Develop Reference

Java regex: Replace all characters with `+` except instances of a given string

I have the following problem which states
Replace all characters in a string with + symbol except instances of the given string in the method
so for example if the string given was abc123efg and they want me to replace every character except every instance of 123 then it would become +++123+++.
I figured a regular expression is probably the best for this and I came up with this.
str.replaceAll("[^str]","+")
where str is a variable, but its not letting me use the method without putting it in quotations. If I just want to replace the variable string str how can I do that? I ran it with the string manually typed and it worked on the method, but can I just input a variable?
as of right now I believe its looking for the string "str" and not the variable string.
Here is the output its right for so many cases except for two :(
List of open test cases:
plusOut("12xy34", "xy") → "++xy++"
plusOut("12xy34", "1") → "1+++++"
plusOut("12xy34xyabcxy", "xy") → "++xy++xy+++xy"
plusOut("abXYabcXYZ", "ab") → "ab++ab++++"
plusOut("abXYabcXYZ", "abc") → "++++abc+++"
plusOut("abXYabcXYZ", "XY") → "++XY+++XY+"
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
plusOut("--++ab", "++") → "++++++"
plusOut("aaxxxxbb", "xx") → "++xxxx++"
plusOut("123123", "3") → "++3++3"

Looks like this is the plusOut problem on CodingBat.
I had 3 solutions to this problem, and wrote a new streaming solution just for fun.
Solution 1: Loop and check
Create a StringBuilder out of the input string, and check for the word at every position. Replace the character if doesn't match, and skip the length of the word if found.
public String plusOut(String str, String word) {
StringBuilder out = new StringBuilder(str);
for (int i = 0; i < out.length(); ) {
if (!str.startsWith(word, i))
out.setCharAt(i++, '+');
else
i += word.length();
}
return out.toString();
}
This is probably the expected answer for a beginner programmer, though there is an assumption that the string doesn't contain any astral plane character, which would be represented by 2 char instead of 1.
Solution 2: Replace the word with a marker, replace the rest, then restore the word
public String plusOut(String str, String word) {
return str.replaceAll(java.util.regex.Pattern.quote(word), "#").replaceAll("[^#]", "+").replaceAll("#", word);
}
Not a proper solution since it assumes that a certain character or sequence of character doesn't appear in the string.
Note the use of Pattern.quote to prevent the word being interpreted as regex syntax by replaceAll method.
Solution 3: Regex with \G
public String plusOut(String str, String word) {
word = java.util.regex.Pattern.quote(word);
return str.replaceAll("\\G((?:" + word + ")*+).", "$1+");
}
Construct regex \G((?:word)*+)., which does more or less what solution 1 is doing:
\G makes sure the match starts from where the previous match leaves off
((?:word)*+) picks out 0 or more instance of word - if any, so that we can keep them in the replacement with $1. The key here is the possessive quantifier *+, which forces the regex to keep any instance of the word it finds. Otherwise, the regex will not work correctly when the word appear at the end of the string, as the regex backtracks to match .
. will not be part of any word, since the previous part already picks out all consecutive appearances of word and disallow backtrack. We will replace this with +
Solution 4: Streaming
public String plusOut(String str, String word) {
return String.join(word,
Arrays.stream(str.split(java.util.regex.Pattern.quote(word), -1))
.map((String s) -> s.replaceAll("(?s:.)", "+"))
.collect(Collectors.toList()));
}
The idea is to split the string by word, do the replacement on the rest, and join them back with word using String.join method.
Same as above, we need Pattern.quote to avoid split interpreting the word as regex. Since split by default removes empty string at the end of the array, we need to use -1 in the second parameter to make split leave those empty strings alone.
Then we create a stream out of the array and replace the rest as strings of +. In Java 11, we can use s -> String.repeat(s.length()) instead.
The rest is just converting the Stream to an Iterable (List in this case) and joining them for the result

This is a bit trickier than you might initially think because you don't just need to match characters, but the absence of specific phrase - a negated character set is not enough. If the string is 123, you would need:
(?<=^|123)(?!123).*?(?=123|$)
https://regex101.com/r/EZWMqM/1/
That is - lookbehind for the start of the string or "123", make sure the current position is not followed by 123, then lazy-repeat any character until lookahead matches "123" or the end of the string. This will match all characters which are not in a "123" substring. Then, you need to replace each character with a +, after which you can use appendReplacement and a StringBuffer to create the result string:
String inputPhrase = "123";
String inputStr = "abc123efg123123hij";
StringBuffer resultString = new StringBuffer();
Pattern regex = Pattern.compile("(?<=^|" + inputPhrase + ")(?!" + inputPhrase + ").*?(?=" + inputPhrase + "|$)");
Matcher m = regex.matcher(inputStr);
while (m.find()) {
String replacement = m.group(0).replaceAll(".", "+");
m.appendReplacement(resultString, replacement);
}
m.appendTail(resultString);
System.out.println(resultString.toString());
Output:
+++123+++123123+++
Note that if the inputPhrase can contain character with a special meaning in a regular expression, you'll have to escape them first before concatenating into the pattern.

You can do it in one line:
input = input.replaceAll("((?:" + str + ")+)?(?!" + str + ").((?:" + str + ")+)?", "$1+$2");
This optionally captures "123" either side of each character and puts them back (a blank if there's no "123"):

So instead of coming up with a regular expression that matches the absence of a string. We might as well just match the selected phrase and append + the number of skipped characters.
StringBuilder sb = new StringBuilder();
Matcher m = Pattern.compile(Pattern.quote(str)).matcher(input);
while (m.find()) {
for (int i = 0; i < m.start(); i++) sb.append('+');
sb.append(str);
}
int remaining = input.length() - sb.length();
for (int i = 0; i < remaining; i++) {
sb.append('+');
}

Absolutely just for the fun of it, a solution using CharBuffer (unexpectedly it took a lot more that I initially hoped for):
private static String plusOutCharBuffer(String input, String match) {
int size = match.length();
CharBuffer cb = CharBuffer.wrap(input.toCharArray());
CharBuffer word = CharBuffer.wrap(match);
int x = 0;
for (; cb.remaining() > 0;) {
if (!cb.subSequence(0, size < cb.remaining() ? size : cb.remaining()).equals(word)) {
cb.put(x, '+');
cb.clear().position(++x);
} else {
cb.clear().position(x = x + size);
}
}
return cb.clear().toString();
}

To make this work you need a beast of a pattern. Let's say you you are operating on the following test case as an example:
plusOut("abXYxyzXYZ", "XYZ") → "+++++++XYZ"
What you need to do is build a series of clauses in your pattern to match a single character at a time:
Any character that is NOT "X", "Y" or "Z" -- [^XYZ]
Any "X" not followed by "YZ" -- X(?!YZ)
Any "Y" not preceded by "X" -- (?<!X)Y
Any "Y" not followed by "Z" -- Y(?!Z)
Any "Z" not preceded by "XY" -- (?<!XY)Z
An example of this replacement can be found here: https://regex101.com/r/jK5wU3/4
Here is an example of how this might work (most certainly not optimized, but it works):
import java.util.regex.Pattern;
public class Test {
public static void plusOut(String text, String exclude) {
StringBuilder pattern = new StringBuilder("");
for (int i=0; i<exclude.length(); i++) {
Character target = exclude.charAt(i);
String prefix = (i > 0) ? exclude.substring(0, i) : "";
String postfix = (i < exclude.length() - 1) ? exclude.substring(i+1) : "";
// add the look-behind (?<!X)Y
if (!prefix.isEmpty()) {
pattern.append("(?<!").append(Pattern.quote(prefix)).append(")")
.append(Pattern.quote(target.toString())).append("|");
}
// add the look-ahead X(?!YZ)
if (!postfix.isEmpty()) {
pattern.append(Pattern.quote(target.toString()))
.append("(?!").append(Pattern.quote(postfix)).append(")|");
}
}
// add in the other character exclusion
pattern.append("[^" + Pattern.quote(exclude) + "]");
System.out.println(text.replaceAll(pattern.toString(), "+"));
}
public static void main(String [] args) {
plusOut("12xy34", "xy");
plusOut("12xy34", "1");
plusOut("12xy34xyabcxy", "xy");
plusOut("abXYabcXYZ", "ab");
plusOut("abXYabcXYZ", "abc");
plusOut("abXYabcXYZ", "XY");
plusOut("abXYxyzXYZ", "XYZ");
plusOut("--++ab", "++");
plusOut("aaxxxxbb", "xx");
plusOut("123123", "3");
}
}
UPDATE: Even this doesn't quite work because it can't deal with exclusions that are just repeated characters, like "xx". Regular expressions are most definitely not the right tool for this, but I thought it might be possible. After poking around, I'm not so sure a pattern even exists that might make this work.

The problem in your solution that you put a set of instance string str.replaceAll("[^str]","+") which it will exclude any character from the variable str and that will not solve your problem
EX: when you try str.replaceAll("[^XYZ]","+") it will exclude any combination of character X , character Y and character Z from your replacing method so you will get "++XY+++XYZ".
Actually you should exclude a sequence of characters instead in str.replaceAll.
You can do it by using capture group of characters like (XYZ) then use a negative lookahead to match a string which does not contain characters sequence : ^((?!XYZ).)*$
Check this solution for more info about this problem but you should know that it may be complicated to find regular expression to do that directly.
I have found two simple solutions for this problem :
Solution 1:
You can implement a method to replace all characters with '+' except the instance of given string:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
for(int i = 0; i < str.length(); i++){
// exclude any instance string of exWord from replacing process in str
if(str.substring(i, str.length()).indexOf(exWord) + i == i){
i = i + exWord.length()-1;
}
else{
str = str.substring(0,i) + "+" + str.substring(i+1);//replace each character with '+' symbol
}
}
Note : str.substring(i, str.length()).indexOf(exWord) + i this if statement will exclude any instance string of exWord from replacing process in str.
Output:
+++++++XYZ
Solution 2:
You can try this Approach using ReplaceAll method and it doesn't need any complex regular expression:
String exWord = "XYZ";
String str = "abXYxyzXYZ";
str = str.replaceAll(exWord,"*"); // replace instance string with * symbol
str = str.replaceAll("[^*]","+"); // replace all characters with + symbol except *
str = str.replaceAll("\\*",exWord); // replace * symbol with instance string
Note : This solution will work only if your input string str doesn't contain any * symbol.
Also you should escape any character with a special meaning in a regular expression in phrase instance string exWord like : exWord = "++".

How to remove single and double quotes at both ends of a string

I would like to remove single or double quotes from both ends of a string. The string may contain additional quotes or/and double quotes which shall remain untouched - so removeAll() is not an option.
String one = "\"some string\"";
String two = "'some \"other string\"'";
// expected result
// some string
// some "other string"
What I tried so far:
two = two.replace("/^[\"\'])|([\"\']$/g", "");
The following would work but there must be a much more elegant way to achieve this..
if ((one != null && one.length() > 1) && ((one.startsWith("\"") && one.endsWith("\"")) ||
(one.startsWith("\'") && one.endsWith("\'")))) {
one = one.substring(1, one.length() - 1);
}
Any ideas?
Update / clarification
My use case is the command line interface of an app, where the user can also drag files/paths into, instead of typing them.
Under Windows the dragged files are beeing surrounded by double quotes, under Linux with single quotes. All I want to do is get rid of them. So in my use case the quotes are always symetric (they match).
But I can perfectly live with a solution, which would strip them even if they wouldn't match, because they always do

Option 1: Removing all single and double quotes from start and end
You can use replaceAll which accepts a regular expression - replace doesn't - and do it twice, once for quotes at the start of the string and once for quotes at the end:
public class Test {
public static void main(String[] args) {
String text = "\"'some \"other string\"'";
String trimmed = text
.replaceAll("^['\"]*", "")
.replaceAll("['\"]*$", "");
System.out.println(trimmed);
}
}
The ^ in the first replacement anchors the quotes to the start of the string; the $ in the second anchors the quotes to the end of the string.
Note that this doesn't try to "match" quotes at all, unlike your later code.
Option 2: Removing a single quote character from start and end, if they match
String trimmed = text.replaceAll("^(['\"])(.*)\\1$", "$2");
This will trim exactly one character from the start and end, if they both match. Sample:
public class Test {
public static void main(String[] args) {
trim("\"foo\"");
trim("'foo'");
trim("\"bar'");
trim("'bar\"");
trim("\"'baz'\"");
}
static void trim(String text) {
String trimmed = text.replaceAll("^(['\"])(.*)\\1$", "$2");
System.out.println(text + " => " + trimmed);
}
}
Output:
"foo" => foo
'foo' => foo
"bar' => "bar'
'bar" => 'bar"
"'baz'" => 'baz'

To complete Jon Skeet response, if you want to remove quotes only if there is one on the beginning AND one on the end you can do :
public String removeQuotes(String str) {
Pattern pattern = Pattern.compile("^['\"](.*)['\"]$");
Matcher matcher = pattern.matcher(str);
if (matcher.find()) {
return matcher.group(1);
} else {
return str;
}
}

if you are looking in javascript try this :
function t(k){
var l="\"\'"; //you can add more chars here.
if (l.indexOf(k[0])>-1) {
return t(k.substr(1,k.length));
} else if (l.indexOf(k[k.length-1])>-1) {
return t(k.substr(0,k.length-1));
} else {
return k;
}
}

One possible way with using replaceFirst():
String one = "\"some string\"";
System.out.println("one: " + one);
one = one.replaceFirst("\"", "");
String reversed = new StringBuilder(one).reverse().toString();
one = one.replaceFirst("\"", "");
one = new StringBuilder(reversed).reverse().toString();
System.out.println("result: " + one);

Regex for int Array

I need to validate a string argument if it can be converted to an int Array.
String pattern = "(\\d+[,])+";
String test = "18,32,5,8,10";
test2.matches(pattern2) //returns false as i requires , in the end
Is there any way to ignore last ' , '

Use a group construct to specify that digits should be followed by (, digits) ...
\\d+(?:,\\d+)+

Regex for both array and single element
(\d|\[(\d|,\s*)*])

This regex will work for you (checks for a "valid array") :
public static void main(String[] args) {
String s = "18,32,5,8,10";
System.out.println(s.matches("(?!.*,$)(?!,.*$)(\\d+(?:,|$))+"));
}
Checks and fails for:
multiple continuous commas
comma at beginning
comma at end

You can try
Pattern = "^\d+(,\d+)*$"
text = "10,5,10" (TRUE)
text = "10,5,10," (FALSE)
text = "10,5,10 " (FALSE)

Since I don't know how to use regex and if I were in your place then this would have been my way to do so
String test = "18,32,5,8,10";
String str[]=test.split(",");
int ar[] = new int[str.length];
for(int i = 0; i<ar.length; i++){
ar[i] = Integer.parseInt(str[i]);
}
The problem in this code if any I can see is this that call to parseInt() method must be wrapped in try-catch as it can throw NumberFormatException if your string contains value other than digit and comma(,).

Remove trailing substring from String in Java

I am looking to remove parts of a string if it ends in a certain string.
An example would be to take this string: "am.sunrise.ios#2x.png"
And remove the #2x.png so it looks like: "am.sunrise.ios"
How would I go about checking to see if the end of a string contains "#2x.png" and remove it?

You could check the lastIndexOf, and if it exists in the string, use substring to remove it:
String str = "am.sunrise.ios#2x.png";
String search = "#2x.png";
int index = str.lastIndexOf(search);
if (index > 0) {
str = str.substring(0, index);
}

Assuming you have a string initialized as String file = "am.sunrise.ios#2x.png";.
if(file.endsWith("#2x.png"))
file = file.substring(0, file.lastIndexOf("#2x.png"));
The endsWith(String) method returns a boolean determining if the string has a certain suffix. Depending on that you can replace the string with a substring of itself starting with the first character and ending before the index of the character that you are trying to remove.

private static String removeSuffixIfExists(String key, String suffix) {
return key.endswith(suffix)
? key.substring(0, key.length() - suffix.length())
: key;
}
}
String suffix = "#2x.png";
String key = "am.sunrise.ios#2x.png";
String output = removeSuffixIfExists(key, suffix);

public static void main(String [] args){
String word = "am.sunrise.ios#2x.png";
word = word.replace("#2x.png", "");
System.out.println(word);
}

If you want to generally remove entire content of string from # till end you can use
yourString = yourString.replaceAll("#.*","");
where #.* is regex (regular expression) representing substring starting with # and having any character after it (represented by .) zero or more times (represented by *).
In case there will be no #xxx part your string will be unchanged.
If you want to change only this particular substring #2x.png (and not substirng like #3x.png) while making sure that it is placed at end of your string you can use
yourString = yourString.replaceAll("#2x\\.png$","");
where
$ represents end of string
\\. represents . literal (we need to escape it since like shown earlier . is metacharacter representing any character)

Since I was trying to do this on an ArrayList of items similarly styled I ended up using the following code:
for (int image = 0; image < IBImages.size(); image++) {
IBImages.set(image, IBImages.get(image).split("~")[0].split("#")[0].split(".png")[0]);
}
If I have a list of images with the names
[am.sunrise.ios.png, am.sunrise.ios#2x.png, am.sunrise.ios#3x.png, am.sunrise.ios~ipad.png, am.sunrise.ios~ipad#2x.png]
This allows me to split the string into 2 parts.
For example, "am.sunrise.ios~ipad.png" will be split into "am.sunrise.ios" and "~ipad.png" if I split on "~". I can just get the first part back by referencing [0]. Therefore I get what I'm looking for in one line of code.
Note that image is "am.sunrise.ios~ipad.png"

You could use String.split():
public static void main(String [] args){
String word = "am.sunrise.ios#2x.png";
String[] parts = word.split("#");
if (parts.length == 2) {
System.out.println("looks like user#host...");
System.out.println("User: " + parts[0]);
System.out.println("Host: " + parts[1]);
}
}
Then you haven an array of Strings, where the first element contains the part before "#" and the second element the part after the "#".

Combining the answers 1 and 2:
String str = "am.sunrise.ios#2x.png";
String search = "#2x.png";
if (str.endsWith(search)) {
str = str.substring(0, str.lastIndexOf(search));
}

Remove all occurrences of char from string

I can use this:
String str = "TextX Xto modifyX";
str = str.replace('X','');//that does not work because there is no such character ''
Is there a way to remove all occurrences of character X from a String in Java?
I tried this and is not what I want: str.replace('X',' '); //replace with space

Try using the overload that takes CharSequence arguments (eg, String) rather than char:
str = str.replace("X", "");

Using
public String replaceAll(String regex, String replacement)
will work.
Usage would be str.replace("X", "");.
Executing
"Xlakjsdf Xxx".replaceAll("X", "");
returns:
lakjsdf xx

If you want to do something with Java Strings, Commons Lang StringUtils is a great place to look.
StringUtils.remove("TextX Xto modifyX", 'X');

String test = "09-09-2012";
String arr [] = test.split("-");
String ans = "";
for(String t : arr)
ans+=t;
This is the example for where I have removed the character - from the String.

Hello Try this code below
public class RemoveCharacter {
public static void main(String[] args){
String str = "MXy nameX iXs farXazX";
char x = 'X';
System.out.println(removeChr(str,x));
}
public static String removeChr(String str, char x){
StringBuilder strBuilder = new StringBuilder();
char[] rmString = str.toCharArray();
for(int i=0; i<rmString.length; i++){
if(rmString[i] == x){
} else {
strBuilder.append(rmString[i]);
}
}
return strBuilder.toString();
}
}

I like using RegEx in this occasion:
str = str.replace(/X/g, '');
where g means global so it will go through your whole string and replace all X with '';
if you want to replace both X and x, you simply say:
str = str.replace(/X|x/g, '');
(see my fiddle here: fiddle)

Use replaceAll instead of replace
str = str.replaceAll("X,"");
This should give you the desired answer.

Evaluation of main answers with a performance benchmark which confirms concerns that the current chosen answer makes costly regex operations under the hood
To date the provided answers come in 3 main styles (ignoring the JavaScript answer ;) ):
Use String.replace(charsToDelete, ""); which uses regex under the hood
Use Lambda
Use simple Java implementation
In terms of code size clearly the String.replace is the most terse. The simple Java implementation is slightly smaller and cleaner (IMHO) than the Lambda (don't get me wrong - I use Lambdas often where they are appropriate)
Execution speed was, in order of fastest to slowest: simple Java implementation, Lambda and then String.replace() (that invokes regex).
By far the fastest implementation was the simple Java implementation tuned so that it preallocates the StringBuilder buffer to the max possible result length and then simply appends chars to the buffer that are not in the "chars to delete" string. This avoids any reallocates that would occur for Strings > 16 chars in length (the default allocation for StringBuilder) and it avoids the "slide left" performance hit of deleting characters from a copy of the string that occurs is the Lambda implementation.
The code below runs a simple benchmark test, running each implementation 1,000,000 times and logs the elapsed time.
The exact results vary with each run but the order of performance never changes:
Start simple Java implementation
Time: 157 ms
Start Lambda implementation
Time: 253 ms
Start String.replace implementation
Time: 634 ms
The Lambda implementation (as copied from Kaplan's answer) may be slower because it performs a "shift left by one" of all characters to the right of the character being deleted. This would obviously get worse for longer strings with lots of characters requiring deletion. Also there might be some overhead in the Lambda implementation itself.
The String.replace implementation, uses regex and does a regex "compile" at each call. An optimization of this would be to use regex directly and cache the compiled pattern to avoid the cost of compiling it each time.
package com.sample;
import java.util.function.BiFunction;
import java.util.stream.IntStream;
public class Main {
static public String deleteCharsSimple(String fromString, String charsToDelete)
{
StringBuilder buf = new StringBuilder(fromString.length()); // Preallocate to max possible result length
for(int i = 0; i < fromString.length(); i++)
if (charsToDelete.indexOf(fromString.charAt(i)) < 0)
buf.append(fromString.charAt(i)); // char not in chars to delete so add it
return buf.toString();
}
static public String deleteCharsLambda(String fromString1, String charsToDelete)
{
BiFunction<String, String, String> deleteChars = (fromString, chars) -> {
StringBuilder buf = new StringBuilder(fromString);
IntStream.range(0, buf.length()).forEach(i -> {
while (i < buf.length() && chars.indexOf(buf.charAt(i)) >= 0)
buf.deleteCharAt(i);
});
return (buf.toString());
};
return deleteChars.apply(fromString1, charsToDelete);
}
static public String deleteCharsReplace(String fromString, String charsToDelete)
{
return fromString.replace(charsToDelete, "");
}
public static void main(String[] args)
{
String str = "XXXTextX XXto modifyX";
String charsToDelete = "X"; // Should only be one char as per OP's requirement
long start, end;
System.out.println("Start simple");
start = System.currentTimeMillis();
for (int i = 0; i < 1000000; i++)
deleteCharsSimple(str, charsToDelete);
end = System.currentTimeMillis();
System.out.println("Time: " + (end - start));
System.out.println("Start lambda");
start = System.currentTimeMillis();
for (int i = 0; i < 1000000; i++)
deleteCharsLambda(str, charsToDelete);
end = System.currentTimeMillis();
System.out.println("Time: " + (end - start));
System.out.println("Start replace");
start = System.currentTimeMillis();
for (int i = 0; i < 1000000; i++)
deleteCharsReplace(str, charsToDelete);
end = System.currentTimeMillis();
System.out.println("Time: " + (end - start));
}
}

You will need to put the characters needs to be removed inside the square brackets during the time of replacement. The example code will be as following:
String s = "$116.42".replaceAll("[$]", "");

here is a lambda function which removes all characters passed as string
BiFunction<String,String,String> deleteChars = (fromString, chars) -> {
StringBuilder buf = new StringBuilder( fromString );
IntStream.range( 0, buf.length() ).forEach( i -> {
while( i < buf.length() && chars.indexOf( buf.charAt( i ) ) >= 0 )
buf.deleteCharAt( i );
} );
return( buf.toString() );
};
String str = "TextX XYto modifyZ";
deleteChars.apply( str, "XYZ" ); // –> "Text to modify"
This solution takes into acount that the resulting String – in difference to replace() – never becomes larger than the starting String when removing characters. So it avoids the repeated allocating and copying while appending character-wise to the StringBuilder as replace() does.
Not to mention the pointless generation of Pattern and Matcher instances in replace() that are never needed for removal.
In difference to replace() this solution can delete several characters in one swoop.

…another lambda
copying a new string from the original, but leaving out the character that is to delete
String text = "removing a special character from a string";
int delete = 'e';
int[] arr = text.codePoints().filter( c -> c != delete ).toArray();
String rslt = new String( arr, 0, arr.length );
gives: rmoving a spcial charactr from a string

package com.acn.demo.action;
public class RemoveCharFromString {
static String input = "";
public static void main(String[] args) {
input = "abadbbeb34erterb";
char token = 'b';
removeChar(token);
}
private static void removeChar(char token) {
// TODO Auto-generated method stub
System.out.println(input);
for (int i=0;i<input.length();i++) {
if (input.charAt(i) == token) {
input = input.replace(input.charAt(i), ' ');
System.out.println("MATCH FOUND");
}
input = input.replaceAll(" ", "");
System.out.println(input);
}
}
}

You can use str = str.replace("X", ""); as mentioned before and you will be fine. For your information '' is not an empty (or a valid) character but '\0' is.
So you could use str = str.replace('X', '\0'); instead.