf(';', csvLine.indexOf(';', csvLine.indexOf(';') + 1) + 1);
//Retrieve string from the char after the third semi-colon
String truncatedLine = csvLine.substring(pos + 1);
//Replace ";" by "::" on our substring
truncatedLine = truncatedLine.replaceAll(";", "::");
//Then concat the first part of csvLine with the second
String result = csvLine.substring(0, pos + 1).concat(truncatedLine);
System.out.println(result); //Print => Firstname;Lastname;Adress;product1::product2::product3::product4
Poor input control and performance but we don't have any loops :)

If I have understood what you want, try this.
First search se position of the third semicolon:
String csvContent = "Firstname;Lastname;Adress;product1;product2;product3;product4;";
int i = 0;
int index= 0;
while(i < 4){
index = csvContent.indexOf(';', (index + 1));
i++;
}//index = position of the thrid semicolon
Second, cut your CSV content at the index position.
String tmp1 = csvContent.substring(0, index);
String tmp2 = csvContent.substring(index, csvContent.length());
Thrid, replace all ';' by '::':
tmp2 = tmp2.replaceAll(";", "::");
Finaly, rebuild your file content:
csvContent = tmp1 + tmp2;

int i = 0;
int pos = 0;
while (i < 3) {
pos = string.indexOf(';', pos+1);
i++;
}
String newString = string.substring(0, pos) +";"+ (string.substring(pos + 1, string.length()).replace(";", "::"));

how about a regex solution?
Pattern pattern = Pattern.compile("(.*?;.*?;.*?;)(.*)");
Matcher match = pattern.matcher(str);
if(match.matches()) {
String firstThree = match.group(1);
String rest = match.group(2);
rest = rest.replace(";", "::");
return firstThree + rest;
}

How do you add a newline character in a string at specific indices?

I have a string:
String testString= "For the time being, programming is a consumer job, assembly line coding is the norm, and what little exciting stuff is being performed is not going to make it compared to the mass-marketed cräp sold by those who think they can surf on the previous half-century's worth of inventions forever"
like this: For the time being, programmi \n........\n.......\n
After each length of 20 characters in this string, I want to put a newline character \n for display in a TextView in Android.

You must have to use regex for achieve your task its fast and efficient. Try below code:-
String str = "....";
String parsedStr = str.replaceAll("(.{20})", "$1\n");
The (.{20}) will capture a group of 20 characters. The $1 in the second will put the content of the group. The \n will be then appended to the 20 characters which have been just matched.

How about something like that?
String s = "...whateverstring...";
for(int i = 0; i < s.length(); i += 20) {
s = new StringBuffer(s).insert(i, "\n").toString();
}

I know there is a technically better solution to use a StringBuffer and the insert method for that class, or even regex, but I'll show you a different algorithmic approach using String#substring:
String s = "12345678901234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890123456789";
int offset = 0; // each time you add a new character, string has "shifted"
for (int i = 20; i + offset < s.length(); i += 20) {
// take first part of string, add a new line, and then add second part
s = s.substring(0, i + offset) + "\n" + s.substring(i + offset);
offset++;
}
System.out.println(s);
The result is this:
12345678901234567890
12345678901234567890
12345678901234567890
12345678901234567890
12345678901234567890
1234567890123456789

StringBuilder sb = new StringBuilder();
int done = 0;
while( done < s.length() ){
int todo = done + 20 < s.length() ? 20 : s.length() - done;
sb.append( s.substring( done, done + todo ) ).append( '\n' );
done += todo;
}
String result = sb.toString();
This also appends a newline at the end, but you can modify it easily to avoid that.

Getting the last integer from a string

Say I have a string 3 + 4 = 7 or 23 - 4 = 19. I'm wondering how I would go about getting the last integer from the string so that it can be compared to another int.
Ive used this code:
int first = Integer.parseInt(str.substring(0, str.indexOf(" ")));
int second = Integer.parseInt(result.substring(result.indexOf('+')+1 , result.indexOf('=')));
to get the first and second integers for computation, but I can't seem to get the last integer in string when I'm using the method above. Is it just something I'm doing wrong or is there something else I need to do?
I also cannot use arrays, regex, try/catch, or a SystemTokenizer.
And my strings will not always be in the a + b = c format with the spaces or a single digit integer.
the strings are taken from a text file that is imported by the scanner. Here's a sample of the equations in the file:
11 - 2 = 9
12 - 1 = 11
7 + 1 = 8
7 - 3 = 4
7 + 2 = 9
14 + 4 = 18

If all of your strings are in this format - a + b = c, then there is a simple way -
String[] parts = str.split("=");
String lastDigit = parts[parts.length - 1].trim();
int last = Integer.parseInt(lastDigit);
Check out the docs.
Update
As per your requirement of not wanting to use regex, you can do something like this -
public int getLastDigit(String expression) {
StringBuffer result = new StringBuffer();
for (int i = expression.length() - 1; i >= 0; i--) {
char c = expression.charAt(i);
if (c == '=') {
break;
}
else {
result.append(c);
}
}
return Integer.parseInt(result.reverse()
.toString()
.trim());
}
int last = getLastDigit("1 + 3 = 14");
System.out.println(last); // will print 14

This will work a lot easier:
String[] tokens = str.Split(" ");
int first = Integer.parseInt(tokens[0]);
int second = Integer.parseInt(tokens[1]);
int last = Integer.parseInt(tokens[3]);
Otherwise without arrays, use lastIndexOf:
int last = Integer.parseInt(str.substring(str.lastIndexOf(" "), str.length()));

simple use split and get index[1] to get last integer in string
just split with "="

Does this work?
int last = Integer.parseInt(str.substring(str.indexOf("=") + 1, str.length));
The idea is to get the index of the "=", and then parse the substring of there to the end.
I did "str.indexOf("=") + 1" so that it didn't include the "sign".

String [] tokens = str.split("=");
String finalDigit = tokens[tokens.length-1];

You would need trim the extracted digit, other wise it is possible to throw
Exception in thread "main" java.lang.NumberFormatException: For input
string: " 7"
To extract the last digit, you may follow the below code
String str1="3 + 4 = 7";
String str2= str1.substring(str1.lastIndexOf("=")+1).trim();
int i = Integer.parseInt(str2);

import java.io.*;
class GetString{
publice static void main(String []args)
{
string str,int x;
Scanner s = new Scanner(input).useDelimiter("\n"); //take each line
while((str=s.next())!=NULL)
{
str.trim();
int first = Integer.parseInt(str.substring(0, str.indexOf(" ")));
//Depending on weather the sign is '+' or '-' it will take the index of that
int second = Integer.parseInt(str.substring((x=str.indexOf('+')>-1?
str.indexOf('+'):str.indexOf('-'))+1 , result.indexOf('=')));
int Third= Integer.parseInt(str.substring(str.indexOf("="), str.Length-1))
}
s.close();
}
}
Its a general idea not a perfect code.Edits and suggestions are welcome.