For example the name Donald trump (12 character) brings up the error string index out of range 7 (where the space is found) even though the name Donald trump is longer.
package test;
import javax.swing.JOptionPane;
public class Usernamesubstring {
public static void main(String[] args) {
String fullname = JOptionPane.showInputDialog("What is your full name");
int breakbetween = fullname.lastIndexOf(" ");
String firstnamess = fullname.substring(breakbetween - 3, breakbetween);
int length = fullname.length();
String lastnamess = fullname.substring(length - 3, length);
String firstnamec = firstnamess.substring(0, 0);
String lastnamec = lastnamess.substring(breakbetween + 1, breakbetween + 1 );
firstnamec = firstnamec.toUpperCase();
lastnamec = lastnamec.toUpperCase();
String firstname = firstnamess.substring(1,3);
String lastname = firstnamess.substring(1,3);
firstname = firstnamec + firstname;
lastname = lastnamec + lastname;
System.out.println(firstname + lastname);
}
}
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 7
at java.lang.String.substring(String.java:1963)
at test.Usernamesubstring.main(Usernamesubstring.java:14)
You've made it more complicated than it needs to be. A simple solution can be made using String.split (which divides a string into an array of smaller strings based on a delimiter, e.g. "Donald Trump".split(" ") == {"Donald", "Trump"})
Full Code
class Usernamesubstring // change that since it no longer uses substrings
{
public static void main (String[] args)
{
String fullName = "Donald Trump";
String[] parts = fullName.split(" ");
String firstName = parts[0]; // first item before the space
String lastName = parts[parts.length - 1]; // last item in the array
System.out.println(firstName + " " + lastName);
}
}
sometimes independent of your indexes
String fullName = "Donald Trump";
String[] result = fullName.split (" ");
in result you will find now
result [0] ==> Donald
result [1] ==> Trump
isn't that a little easier for your project?
Your error shoul be in the line String lastnamec = lastnamess.substring(breakbetween + 1, breakbetween + 1 ); as lastnamess is a string of lenght 3 from fullname.substring(length - 3, length); and breakbetween is greater then 3 for "Donald Trump", where space is character 6.
You should simpify your code a bit, it makes it easier to read and find the problems.
tl;dr: The exception occurs when you try to access a String at an index which exceeds it's length or is just not contained in the string (negative values).
Regarding your approach: It's usually not a good idea to prompt a name in full because people tend to input weird stuff or mix up the order. Better prompt for first and last name separately.
Assuming someone input his name with Firstname Lastname you wouldn't have to make such a substring mess, Java has some nice features:
String name = "Mario Peach Bowser";
name = name.trim();
String[] parts = name.split(" ");
String lastname = parts[parts.length-1];
String firstname = name.replace(lastname, "").trim();
System.out.println("Hello "+firstname+", your last name is: "+lastname);
In this case I am using the trim() function to remove whitespaces at the start and end and just split the string when a white space occurs. Since people can have some middle names and stuff, I just replace the last name out of the raw input string, call trim() on it again and you have everything extracted.
If you really want a substring approach, the following would work:
String lastname = name.substring(name.lastIndexOf(" ")).trim();
String firstname = name.substring(0,name.lastIndexOf(" ")).trim();
You usually don't store the index variables. But each variant would need some sort of error check, you can either use try{} and catch() or check the String before parsing.
Only these lines are required.
String[] nameArr = fullname.split(" ");
String lastN = nameArr[nameArr.length - 1];
int lastIndexOf = fullname.lastIndexOf(lastN);
String firstN = fullname.substring(0, lastIndexOf);
System.out.println(firstN + " " + lastN);
This is my string
1 AP(PEAR + ANC)E (CAN anag)
14 EN(T)ICE (NIECE anag)
15 CHIC ("SHEIK" hom)
The string has an answer hidden,
The first string has an answer APPEARANCE and second string has ENTICE
I must extract those particular answer alone from the strings.
I tried to extract those words from it by
String input = "AP(PEAR + ANC)E (CAN anag)";;
String output = input.substring(0, input.indexOf(' '));
System.out.println(output);
Output:
AP(PEAR
As you can see, there is a space after R so the sub-string stops there and so the output. But Is there any way to read till the character 'E'(ie. end of string 'APPEARANCE') ? I want to stop reading if there is a space and if the next character is '(' .
I have another type of string in the same program "EN(T)ICE (NIECE anag)"
String input = "EN(T)ICE (NIECE anag)";
String output = input.substring(0, input.indexOf(' '));
System.out.println(output);
Output:
EN(T)ICE
There is a space after the character 'E' so it successfully gave the full output. Is there any way to get output like this for the first string. ? Any help would be great !!
Use replaceAll instead of indexOf and substring.
String[] inputs = {
"AP(PEAR + ANC)E (CAN anag)",
"EN(T)ICE (NIECE anag)",
"CHIC (\"SHEIK\" hom)"};
for (String s : inputs) {
String output = s.replaceAll(" \\(.*|[()+\\s]", "");
System.out.println(output);
}
result:
APPEARANCE
ENTICE
CHIC
You can try this splitting the string on the index of " (":
String input1 = "EN(T)ICE (NIECE anag)";
String input2 = "AP(PEAR + ANC)E (CAN anag)";
String input3 = "CHIC (\"SHEIK\" hom)";
System.out.println(extract(input1));
System.out.println(extract(input2));
System.out.println(extract(input3));
public static String extract(String s){
return s.split(" \\(.*")[0]
.replace("(", "")
.replace(")", "")
.replace(" + ", "");
}
Will produce :
ENTICE
APPEARANCE
CHIC
Split on the " (";
like so:
String input = "AP(PEAR + ANC)E (CAN anag)";
System.out.println(input.split(" \\(.*")[0]);
Just replace the String input with new values!
Can't you just replace the " + " with nothing, and then split by " " (space)?
Something like this:
private static String convertInputToOutput(final String input) {
String[] splittedArray = input.replaceAll(" \\+ ", "").split(" ");
return splittedArray[0];
}
public static void main(final String[] args) {
System.out.println(convertInputToOutput("AP(PEAR + ANC)E (CAN anag)"));
System.out.println(convertInputToOutput("EN(T)ICE (NIECE anag)"));
System.out.println(convertInputToOutput("CHIC (\"SHEIK\" hom)"));
}
Output:
AP(PEARANC)E
EN(T)ICE
CHIC
If you want output without parenthesis, also use a replaceAll for the parenthesis:
private static String convertInputToOutput(final String input) {
String[] splittedArray = input.replaceAll(" \\+ ", "").split(" ");
return splittedArray[0].replaceAll("(", "").replaceAll(")", "");
}
Output:
APPEARANCE
ENTICE
CHIC
What would be the best way to split this string directly after the CN= to store both the first and last name in separate fields as shown below?
String distinguisedName = "CN=Paul M. Sebula,OU=BBB,OU=Users,OU=TIES Project,DC=SPHQTest,DC=na,DC=BBBBBB,DC=com"
String firstName"Paul"
String lastName="Sebula"
Don't re-invent the wheel. Assuming these are well-formed DN's, see the accepted answer on this question for how to parse without directly writing your own regex: Parsing the CN out of a certificate DN
Once you've extracted the CN, then you can apply some of the other parsing techniques suggested (use the Java StringTokenizer or the String.split() method as others here have suggested if it's known to be separated only by spaces). That assumes that you can make assumptions (eg. the first element in the resulting array is the firstName,the last element is the lastName and everything in between is middle names / initials) about the CN format.
You can use split:
String distinguisedName = "CN=Paul Sebula,OU=BAE,OU=Users,OU=TIES Project,DC=SPHQTest,DC=na,DC=baesystems,DC=com";
String[] names = distinguisedName.split(",")[0].split("=")[1].split(" ");
String firstName = names[0];
String lastName= names.length > 2 ? names[names.length-1] : names[1];
System.out.println(firstName + " " + lastName);
See IDEONE demo, output: Paul Sebula.
This also accounts for just 2 names (first and last only). Note how last name is accessed it being the last item in the array.
public static void main(String[] args) {
String distinguisedName = "CN=Paul M. Sebula,OU=BBB,OU=Users,OU=TIES Project,DC=SPHQTest,DC=na,DC=BBBBBB,DC=com";
String splitResult[]=distinguisedName.split(",")[0].split("=");
String resultTwo[]=splitResult[1].split("\\.");
String firstName=resultTwo[0].split(" ")[0].trim();
String lastName=resultTwo[1].trim();
System.out.println(firstName);
System.out.println(lastName);
}
output
Paul
Sebula
String distinguisedName = "CN=Paul M. Sebula,OU=BBB,OU=Users,OU=TIES Project,DC=SPHQTest,DC=na,DC=BBBBBB,DC=com"
String[] commaSplit = distinguisedName.split(',');
String[] whitespaceSplit = commaSplit[0].split(' ');
String firstName = whitespaceSplit[0].substring(3);
String lastName = whiteSpaceSplit[2];
In steps:
String distinguisedName = "CN=Paul M. Sebula,OU=BBB,OU=Users,OU=TIES Project,DC=SPHQTest,DC=na,DC=BBBBBB,DC=com";
String fullName = distinguisedName.substring(3, distinguisedName.indexOf(','));
String[] nameParts = fullName.split(" ");
String firstName = nameParts[0];
String lastName = nameParts[nameParts.length-1];
This will work for cases where the middle name/initial are not present as well.
Looking for quick, simple way in Java to change this string
" hello there "
to something that looks like this
"hello there"
where I replace all those multiple spaces with a single space, except I also want the one or more spaces at the beginning of string to be gone.
Something like this gets me partly there
String mytext = " hello there ";
mytext = mytext.replaceAll("( )+", " ");
but not quite.
Try this:
String after = before.trim().replaceAll(" +", " ");
See also
String.trim()
Returns a copy of the string, with leading and trailing whitespace omitted.
regular-expressions.info/Repetition
No trim() regex
It's also possible to do this with just one replaceAll, but this is much less readable than the trim() solution. Nonetheless, it's provided here just to show what regex can do:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$|( )+", "$1")
);
}
There are 3 alternates:
^_+ : any sequence of spaces at the beginning of the string
Match and replace with $1, which captures the empty string
_+$ : any sequence of spaces at the end of the string
Match and replace with $1, which captures the empty string
(_)+ : any sequence of spaces that matches none of the above, meaning it's in the middle
Match and replace with $1, which captures a single space
See also
regular-expressions.info/Anchors
You just need a:
replaceAll("\\s{2,}", " ").trim();
where you match one or more spaces and replace them with a single space and then trim whitespaces at the beginning and end (you could actually invert by first trimming and then matching to make the regex quicker as someone pointed out).
To test this out quickly try:
System.out.println(new String(" hello there ").trim().replaceAll("\\s{2,}", " "));
and it will return:
"hello there"
Use the Apache commons StringUtils.normalizeSpace(String str) method. See docs here
This worked perfectly for me : sValue = sValue.trim().replaceAll("\\s+", " ");
trim() method removes the leading and trailing spaces and using replaceAll("regex", "string to replace") method with regex "\s+" matches more than one space and will replace it with a single space
myText = myText.trim().replaceAll("\\s+"," ");
The following code will compact any whitespace between words and remove any at the string's beginning and end
String input = "\n\n\n a string with many spaces, \n"+
" a \t tab and a newline\n\n";
String output = input.trim().replaceAll("\\s+", " ");
System.out.println(output);
This will output a string with many spaces, a tab and a newline
Note that any non-printable characters including spaces, tabs and newlines will be compacted or removed
For more information see the respective documentation:
String#trim() method
String#replaceAll(String regex, String replacement) method
For information about Java's regular expression implementation see the documentation of the Pattern class
"[ ]{2,}"
This will match more than one space.
String mytext = " hello there ";
//without trim -> " hello there"
//with trim -> "hello there"
mytext = mytext.trim().replaceAll("[ ]{2,}", " ");
System.out.println(mytext);
OUTPUT:
hello there
To eliminate spaces at the beginning and at the end of the String, use String#trim() method. And then use your mytext.replaceAll("( )+", " ").
You can first use String.trim(), and then apply the regex replace command on the result.
Try this one.
Sample Code
String str = " hello there ";
System.out.println(str.replaceAll("( +)"," ").trim());
OUTPUT
hello there
First it will replace all the spaces with single space. Than we have to supposed to do trim String because Starting of the String and End of the String it will replace the all space with single space if String has spaces at Starting of the String and End of the String So we need to trim them. Than you get your desired String.
String blogName = "how to do in java . com";
String nameWithProperSpacing = blogName.replaceAll("\\\s+", " ");
trim()
Removes only the leading & trailing spaces.
From Java Doc,
"Returns a string whose value is this string, with any leading and trailing whitespace removed."
System.out.println(" D ev Dum my ".trim());
"D ev Dum my"
replace(), replaceAll()
Replaces all the empty strings in the word,
System.out.println(" D ev Dum my ".replace(" ",""));
System.out.println(" D ev Dum my ".replaceAll(" ",""));
System.out.println(" D ev Dum my ".replaceAll("\\s+",""));
Output:
"DevDummy"
"DevDummy"
"DevDummy"
Note: "\s+" is the regular expression similar to the empty space character.
Reference : https://www.codedjava.com/2018/06/replace-all-spaces-in-string-trim.html
In Kotlin it would look like this
val input = "\n\n\n a string with many spaces, \n"
val cleanedInput = input.trim().replace(Regex("(\\s)+"), " ")
A lot of correct answers been provided so far and I see lot of upvotes. However, the mentioned ways will work but not really optimized or not really readable.
I recently came across the solution which every developer will like.
String nameWithProperSpacing = StringUtils.normalizeSpace( stringWithLotOfSpaces );
You are done.
This is readable solution.
You could use lookarounds also.
test.replaceAll("^ +| +$|(?<= ) ", "");
OR
test.replaceAll("^ +| +$| (?= )", "")
<space>(?= ) matches a space character which is followed by another space character. So in consecutive spaces, it would match all the spaces except the last because it isn't followed by a space character. This leaving you a single space for consecutive spaces after the removal operation.
Example:
String[] tests = {
" x ", // [x]
" 1 2 3 ", // [1 2 3]
"", // []
" ", // []
};
for (String test : tests) {
System.out.format("[%s]%n",
test.replaceAll("^ +| +$| (?= )", "")
);
}
See String.replaceAll.
Use the regex "\s" and replace with " ".
Then use String.trim.
String str = " hello world"
reduce spaces first
str = str.trim().replaceAll(" +", " ");
capitalize the first letter and lowercase everything else
str = str.substring(0,1).toUpperCase() +str.substring(1,str.length()).toLowerCase();
you should do it like this
String mytext = " hello there ";
mytext = mytext.replaceAll("( +)", " ");
put + inside round brackets.
String str = " this is string ";
str = str.replaceAll("\\s+", " ").trim();
This worked for me
scan= filter(scan, " [\\s]+", " ");
scan= sac.trim();
where filter is following function and scan is the input string:
public String filter(String scan, String regex, String replace) {
StringBuffer sb = new StringBuffer();
Pattern pt = Pattern.compile(regex);
Matcher m = pt.matcher(scan);
while (m.find()) {
m.appendReplacement(sb, replace);
}
m.appendTail(sb);
return sb.toString();
}
The simplest method for removing white space anywhere in the string.
public String removeWhiteSpaces(String returnString){
returnString = returnString.trim().replaceAll("^ +| +$|( )+", " ");
return returnString;
}
check this...
public static void main(String[] args) {
String s = "A B C D E F G\tH I\rJ\nK\tL";
System.out.println("Current : "+s);
System.out.println("Single Space : "+singleSpace(s));
System.out.println("Space count : "+spaceCount(s));
System.out.format("Replace all = %s", s.replaceAll("\\s+", ""));
// Example where it uses the most.
String s = "My name is yashwanth . M";
String s2 = "My nameis yashwanth.M";
System.out.println("Normal : "+s.equals(s2));
System.out.println("Replace : "+s.replaceAll("\\s+", "").equals(s2.replaceAll("\\s+", "")));
}
If String contains only single-space then replace() will not-replace,
If spaces are more than one, Then replace() action performs and removes spacess.
public static String singleSpace(String str){
return str.replaceAll(" +| +|\t|\r|\n","");
}
To count the number of spaces in a String.
public static String spaceCount(String str){
int i = 0;
while(str.indexOf(" ") > -1){
//str = str.replaceFirst(" ", ""+(i++));
str = str.replaceFirst(Pattern.quote(" "), ""+(i++));
}
return str;
}
Pattern.quote("?") returns literal pattern String.
My method before I found the second answer using regex as a better solution. Maybe someone needs this code.
private String replaceMultipleSpacesFromString(String s){
if(s.length() == 0 ) return "";
int timesSpace = 0;
String res = "";
for (int i = 0; i < s.length(); i++) {
char c = s.charAt(i);
if(c == ' '){
timesSpace++;
if(timesSpace < 2)
res += c;
}else{
res += c;
timesSpace = 0;
}
}
return res.trim();
}
Stream version, filters spaces and tabs.
Stream.of(str.split("[ \\t]")).filter(s -> s.length() > 0).collect(Collectors.joining(" "))
I know replaceAll method is much easier but I wanted to post this as well.
public static String removeExtraSpace(String input) {
input= input.trim();
ArrayList <String> x= new ArrayList<>(Arrays.asList(input.split("")));
for(int i=0; i<x.size()-1;i++) {
if(x.get(i).equals(" ") && x.get(i+1).equals(" ")) {
x.remove(i);
i--;
}
}
String word="";
for(String each: x)
word+=each;
return word;
}
String myText = " Hello World ";
myText = myText.trim().replace(/ +(?= )/g,'');
// Output: "Hello World"
string.replaceAll("\s+", " ");
If you already use Guava (v. 19+) in your project you may want to use this:
CharMatcher.whitespace().trimAndCollapseFrom(input, ' ');
or, if you need to remove exactly SPACE symbol ( or U+0020, see more whitespaces) use:
CharMatcher.anyOf(" ").trimAndCollapseFrom(input, ' ');
public class RemoveExtraSpacesEfficient {
public static void main(String[] args) {
String s = "my name is mr space ";
char[] charArray = s.toCharArray();
char prev = s.charAt(0);
for (int i = 0; i < charArray.length; i++) {
char cur = charArray[i];
if (cur == ' ' && prev == ' ') {
} else {
System.out.print(cur);
}
prev = cur;
}
}
}
The above solution is the algorithm with the complexity of O(n) without using any java function.
Please use below code
package com.myjava.string;
import java.util.StringTokenizer;
public class MyStrRemoveMultSpaces {
public static void main(String a[]){
String str = "String With Multiple Spaces";
StringTokenizer st = new StringTokenizer(str, " ");
StringBuffer sb = new StringBuffer();
while(st.hasMoreElements()){
sb.append(st.nextElement()).append(" ");
}
System.out.println(sb.toString().trim());
}
}