I have the string like:
TEST
FURNITURE-34_TEST>
My requirement is to remove all those junk characters from the above string.
so my expected output will be:
TEST FURNITURE-34_TEST
I have tried the below code
public static String removeUnPrintableChars(String str) {
if (str != null) {
str = str.replaceAll("[^\\x00-\\x7F]", "");
str = str.replaceAll("[\\p{Cntrl}&&[^\r\n\t]]", "");
str = str.replaceAll("\\p{C}", "");
str = str.replaceAll("\\P{Print}", "");
str = str.substring(0, Math.min(256, str.length()));
str = str.trim();
if (str.isEmpty()) {
str = null;
}
}
return str;
}
But it does nothing. Instead of finding and replacing each character as empty, can anyone please help me with the generic solution to replace those kinds of characters from the string?
Simple way to split a string :
public class Trim {
public static void main(String[] args) {
String myString = "TEST FURNITURE-34_TEST&"
+ "amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;amp;#38;amp;amp;"
+ "#38;amp;#38;gt;";
String[] parts = myString.split("&");
String part1 = parts[0];
System.out.println(parts[0]);
}
}
Link to original thread :
How to split a string in Java
The sample strings you are presenting (within your post and in comments) are rather ridiculous and in my opinion, whatever is generating them should be burned....twice.
Try the following method on your string(s). Add whatever you like to have removed from the input string by adding it to the 2D removableItems String Array. This 2D array contains preparation strings for the String#replaceAll() method. The first element of each row contains a Regular Expression (regex) of a particular string item to replace and the second element of each row contains the string item to replace the found items with.
public static String cleanString(String inputString) {
String[][] removableItems = {
{"(&?amp;){1,}", " "},
{"(#38);?", ""},
{"gt;", ""}, {"lt;", ""}
};
String desiredString = inputString;
for (int i = 0; i < removableItems.length; i++) {
desiredString = desiredString.replaceAll(removableItems[i][0],
removableItems[i][1]).trim();
}
return desiredString;
}
You can use this method. This is work with marking word boundaries.
public static String removeUnPrintableChars(String str) {
if(str != null){
str = str.replaceAll("(\\b&?\\w+;#?)", "");
}
return str;
}
I have a string like 23.Piano+trompet, and i wanted to remove the 23. part from the string using this function:
private String removeSignsFromName(String name) {
name = name.replaceAll(" ", "");
name = name.replaceAll(".", "");
return name.replaceAll("\\^([0-9]+)", "");
}
But it doesn't do it. Also, there is no error in runtime.
The following replaces all whitespace characters (\\s), dots (\\.), and digits (\\d) with "":
name.replaceAll("^[\\s\\.\\d]+", "");
what if I want to replace the + with _?
name.replaceAll("^[\\s\\.\\d]+", "").replaceAll("\\+", "_");
You don't need to escape the ^, you can use \\d+ to match multiple digits, and \\. for a literal dot and you don't need multiple calls to replaceAll. For example,
private static String removeSignsFromName(String name) {
return name.replaceAll("^\\d+\\.", "");
}
Which I tested like
public static void main(String[] args) {
System.out.println(removeSignsFromName("23.Piano+trompet"));
}
And got
Piano+trompet
Two problems:
The . in the second replaceAll should be escaped:
name=name.replaceAll("\\.", "");
The ^ in the third one should NOT be escaped:
return name.replaceAll("^([0-9]+)", "");
Oh! and the parentheses are useless since you don't use the captured string.
How about this：
public static String removeNumOfStr(String str) {
if (str == null) {
return null;
}
char[] ch = str.toCharArray();
int length = ch.length;
StringBuilder sb = new StringBuilder();
int i = 0;
while (i < length) {
if (Character.isDigit(ch[i])) {
i++;
} else {
sb.append(ch[i]);
i++;
}
}
return sb.toString();
}
return name.replaceFirst("^\\d+\\.", "");
public static void removenum(String str){
char[] arr=str.toCharArray();
String s="";
for(char ch:arr){
if(!(ch>47 & ch<57)){
s=s+ch;
}
}
System.out.println(s);
}
How can I trim the leading or trailing characters from a string in java?
For example, the slash character "/" - I'm not interested in spaces, and am looking to trim either leading or trailing characters at different times.
You could use
Leading:
System.out.println("//test/me".replaceAll("^/+", ""));
Trailing:
System.out.println("//test/me//".replaceAll("/+$", ""));
You can use Apache StringUtils.stripStart to trim leading characters, or StringUtils.stripEnd to trim trailing characters.
For example:
System.out.println(StringUtils.stripStart("//test/me", "/"));
will output:
test/me
Note that if for some reason you can't use the whole StringUtils library, you could just rip out the relevant parts, as detailed here:
Trim with Character, String, or Regex
If run-time is not a big issue for you, then this code will prove really helpful.
public class StringTrimmer {
public static String trim(String string, char ch){
return trim(string, ch, ch);
}
public static String trim(String string, char leadingChar, char trailingChar){
return string.replaceAll("^["+leadingChar+"]+|["+trailingChar+"]+$", "");
}
public static String trim(String string, String regex){
return trim(string, regex, regex);
}
public static String trim(String string, String leadingRegex, String trailingRegex){
return string.replaceAll("^("+leadingRegex+")+|("+trailingRegex+")+$", "");
}
// test
public static void main(String[] args) {
System.out.println(trim("110100", '1', '0')); // outputs: 01
System.out.println(trim("**Aa0*#**", '*')); // outputs: Aa0*#
System.out.println(trim("123##22222", "12", "22")); // outputs: 3##2
System.out.println(trim("101101##10101", "101")); // outputs: ##10
System.out.println(trim("123##abcde", "\\d", "[c-e]")); // outputs: ##ab
}
}
You could use a simple iteration if you want to remove the leading characters from a string :
String removeLeadingChar(String s, char c) {
int i;
for(i = 0; i < s.length() && s.charAt(i) == c; ++i);
return s.substring(i);
}
same logic applies if you want to remove any trailing char.
For those using Spring:
https://docs.spring.io/spring-framework/docs/current/javadoc-api/org/springframework/util/StringUtils.html#trimTrailingCharacter-java.lang.String-char-
import org.springframework.util.StringUtils;
public void setFileName(String fileName) {
// If the extension does not exist trim the trailing period
this.fileName = StringUtils.trimTrailingCharacter(fileName,'.');
}
My version of trimming leading and/or trailing String s from String str. Both arguments may be null. When str does not has leading and/or trailing s, it is not changed.
String trim(String str, String s) {
String res = s == null ? str : str == null ? null : s.length() >= str.length() ? str : str.replaceFirst(s, "");
if ((res != null) && (s != null) && (res.length() >= s.length())) {
return res.substring(res.length() - s.length(), res.length()).equals(s) ? res.substring(0, res.length() - s.length()) : res;
}
return res;
}
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
String replace method is not working
public static String capitalise(String str)
{
if (str != null || !"".equals(str))
{
char chr=str.charAt(0);
String check= Character.toString(chr);
String check1= check.toUpperCase();
char chr1=check1.charAt(0);
str=str.replace(chr, chr1);
return str;
}
else
{
System.out.println("Not a valid String");
}
return str;
}
i want to uppercase only the first character but it capitalize the first character where it comes for eg before upeercase string = shashank after it becomes ShaShank...what should i do for it .
str.replace(chr, chr1); replaces all occurences of the char in that string. It will change all s to S in your case.
User substring to concatenate the upper-cased first char with the rest of the string.
Here's a simple solution for the additional question in your comment. It will uppercase each character after a single(!) space. You may want to enhance it to allow mulitple spaces or multiple whitechars in general.
public static void main(final String[] args) {
String s = "some words";
StringBuilder result = new StringBuilder();
boolean capitalizeNextLetter = true;
for (char c : s.toCharArray()) {
if (capitalizeNextLetter) {
result.append(Character.toUpperCase(c));
capitalizeNextLetter = false;
} else {
if (c == ' ') {
capitalizeNextLetter = true;
}
result.append(c);
}
}
System.out.println(result.toString());
}
How about something like:
public static String capitalize(String str) {
return str.substring(0, 1).toUpperCase() + str.substring(1);
}
Why not simply do following
str= Character.toUpperCase(str.charAt(0)) + str.substring(1)
In your code replace method will replace all 's' with 'S'.
Because char contains 's';
Try this answer
public static String capitalise(String str)
{
String test = str.substring(0,1);
test=test.toUpperCase();
System.out.println(test);
String new1= test+str.substring(1,str.length());
return new1;
}
try this
public static String capitalise(String str)
{
if (str != null || !"".equals(str))
{
char chr = str.charAt(0);
String capsString = Character.toString(chr).toUpperCase() + str.substring(1);
return capsString;
}
else
{
System.out.println("Not a valid String");
}
return str;
}
Use ReplaceFirst method.
str=str.replaceFirst(check, chreck1);
replaceFirst
Also, you should change your if condition to following:
if (str != null && !"".equals(str))
This way empty string won't be processed.
Try this.just change one line.
public static String capitalise(String str)
{
if (str != null || !"".equals(str))
{
char chr=str.charAt(0);
String check= Character.toString(chr);
String check1= check.toUpperCase();
str=check1+str.substring(1);
return str;
}
else
{
System.out.println("Not a valid String");
}
return str;
}
I want to remove the last character from a string. I've tried doing this:
public String method(String str) {
if (str.charAt(str.length()-1)=='x'){
str = str.replace(str.substring(str.length()-1), "");
return str;
} else{
return str;
}
}
Getting the length of the string - 1 and replacing the last letter with nothing (deleting it), but every time I run the program, it deletes middle letters that are the same as the last letter.
For example, the word is "admirer"; after I run the method, I get "admie." I want it to return the word admire.
replace will replace all instances of a letter. All you need to do is use substring():
public String method(String str) {
if (str != null && str.length() > 0 && str.charAt(str.length() - 1) == 'x') {
str = str.substring(0, str.length() - 1);
}
return str;
}
Why not just one liner?
public static String removeLastChar(String str) {
return removeLastChars(str, 1);
}
public static String removeLastChars(String str, int chars) {
return str.substring(0, str.length() - chars);
}
Full Code
public class Main {
public static void main (String[] args) throws java.lang.Exception {
String s1 = "Remove Last CharacterY";
String s2 = "Remove Last Character2";
System.out.println("After removing s1==" + removeLastChar(s1) + "==");
System.out.println("After removing s2==" + removeLastChar(s2) + "==");
}
public static String removeLastChar(String str) {
return removeLastChars(str, 1);
}
public static String removeLastChars(String str, int chars) {
return str.substring(0, str.length() - chars);
}
}
Demo
Since we're on a subject, one can use regular expressions too
"aaabcd".replaceFirst(".$",""); //=> aaabc
The described problem and proposed solutions sometimes relate to removing separators. If this is your case, then have a look at Apache Commons StringUtils, it has a method called removeEnd which is very elegant.
Example:
StringUtils.removeEnd("string 1|string 2|string 3|", "|");
Would result in:
"string 1|string 2|string 3"
public String removeLastChar(String s) {
if (s == null || s.length() == 0) {
return s;
}
return s.substring(0, s.length()-1);
}
Don't try to reinvent the wheel, while others have already written libraries to perform string manipulation:
org.apache.commons.lang3.StringUtils.chop()
In Kotlin you can used dropLast() method of the string class.
It will drop the given number from string, return a new string
var string1 = "Some Text"
string1 = string1.dropLast(1)
Use this:
if(string.endsWith("x")) {
string= string.substring(0, string.length() - 1);
}
if (str.endsWith("x")) {
return str.substring(0, str.length() - 1);
}
return str;
For example, the word is "admirer"; after I run the method, I get "admie." I want it to return the word admire.
In case you're trying to stem English words
Stemming is the process for reducing inflected (or sometimes derived) words to their stem, base or root form—generally a written word form.
...
A stemmer for English, for example, should identify the string "cats" (and possibly "catlike", "catty" etc.) as based on the root "cat", and "stemmer", "stemming", "stemmed" as based on "stem". A stemming algorithm reduces the words "fishing", "fished", "fish", and "fisher" to the root word, "fish".
Difference between Lucene stemmers: EnglishStemmer, PorterStemmer, LovinsStemmer outlines some Java options.
As far as the readability is concerned, I find this to be the most concise
StringUtils.substring("string", 0, -1);
The negative indexes can be used in Apache's StringUtils utility.
All negative numbers are treated from offset from the end of the string.
string = string.substring(0, (string.length() - 1));
I'm using this in my code, it's easy and simple.
it only works while the String is > 0.
I have it connected to a button and inside the following if statement
if (string.length() > 0) {
string = string.substring(0, (string.length() - 1));
}
public String removeLastChar(String s) {
if (!Util.isEmpty(s)) {
s = s.substring(0, s.length()-1);
}
return s;
}
removes last occurence of the 'xxx':
System.out.println("aaa xxx aaa xxx ".replaceAll("xxx([^xxx]*)$", "$1"));
removes last occurrence of the 'xxx' if it is last:
System.out.println("aaa xxx aaa ".replaceAll("xxx\\s*$", ""));
you can replace the 'xxx' on what you want but watch out on special chars
Look to StringBuilder Class :
StringBuilder sb=new StringBuilder("toto,");
System.out.println(sb.deleteCharAt(sb.length()-1));//display "toto"
// creating StringBuilder
StringBuilder builder = new StringBuilder(requestString);
// removing last character from String
builder.deleteCharAt(requestString.length() - 1);
How can a simple task be made complicated. My solution is:
public String removeLastChar(String s) {
return s[0..-1]
}
or
public String removeLastChar(String s) {
if (s.length() > 0) {
return s[0..-1]
}
return s
}
// Remove n last characters
// System.out.println(removeLast("Hello!!!333",3));
public String removeLast(String mes, int n) {
return mes != null && !mes.isEmpty() && mes.length()>n
? mes.substring(0, mes.length()-n): mes;
}
// Leave substring before character/string
// System.out.println(leaveBeforeChar("Hello!!!123", "1"));
public String leaveBeforeChar(String mes, String last) {
return mes != null && !mes.isEmpty() && mes.lastIndexOf(last)!=-1
? mes.substring(0, mes.lastIndexOf(last)): mes;
}
A one-liner answer (just a funny alternative - do not try this at home, and great answers already given):
public String removeLastChar(String s){return (s != null && s.length() != 0) ? s.substring(0, s.length() - 1): s;}
Most answers here forgot about surrogate pairs.
For instance, the character 𝕫 (codepoint U+1D56B) does not fit into a single char, so in order to be represented, it must form a surrogate pair of two chars.
If one simply applies the currently accepted answer (using str.substring(0, str.length() - 1), one splices the surrogate pair, leading to unexpected results.
One should also include a check whether the last character is a surrogate pair:
public static String removeLastChar(String str) {
Objects.requireNonNull(str, "The string should not be null");
if (str.isEmpty()) {
return str;
}
char lastChar = str.charAt(str.length() - 1);
int cut = Character.isSurrogate(lastChar) ? 2 : 1;
return str.substring(0, str.length() - cut);
}
Java 8
import java.util.Optional;
public class Test
{
public static void main(String[] args) throws InterruptedException
{
System.out.println(removeLastChar("test-abc"));
}
public static String removeLastChar(String s) {
return Optional.ofNullable(s)
.filter(str -> str.length() != 0)
.map(str -> str.substring(0, str.length() - 1))
.orElse(s);
}
}
Output : test-ab
public String removeLastCharacter(String str){
String result = null;
if ((str != null) && (str.length() > 0)) {
return str.substring(0, str.length() - 1);
}
else{
return "";
}
}
if we want to remove file extension of the given file,
** Sample code
public static String removeNCharactersFromLast(String str,int n){
if (str != null && (str.length() > 0)) {
return str.substring(0, str.length() - n);
}
return "";
}
For kotlin check out
string.dropLast(1)
if you have special character like ; in json just use String.replace(";", "") otherwise you must rewrite all character in string minus the last.
Why not use the escape sequence ... !
System.out.println(str + '\b');
Life is much easier now . XD ! ~ A readable one-liner
How to make the char in the recursion at the end:
public static String removeChar(String word, char charToRemove)
{
String char_toremove=Character.toString(charToRemove);
for(int i = 0; i < word.length(); i++)
{
if(word.charAt(i) == charToRemove)
{
String newWord = word.substring(0, i) + word.substring(i + 1);
return removeChar(newWord,charToRemove);
}
}
System.out.println(word);
return word;
}
for exemple:
removeChar ("hello world, let's go!",'l') → "heo word, et's go!llll"
removeChar("you should not go",'o') → "yu shuld nt goooo"
Here's an answer that works with codepoints outside of the Basic Multilingual Plane (Java 8+).
Using streams:
public String method(String str) {
return str.codePoints()
.limit(str.codePoints().count() - 1)
.mapToObj(i->new String(Character.toChars(i)))
.collect(Collectors.joining());
}
More efficient maybe:
public String method(String str) {
return str.isEmpty()? "": str.substring(0, str.length() - Character.charCount(str.codePointBefore(str.length())));
}
just replace the condition of "if" like this:
if(a.substring(a.length()-1).equals("x"))'
this will do the trick for you.
Suppose total length of my string=24
I want to cut last character after position 14 to end, mean I want starting 14 to be there.
So I apply following solution.
String date = "2019-07-31T22:00:00.000Z";
String result = date.substring(0, date.length() - 14);
I had to write code for a similar problem. One way that I was able to solve it used a recursive method of coding.
static String removeChar(String word, char charToRemove)
{
for(int i = 0; i < word.lenght(); i++)
{
if(word.charAt(i) == charToRemove)
{
String newWord = word.substring(0, i) + word.substring(i + 1);
return removeChar(newWord, charToRemove);
}
}
return word;
}
Most of the code I've seen on this topic doesn't use recursion so hopefully I can help you or someone who has the same problem.