Replacing multiple char from a string in java - java

I have a PHP script <?=str_replace(array('(',')','-',' ','.'), "", $rs["hq_tel"])?> this is a string replace function that take array of chars and replace them if find any of the char in string. Is there any java equivalent of the function. I found some ways but some are using loop and some repeating the statements but not found any single line solution like this in java.
Thanks in advance.

You can use a regex like this:
//char1, char2 will be replaced by the replacement String. You can add more characters if you want!
String.replaceAll("[char1char2]", "replacement");
where the first parameter is the regex and the second parameter is the replacement.
Refer the docs on how to escape special characters(in case you need to!).

your solution is here..
Replace all special character
str.replaceAll("[^\\dA-Za-z ]", "");
Replace specific special character
str.replaceAll("[()?:!.,;{}]+", " ");

If you don't know about regex you can use something more elaborated:
private static ArrayList<Character> special = new ArrayList<Character>(Arrays.asList('(', ')', '-', ' ', '.'));
public static void main(String[] args) {
String test = "Hello(how-are.you ?";
String outputText = "";
for (int i = 0; i < test.length(); i++) {
Character c = new Character(test.charAt(i));
if (!special.contains(c))
outputText += c;
else
outputText += "";
}
System.out.println(outputText);
}
Output:
Hellohowareyou?
EDIT (without loop but with regex):
public static void main(String[] args) {
String test = "Hello(how-are.you ?)";
String outputText = test.replaceAll("[()-. ]+", "");
System.out.println(outputText);
}

String.replaceAll(String regex, String replacement)

Related

How to split a string in JAVA with two different seperators? [duplicate]

I want to split the string "004-034556" into two strings by the delimiter "-":
part1 = "004";
part2 = "034556";
That means the first string will contain the characters before '-', and the second string will contain the characters after '-'.
I also want to check if the string has '-' in it.
Use the appropriately named method String#split().
String string = "004-034556";
String[] parts = string.split("-");
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556
Note that split's argument is assumed to be a regular expression, so remember to escape special characters if necessary.
there are 12 characters with special meanings: the backslash \, the caret ^, the dollar sign $, the period or dot ., the vertical bar or pipe symbol |, the question mark ?, the asterisk or star *, the plus sign +, the opening parenthesis (, the closing parenthesis ), and the opening square bracket [, the opening curly brace {, These special characters are often called "metacharacters".
For instance, to split on a period/dot . (which means "any character" in regex), use either backslash \ to escape the individual special character like so split("\\."), or use character class [] to represent literal character(s) like so split("[.]"), or use Pattern#quote() to escape the entire string like so split(Pattern.quote(".")).
String[] parts = string.split(Pattern.quote(".")); // Split on the exact string.
To test beforehand if the string contains certain character(s), just use String#contains().
if (string.contains("-")) {
// Split it.
} else {
throw new IllegalArgumentException("String " + string + " does not contain -");
}
Note, this does not take a regular expression. For that, use String#matches() instead.
If you'd like to retain the split character in the resulting parts, then make use of positive lookaround. In case you want to have the split character to end up in left hand side, use positive lookbehind by prefixing ?<= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?<=-)");
String part1 = parts[0]; // 004-
String part2 = parts[1]; // 034556
In case you want to have the split character to end up in right hand side, use positive lookahead by prefixing ?= group on the pattern.
String string = "004-034556";
String[] parts = string.split("(?=-)");
String part1 = parts[0]; // 004
String part2 = parts[1]; // -034556
If you'd like to limit the number of resulting parts, then you can supply the desired number as 2nd argument of split() method.
String string = "004-034556-42";
String[] parts = string.split("-", 2);
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556-42
An alternative to processing the string directly would be to use a regular expression with capturing groups. This has the advantage that it makes it straightforward to imply more sophisticated constraints on the input. For example, the following splits the string into two parts, and ensures that both consist only of digits:
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class SplitExample
{
private static Pattern twopart = Pattern.compile("(\\d+)-(\\d+)");
public static void checkString(String s)
{
Matcher m = twopart.matcher(s);
if (m.matches()) {
System.out.println(s + " matches; first part is " + m.group(1) +
", second part is " + m.group(2) + ".");
} else {
System.out.println(s + " does not match.");
}
}
public static void main(String[] args) {
checkString("123-4567");
checkString("foo-bar");
checkString("123-");
checkString("-4567");
checkString("123-4567-890");
}
}
As the pattern is fixed in this instance, it can be compiled in advance and stored as a static member (initialised at class load time in the example). The regular expression is:
(\d+)-(\d+)
The parentheses denote the capturing groups; the string that matched that part of the regexp can be accessed by the Match.group() method, as shown. The \d matches and single decimal digit, and the + means "match one or more of the previous expression). The - has no special meaning, so just matches that character in the input. Note that you need to double-escape the backslashes when writing this as a Java string. Some other examples:
([A-Z]+)-([A-Z]+) // Each part consists of only capital letters
([^-]+)-([^-]+) // Each part consists of characters other than -
([A-Z]{2})-(\d+) // The first part is exactly two capital letters,
// the second consists of digits
Use:
String[] result = yourString.split("-");
if (result.length != 2)
throw new IllegalArgumentException("String not in correct format");
This will split your string into two parts. The first element in the array will be the part containing the stuff before the -, and the second element in the array will contain the part of your string after the -.
If the array length is not 2, then the string was not in the format: string-string.
Check out the split() method in the String class.
This:
String[] out = string.split("-");
should do the thing you want. The string class has many method to operate with a string.
// This leaves the regexes issue out of question
// But we must remember that each character in the Delimiter String is treated
// like a single delimiter
public static String[] SplitUsingTokenizer(String subject, String delimiters) {
StringTokenizer strTkn = new StringTokenizer(subject, delimiters);
ArrayList<String> arrLis = new ArrayList<String>(subject.length());
while(strTkn.hasMoreTokens())
arrLis.add(strTkn.nextToken());
return arrLis.toArray(new String[0]);
}
With Java 8:
List<String> stringList = Pattern.compile("-")
.splitAsStream("004-034556")
.collect(Collectors.toList());
stringList.forEach(s -> System.out.println(s));
Use org.apache.commons.lang.StringUtils' split method which can split strings based on the character or string you want to split.
Method signature:
public static String[] split(String str, char separatorChar);
In your case, you want to split a string when there is a "-".
You can simply do as follows:
String str = "004-034556";
String split[] = StringUtils.split(str,"-");
Output:
004
034556
Assume that if - does not exists in your string, it returns the given string, and you will not get any exception.
The requirements left room for interpretation. I recommend writing a method,
public final static String[] mySplit(final String s)
which encapsulate this function. Of course you can use String.split(..) as mentioned in the other answers for the implementation.
You should write some unit-tests for input strings and the desired results and behaviour.
Good test candidates should include:
- "0022-3333"
- "-"
- "5555-"
- "-333"
- "3344-"
- "--"
- ""
- "553535"
- "333-333-33"
- "222--222"
- "222--"
- "--4555"
With defining the according test results, you can specify the behaviour.
For example, if "-333" should return in [,333] or if it is an error.
Can "333-333-33" be separated in [333,333-33] or [333-333,33] or is it an error? And so on.
To summarize: there are at least five ways to split a string in Java:
String.split():
String[] parts ="10,20".split(",");
Pattern.compile(regexp).splitAsStream(input):
List<String> strings = Pattern.compile("\\|")
.splitAsStream("010|020202")
.collect(Collectors.toList());
StringTokenizer (legacy class):
StringTokenizer strings = new StringTokenizer("Welcome to EXPLAINJAVA.COM!", ".");
while(strings.hasMoreTokens()){
String substring = strings.nextToken();
System.out.println(substring);
}
Google Guava Splitter:
Iterable<String> result = Splitter.on(",").split("1,2,3,4");
Apache Commons StringUtils:
String[] strings = StringUtils.split("1,2,3,4", ",");
So you can choose the best option for you depending on what you need, e.g. return type (array, list, or iterable).
Here is a big overview of these methods and the most common examples (how to split by dot, slash, question mark, etc.)
You can try like this also
String concatenated_String="hi^Hello";
String split_string_array[]=concatenated_String.split("\\^");
Assuming, that
you don't really need regular expressions for your split
you happen to already use apache commons lang in your app
The easiest way is to use StringUtils#split(java.lang.String, char). That's more convenient than the one provided by Java out of the box if you don't need regular expressions. Like its manual says, it works like this:
A null input String returns null.
StringUtils.split(null, *) = null
StringUtils.split("", *) = []
StringUtils.split("a.b.c", '.') = ["a", "b", "c"]
StringUtils.split("a..b.c", '.') = ["a", "b", "c"]
StringUtils.split("a:b:c", '.') = ["a:b:c"]
StringUtils.split("a b c", ' ') = ["a", "b", "c"]
I would recommend using commong-lang, since usually it contains a lot of stuff that's usable. However, if you don't need it for anything else than doing a split, then implementing yourself or escaping the regex is a better option.
For simple use cases String.split() should do the job. If you use guava, there is also a Splitter class which allows chaining of different string operations and supports CharMatcher:
Splitter.on('-')
.trimResults()
.omitEmptyStrings()
.split(string);
The fastest way, which also consumes the least resource could be:
String s = "abc-def";
int p = s.indexOf('-');
if (p >= 0) {
String left = s.substring(0, p);
String right = s.substring(p + 1);
} else {
// s does not contain '-'
}
String Split with multiple characters using Regex
public class StringSplitTest {
public static void main(String args[]) {
String s = " ;String; String; String; String, String; String;;String;String; String; String; ;String;String;String;String";
//String[] strs = s.split("[,\\s\\;]");
String[] strs = s.split("[,\\;]");
System.out.println("Substrings length:"+strs.length);
for (int i=0; i < strs.length; i++) {
System.out.println("Str["+i+"]:"+strs[i]);
}
}
}
Output:
Substrings length:17
Str[0]:
Str[1]:String
Str[2]: String
Str[3]: String
Str[4]: String
Str[5]: String
Str[6]: String
Str[7]:
Str[8]:String
Str[9]:String
Str[10]: String
Str[11]: String
Str[12]:
Str[13]:String
Str[14]:String
Str[15]:String
Str[16]:String
But do not expect the same output across all JDK versions. I have seen one bug which exists in some JDK versions where the first null string has been ignored. This bug is not present in the latest JDK version, but it exists in some versions between JDK 1.7 late versions and 1.8 early versions.
There are only two methods you really need to consider.
Use String.split for a one-character delimiter or you don't care about performance
If performance is not an issue, or if the delimiter is a single character that is not a regular expression special character (i.e., not one of .$|()[{^?*+\) then you can use String.split.
String[] results = input.split(",");
The split method has an optimization to avoid using a regular expression if the delimeter is a single character and not in the above list. Otherwise, it has to compile a regular expression, and this is not ideal.
Use Pattern.split and precompile the pattern if using a complex delimiter and you care about performance.
If performance is an issue, and your delimiter is not one of the above, you should pre-compile a regular expression pattern which you can then reuse.
// Save this somewhere
Pattern pattern = Pattern.compile("[,;:]");
/// ... later
String[] results = pattern.split(input);
This last option still creates a new Matcher object. You can also cache this object and reset it for each input for maximum performance, but that is somewhat more complicated and not thread-safe.
You can split a string by a line break by using the following statement:
String textStr[] = yourString.split("\\r?\\n");
You can split a string by a hyphen/character by using the following statement:
String textStr[] = yourString.split("-");
public class SplitTest {
public static String[] split(String text, String delimiter) {
java.util.List<String> parts = new java.util.ArrayList<String>();
text += delimiter;
for (int i = text.indexOf(delimiter), j=0; i != -1;) {
String temp = text.substring(j,i);
if(temp.trim().length() != 0) {
parts.add(temp);
}
j = i + delimiter.length();
i = text.indexOf(delimiter,j);
}
return parts.toArray(new String[0]);
}
public static void main(String[] args) {
String str = "004-034556";
String delimiter = "-";
String result[] = split(str, delimiter);
for(String s:result)
System.out.println(s);
}
}
Please don't use StringTokenizer class as it is a legacy class that is retained for compatibility reasons, and its use is discouraged in new code. And we can make use of the split method as suggested by others as well.
String[] sampleTokens = "004-034556".split("-");
System.out.println(Arrays.toString(sampleTokens));
And as expected it will print:
[004, 034556]
In this answer I also want to point out one change that has taken place for split method in Java 8. The String#split() method makes use of Pattern.split, and now it will remove empty strings at the start of the result array. Notice this change in documentation for Java 8:
When there is a positive-width match at the beginning of the input
sequence then an empty leading substring is included at the beginning
of the resulting array. A zero-width match at the beginning however
never produces such empty leading substring.
It means for the following example:
String[] sampleTokensAgain = "004".split("");
System.out.println(Arrays.toString(sampleTokensAgain));
we will get three strings: [0, 0, 4] and not four as was the case in Java 7 and before. Also check this similar question.
One way to do this is to run through the String in a for-each loop and use the required split character.
public class StringSplitTest {
public static void main(String[] arg){
String str = "004-034556";
String split[] = str.split("-");
System.out.println("The split parts of the String are");
for(String s:split)
System.out.println(s);
}
}
Output:
The split parts of the String are:
004
034556
import java.io.*;
public class BreakString {
public static void main(String args[]) {
String string = "004-034556-1234-2341";
String[] parts = string.split("-");
for(int i=0;i<parts.length;i++) {
System.out.println(parts[i]);
}
}
}
You can use Split():
import java.io.*;
public class Splitting
{
public static void main(String args[])
{
String Str = new String("004-034556");
String[] SplittoArray = Str.split("-");
String string1 = SplittoArray[0];
String string2 = SplittoArray[1];
}
}
Else, you can use StringTokenizer:
import java.util.*;
public class Splitting
{
public static void main(String[] args)
{
StringTokenizer Str = new StringTokenizer("004-034556");
String string1 = Str.nextToken("-");
String string2 = Str.nextToken("-");
}
}
Here are two ways two achieve it.
WAY 1: As you have to split two numbers by a special character you can use regex
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class TrialClass
{
public static void main(String[] args)
{
Pattern p = Pattern.compile("[0-9]+");
Matcher m = p.matcher("004-034556");
while(m.find())
{
System.out.println(m.group());
}
}
}
WAY 2: Using the string split method
public class TrialClass
{
public static void main(String[] args)
{
String temp = "004-034556";
String [] arrString = temp.split("-");
for(String splitString:arrString)
{
System.out.println(splitString);
}
}
}
You can simply use StringTokenizer to split a string in two or more parts whether there are any type of delimiters:
StringTokenizer st = new StringTokenizer("004-034556", "-");
while(st.hasMoreTokens())
{
System.out.println(st.nextToken());
}
Check out the split() method in the String class on javadoc.
https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#split(java.lang.String)
String data = "004-034556-1212-232-232";
int cnt = 1;
for (String item : data.split("-")) {
System.out.println("string "+cnt+" = "+item);
cnt++;
}
Here many examples for split string but I little code optimized.
String str="004-034556"
String[] sTemp=str.split("-");// '-' is a delimiter
string1=004 // sTemp[0];
string2=034556//sTemp[1];
I just wanted to write an algorithm instead of using Java built-in functions:
public static List<String> split(String str, char c){
List<String> list = new ArrayList<>();
StringBuilder sb = new StringBuilder();
for (int i = 0; i < str.length(); i++){
if(str.charAt(i) != c){
sb.append(str.charAt(i));
}
else{
if(sb.length() > 0){
list.add(sb.toString());
sb = new StringBuilder();
}
}
}
if(sb.length() >0){
list.add(sb.toString());
}
return list;
}
You can use the method split:
public class Demo {
public static void main(String args[]) {
String str = "004-034556";
if ((str.contains("-"))) {
String[] temp = str.split("-");
for (String part:temp) {
System.out.println(part);
}
}
else {
System.out.println(str + " does not contain \"-\".");
}
}
}
To split a string, uses String.split(regex). Review the following examples:
String data = "004-034556";
String[] output = data.split("-");
System.out.println(output[0]);
System.out.println(output[1]);
Output
004
034556
Note:
This split (regex) takes a regex as an argument. Remember to escape the regex special characters, like period/dot.
String s = "TnGeneral|DOMESTIC";
String a[]=s.split("\\|");
System.out.println(a.toString());
System.out.println(a[0]);
System.out.println(a[1]);
Output:
TnGeneral
DOMESTIC
String s="004-034556";
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)=='-')
{
System.out.println(s.substring(0,i));
System.out.println(s.substring(i+1));
}
}
As mentioned by everyone, split() is the best option which may be used in your case. An alternative method can be using substring().

Replace special character in a String

I have to replace a string whit special character like this: XY_Universit�-WWWWW-ZZZZZ in another one like: XY_Universita-WWWWW-ZZZZZ.
I tried solutions like stringToReplace.replaceAll(".*Universit.*", "Universita"); but it replace all the string with the word ''Universita" and it isn't what i want.
stringToReplace.replaceAll("Universit[^a]", "Universita")
public static void main(String[] args) throws IOException {
String exp = "XY_Universit�-WWWWW-ZZZZZ";
exp = exp.replaceAll("[^\\w\\s\\-_]", "");
System.out.println(exp);
}
output
XY_Universit-WWWWW-ZZZZZ
You have to be lazy :P
use : .*?XY_Universit. to select only the first XY_universit
demo here
Another odd way..
String example = "XY_Universit�-WWWWW-ZZZZZ";
char ch = 65533; //find ascii of special char
System.out.println(example.replace(ch, 'a'));

How to get alphabets only from given albha-numberic word in java?

sorry for this if this is a silly question.but i need to know about this.
If i have a word like alphabets,numeric and special charters. I need to extract alphabets only.No need for numeric and special characters.I need to know is there default function is there in Java to split characters only?
eg.String word="te123##st";
I need test only.
This solution works with accentued/non-ascii caracters :
"te123##st\néàø_".replaceAll("[\\p{Digit}\\p{Punct}\\p{Space}]", "");
try this word.replaceAll("[^a-zA-Z]", "");
This will remove all non alphanumeric characters, but it will still remove accented characters.
String word = "te123##st";
word = word.replaceAll("[^\\p{Alpha}]", "");
// or word = word.replaceAll("[\\P{Alpha}]", "");
See apidoc reference.
try
word = word.replaceAll("\\P{Alpha}", "");
String word = "te123##st";
word = word.replaceAll("[\\W\\d._]", "");
try this:
word = word.replaceAll("[\\d##_]", "");
- I won't make this complicated using Regex, but will use inbuilt Java functionalities to answer this.
- First use subString() method to get the "abcd" part of the String, then use toCharArray() method to break the String into char elements, then use Character class's isDigit() method to know whether its a digit or not.
public class T1 {
public static void main(String[] args){
String s = "te123##st";
String str = s.substring(0,4);
System.out.println(str);
String tempStr = new String();
char[] cArr = str.toCharArray();
for(char a :cArr){
if(Character.isAlphabetic(a)){
System.out.println(a+" is a alphabet");
tempStr = tempStr + a;
}else{
System.out.println(a+" is not a alphabet");
}
}
System.out.println("The extracted String is: "+tempStr);
}
}

Remove a specific word from a string

I'm trying to remove a specific word from a certain string using the function replace() or replaceAll() but these remove all the occurrences of this word even if it's part of another word!
Example:
String content = "is not like is, but mistakes are common";
content = content.replace("is", "");
output: "not like , but mtakes are common"
desired output: "not like , but mistakes are common"
How can I substitute only whole words from a string?
What the heck,
String regex = "\\s*\\bis\\b\\s*";
content = content.replaceAll(regex, "");
Remember you need to use replaceAll(...) to use regular expressions, not replace(...)
\\b gives you the word boundaries
\\s* sops up any white space on either side of the word being removed (if you want to remove this too).
content = content.replaceAll("\\Wis\\W|^is\\W|\\Wis$", "");
You can try replacing " is " by " ". The is with a space before and one after, replaced by a single space.
Update:
To make it work for the first "is" in the sentence, also do another replace of "is " for "". Replacing the first is and the first space, with an empty string.
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
String input = s.nextLine();
char c = s.next().charAt(0);
System.out.println(removeAllOccurrencesOfChar(input, c));
}
public static String removeAllOccurrencesOfChar(String input, char c) {
String r = "";
for (int i = 0; i < input.length(); i ++) {
if (input.charAt(i) != c) r += input.charAt(i);
}
return r;
}
}

How to remove invalid characters from a string?

I have no idea how to remove invalid characters from a string in Java. I'm trying to remove all the characters that are not numbers, letters, or ( ) [ ] . How can I do this?
Thanks
String foo = "this is a thing with & in it";
foo = foo.replaceAll("[^A-Za-z0-9()\\[\\]]", "");
Javadocs are your friend. Regular expressions are also your friend.
Edit:
That being siad, this is only for the Latin alphabet; you can adjust accordingly. \\w can be used for a-zA-Z to denote a "word" character if that works for your case though it includes _.
Using Guava, and almost certainly more efficient (and more readable) than regexes:
CharMatcher desired = CharMatcher.JAVA_DIGIT
.or(CharMatcher.JAVA_LETTER)
.or(CharMatcher.anyOf("()[]"))
.precomputed(); // optional, may improve performance, YMMV
return desired.retainFrom(string);
Try this:
String s = "123abc&^%[]()";
s = s.replaceAll("[^A-Za-z0-9()\\[\\]]", "");
System.out.println(s);
The above will remove characters "&^%" in the sample string, leaving in s only "123abc[]()".
public static void main(String[] args) {
String c = "hjdg$h&jk8^i0ssh6+/?:().,+-#";
System.out.println(c);
Pattern pt = Pattern.compile("[^a-zA-Z0-9/?:().,'+/-]");
Matcher match = pt.matcher(c);
if (!match.matches()) {
c = c.replaceAll(pt.pattern(), "");
}
System.out.println(c);
}
Use this code:
String s = "Test[]"
s = s.replaceAll("[");
s = s.replaceAll("]");
myString.replaceAll("[^\\w\\[\\]\\(\\)]", "");
replaceAll method takes a regex as first parameter and replaces all matches in string. This regex matches all characters which are not digit, letter or underscore (\\w) and braces you need (\\[\\]\\(\\)])
You can remove specials characters from your String/Url or any request parameters you have get from user side
public static String removeSpecialCharacters(String inputString){
final String[] metaCharacters = {"../","\\..","\\~","~/","~"};
String outputString="";
for (int i = 0 ; i < metaCharacters.length ; i++){
if(inputString.contains(metaCharacters[i])){
outputString = inputString.replace(metaCharacters[i],"");
inputString = outputString;
}else{
outputString = inputString;
}
}
return outputString;
}
You can specify the range of characters to keep/remove based on the order of characters in the ASCII table. The regex can use actual characters or character hex codes:
// Example - remove characters outside of the range of "space to tilde".
// 1) using characters
someString.replaceAll("[^ -~]", "");
// 2) using hex codes for "space" and "tilde"
someString.replaceAll("[^\\u0020-\\u007E]", "");

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