Develop Reference

Java PatternSyntaxException when replacing characters in a String object

I am trying to create string of this list without the following character , [] as will I want to replace all two spaces after deleting them.
I have tried the following but I am geting the error in the title.
Simple:
[06:15, 06:45, 07:16, 07:46]
Result should look as this:
06:15 06:45 07:16 07:46
Code:
List<String> value = entry.getValue();
String timeEntries = value.toString();
String after = timeEntries.replaceAll(",", " ");
String after2 = after.replaceAll(" ", " ");
String after3 = after2.replaceAll("[", "");
String after4 = after3.replaceAll("]", "");

replaceAll replaces all occurrences that match a given regular expression. Since you just want to match a simple string, you should use replace instead:
List<String> value = entry.getValue();
String timeEntries = value.toString();
String after = timeEntries.replace(",", " ");
String after2 = after.replace(" ", " ");
String after3 = after2.replace("[", "");
String after4 = after3.replace("]", "");

To answer the main question, if you use replaceAll, make sure your 1st argument is a valid regular expression. For your example, you can actually reduce it to 2 calls to replaceAll, as 2 of the substitutions are identical.
List<String> value = entry.getValue();
String timeEntries = value.toString();
String after = timeEntries.replaceAll("[, ]", " ");
String after2 = after.replaceAll("\\[|\\]", "");
But, it looks like you're just trying to concatenate all the elements of a String list together. It's much more efficient to construct this string directly, by iterating your list and using StringBuilder:
StringBuilder builder = new StringBuilder();
for (String timeEntry: entry.getValue()) {
builder.append(timeEntry);
}
String after = builder.toString();

Splitting a string based on " " and spaces [duplicate]

This question already has answers here:
Regular Expression to Split String based on space and matching quotes in java
(3 answers)
Closed 8 years ago.
I have a String str, which is comprised of several words separated by single spaces.
If I want to create a set or list of strings I can simply call str.split(" ") and I would get I want.
Now, assume that str is a little more complicated, for example it is something like:
str = "hello bonjour \"good morning\" buongiorno";
In this case what is in between " " I want to keep so that my list of strings is:
hello
bonjour
good morning
buongiorno
Clearly, if I used split(" ") in this case it won't work because I'd get
hello
bonjour
"good
morning"
buongiorno
So, how do I get what I want?

You can create a regex that finds every word or words between "".. like:
\w+|(\"\w+(\s\w+)*\")
and search for them with the Pattern and Matcher classes.
ex.
String searchedStr = "";
Pattern pattern = Pattern.compile("\\w+|(\\\"\\w+(\\s\\w+)*\\\")");
Matcher matcher = pattern.matcher(searchedStr);
while(matcher.find()){
String word = matcher.group();
}
Edit: works for every number of words within "" now. XD forgot that

You can do something like below. First split the Sting using "\"" and then split the remaining ones using space" " . The even tokens will be the ones between quotes "".
public static void main(String args[]) {
String str = "hello bonjour \"good morning\" buongiorno";
System.out.println(str);
String[] parts = str.split("\"");
List<String> myList = new ArrayList<String>();
int i = 1;
for(String partStr : parts) {
if(i%2 == 0){
myList.add(partStr);
}
else {
myList.addAll(Arrays.asList(partStr.trim().split(" ")));
}
i++;
}
System.out.println("MyList : " + myList);
}
and the output is
hello bonjour "good morning" buongiorno
MyList : [hello, bonjour, good morning, buongiorno]

You may be able to find a solution using regular expressions, but what I'd do is simply manually write a string breaker.
List<String> splitButKeepQuotes(String s, char splitter) {
ArrayList<String> list = new ArrayList<String>();
boolean inQuotes = false;
int startOfWord = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == splitter && !inQuotes && i != startOfWord) {
list.add(s.substring(startOfWord, i));
startOfWord = i + 1;
}
if (s.charAt(i) == "\"") {
inQuotes = !inQuotes;
}
}
return list;
}

Remove first word from a string in Java

What's the best way to remove the first word from a string in Java?
If I have
String originalString = "This is a string";
I want to remove the first word from it and in effect form two strings -
removedWord = "This"
originalString = "is a string;

Simple.
String o = "This is a string";
System.out.println(Arrays.toString(o.split(" ", 2)));
Output :
[This, is a string]
EDIT:
In line 2 below the values are stored in the arr array. Access them like normal arrays.
String o = "This is a string";
String [] arr = o.split(" ", 2);
arr[0] // This
arr[1] // is a string

You can use substring
removedWord = originalString.substring(0,originalString.indexOf(' '));
originalString = originalString.substring(originalString.indexOf(' ')+1);

This will definitely a good solution
String originalString = "This is a string";
originalString =originalString.replaceFirst("This ", "");

Try this using an index var, I think it's quite efficient :
int spaceIdx = originalString.indexOf(" ");
String removedWord = originalString.substring(0,spaceIdx);
originalString = originalString.substring(spaceIdx);
Prior to JDK 1.7 using below method might be more efficient, especially if you are using long string (see this article).
originalString = new String(originalString.substring(spaceIdx));

For an immediate answer you can use this :
removeWord = originalString.substring(0,originalString.indexOf(' '));
originalString = originalString.substring(originalString.indexOf(' '));

You can check where is the first space character and seperate string.
String full = "Sample Text";
String cut;
int pointToCut = full.indexOf( ' ');
if ( offset > -1)
{
cut = full.substring( space + 1);
}

String str = "This is a string";
String str2=str.substring(str.indexOf(" "));
String str3=str.replaceFirst(str2, "");
String's replaceFirst and substring

also you can use this solution:
static String substringer(String inputString, String remove) {
if (inputString.substring(0, remove.length()).equalsIgnoreCase(remove)) {
return inputString.substring(remove.length()).trim();
}
else {
return inputString.trim();
}
}
Example :
substringer("This is a string", "This");

You can use the StringTokenizer class.

Insert a character before and after all letters in a string in Java

I want to insert a % character before after every letter in a string, but using StringBuilder to make it fast.
For example, if a string is 'AA' then it would be '%A%A%'. If it is 'XYZ' then it would be '%X%Y%Z%'

String foo = "VWXYZ";
foo = "%" + foo.replaceAll("(.)","$1%");
System.out.println(foo);
Output:
%V%W%X%Y%Z%
You don't need a StringBuilder. The compiler will take care of that simple concatenation prior to the regex for you by using one.
Edit in response to comment below:
replaceAll() uses a Regular Expression (regex).
The regex (.) says "match any character, and give me a reference to it" . is a wildcard for any character, the parenthesis create the backreference. The $1 in the second argument says "Use backreference #1 from the match".
replaceAll() keeps running this expression over the whole string replacing each character with itself followed by a percent sign, building a new String which it then returns to you.

Try something like this:
String test = "ABC";
StringBuilder builder = new StringBuilder("");
builder.append("%");
for (char achar : test.toCharArray()) {
builder.append(achar);
builder.append("%");
}
System.out.println(builder.toString());

public static String escape(String s) {
StringBuilder buf = new StringBuilder();
boolean wasLetter = false;
for (char c: s.toCharArray()) {
boolean isLetter = Character.isLetter(c);
if (isLetter && !wasLetter) {
buf.append('%');
}
buf.append(c);
if (isLetter) {
buf.append('%');
}
wasLetter = isLetter;
}
return buf.toString();
}

StringBuilder sb = new StringBuilder("AAAAAAA");
for(int i = sb.length(); i >= 0; i--)
{
sb.insert(i, '%');
}

You may see this.
String s="AAAA";
StringBuilder builder = new StringBuilder();
char[] ch=s.toCharArray();
for(int i=0;i<ch.length;i++)
{
builder.append("%"+ch[i]);
}
builder.append("%");
System.out.println(builder.toString());
Output
%A%A%A%A%

I agree with #Brian Roach to add character to before and after but if you want to add any specific character then do like this
String source = "hello good old world";
StringBuffer res = new StringBuffer();
String[] strArr = tagList.split(" ");
for (String str : strArr) {
char[] stringArray = str.trim().toCharArray();
stringArray[0] = stringArray[0];
str = new String(stringArray);
//here you need to specify your first and last character which you want to set
res.append("#"+ str + "$").append(" ");
}
System.out.println("Result: " + res.toString().trim());
Output :- #hello$ #good$ #old$ #world$

Java how to replace 2 or more spaces with single space in string and delete leading and trailing spaces

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());
}
}