I'm trying to use the String.replaceAll() method with regex to only keep letter characters and ['-_]. I'm trying to do this by replacing every character that is neither a letter nor one of the characters above by an empty string.
So far I have tried something like this (in different variations) which correctly keeps letters but replaces the special characters I want to keep:
current = current.replaceAll("(?=\\P{L})(?=[^\\'-_])", "");
Make it simplier :
current = current.replaceAll("[^a-zA-Z'_-]", "");
Explanation :
Match any char not in a to z, A to Z, ', _, - and replaceAll() method will replace any matched char with nothing.
Tested input : "a_zE'R-z4r#m"
Output : a_zE'R-zrm
You don't need lookahead, just use negated regex:
current = current.replaceAll("[^\\p{L}'_-]+", "");
[^\\p{L}'_-] will match anything that is not a letter (unicode) or single quote or underscore or hyphen.
Your regex is too complicated. Just specify the characters you want to keep, and use ^ to negate, so [^a-z'_-] means "anything but these".
public class Replacer {
public static void main(String[] args) {
System.out.println("with 1234 &*()) -/.,>>?chars".replaceAll("[^\\w'_-]", ""));
}
}
You can try this:
String str = "Se#rbi323a`and_Eur$ope#-t42he-[A%merica]";
str = str.replaceAll("[\\d+\\p{Punct}&&[^-'_\\[\\]]]+", "");
System.out.println("str = " + str);
And it is the result:
str = Serbia'and_Europe-the-[America]
Related
I used a regex expression to remove special characters from name. The expression will remove all letters except English alphabets.
public static void main(String args[]) {
String name = "Özcan Sevim.";
name = name.replaceAll("[^a-zA-Z\\s]", " ").trim();
System.out.println(name);
}
Output:
zcan Sevim
Expected Output:
Özcan Sevim
I get bad result as I did it this way, the right way will be to remove special characters based on ASCII codes so that other letters will not be removed, can someone help me with a regex that would remove only special characters.
You can use \p{IsLatin} or \p{IsAlphabetic}
name = name.replaceAll("[^\\p{IsLatin}]", " ").trim();
Or to remove the punctuation just use \p{Punct} like this :
name = name.replaceAll("\\p{Punct}", " ").trim();
Outputs
Özcan Sevim
take a look at the full list of Summary of regular-expression constructs and use the one which can help you.
Use Guava CharMatcher for that :) It will be easier to read and maintain it.
name = CharMatcher.ASCII.negate().removeFrom(name);
use [\W+] or "[^a-zA-Z0-9]" as regex to match any special characters and also use String.replaceAll(regex, String) to replace the spl charecter with an empty string. remember as the first arg of String.replaceAll is a regex you have to escape it with a backslash to treat em as a literal charcter.
String string= "hjdg$h&jk8^i0ssh6";
Pattern pt = Pattern.compile("[^a-zA-Z0-9]");
Matcher match= pt.matcher(string);
while(match.find())
{
String s= match.group();
string=string.replaceAll("\\"+s, "");
}
System.out.println(string);
I have strings that I need to use regex to replace a specific character. The strings are in the following format:
"abc.edf" : "abc.abc", "ghi.ghk" : "bbb.bbb" , "qwq.tyt" : "ddd.ddd"
I need to replace the periods, '.', that are between the strings in quotes before the colon but not the strings in quotes after the colon and before the comma. Could someone shed some light?
This pattern will match the entire part that you want to touch: "\w{3}\.\w{3}" : "\w{3}\.\w{3}". Since it includes the colon and the values on both side, it won't match ones where there is a comma between the values. Depending on your needs, you may need to change \w to some other character class.
But, as I'm sure you are aware, you don't want to replace the entire string. You only want to replace the one character. There are two ways to do that. You can either use look-aheads and look-behinds to exclude everything else except the period from the resulting match:
Pattern: (?<="\w{3})\.(?=\w{3}" : "\w{3}\.\w{3}")
Replacement: :
Or, if the look-aheads and look-behinds confuse you, you could just capture the whole thing and include the original values from the captured groups in the replacement value:
Pattern: ("\w{3})\.(\w{3}" : "\w{3}\.\w{3}")
Replacement: $1:$2
Try with the following patern: /.(?=[a-z]+)/g
Working regex-demo for substitution # regex101
Java Working Demo:
public class StackOverFlow31520446 {
public static String text;
public static String pattern;
public static String replacement;
static {
text = "\"abc.edf\" : \"123.231\", \"ghi.ghk\" : \"456.678\" , \"qwq.tyt\" : \"141.242\"";
pattern = "\\.(?=[a-z]+)";
replacement = ";";
}
public static String replaceMatches(String text, String pattern, String replacement) {
return text.replaceAll(pattern, replacement);
}
public static void main(String[] args) {
System.out.println(replaceMatches(text, pattern, replacement));
}
}
Not sure what you intend to do with the string but this is a way to
match the contents of the quote's.
The contents are in capture buffer 1.
You could use a callback to replace the dots within the
contents, passing that back within the main replacement function.
Find: "([^"]*\.[^"]*)"(?=\s*:)
Replace: " + func( call to replace dots from capt buff 1 ) + "
Formatted:
" # Open quote
( [^"]* \. [^"]* ) # (1), group 1 - contents
" # Close quote
(?= # Lookahead, must be a colon
\s*
:
)
If would go for a different approach (maybe it is even faster). In your loop over all strings first try if the string matches a number \d*\.?\d* - if not, do the replacement of . with : (without any regexp).
Would that solve your problem?
You can do it without look arounds:
str = str.replaceAll("(\\D)\\.(\\D)", "$1:$2");
should be sufficient for the task.
i have seen to replace "," to "." by using ".$"|",$", but this logic is not working with alphabets.
i need to replace last letter of a word to another letter for all word in string containing EXAMPLE_TEST using java
this is my code
Pattern replace = Pattern.compile("n$");//here got the real problem
matcher2 = replace.matcher(EXAMPLE_TEST);
EXAMPLE_TEST=matcher2.replaceAll("k");
i also tried "//n$" ,"\n$" etc
Please help me to get the solution
input text=>njan ayman
output text=> njak aymak
Instead of the end of string $ anchor, use a word boundary \b
String s = "njan ayman";
s = s.replaceAll("n\\b", "k");
System.out.println(s); //=> "njak aymak"
You can use lookahead and group matching:
String EXAMPLE_TEST = "njan ayman";
s = EXAMPLE_TEST.replaceAll("(n)(?=\\s|$)", "k");
System.out.println("s = " + s); // prints: s = njak aymak
Explanation:
(n) - the matched word character
(?=\\s|$) - which is followed by a space or at the end of the line (lookahead)
The above is only an example! if you want to switch every comma with a period the middle line should be changed to:
s = s.replaceAll("(,)(?=\\s|$)", "\\.");
Here's how I would set it up:
(?=.\b)\w
Which in Java would need to be escaped as following:
(?=.\\b)\\w
It translates to something like "a character (\w) after (?=) any single character (.) at the end of a word (\b)".
String s = "njan ayman aowkdwo wdonwan. wadawd,.. wadwdawd;";
s = s.replaceAll("(?=.\\b)\\w", "");
System.out.println(s); //nja ayma aowkdw wdonwa. wadaw,.. wadwdaw;
This removes the last character of all words, but leaves following non-alphanumeric characters. You can specify only specific characters to remove/replace by changing the . to something else.
However, the other answers are perfectly good and might achieve exactly what you are looking for.
if (word.endsWith("char oldletter")) {
name = name.substring(0, name.length() - 1 "char newletter");
}
public static final String specialChars1= "\\W\\S";
String str2 = str1.replaceAll(specialChars1, "").replace(" ", "+");
public static final String specialChars2 = "`~!##$%^&*()_+[]\\;\',./{}|:\"<>?";
String str2 = str1.replaceAll(specialChars2, "").replace(" ", "+");
Whatever str1 is I want all the characters other than letters and numbers to be removed, and spaces to be replaced by a plus sign (+).
My problem is if I use specialChar1, it does not remove some characters like ;, ', ", and if I am use specialChar2 it gives me an error :
java.util.regex.PatternSyntaxException: Syntax error U_REGEX_MISSING_CLOSE_BRACKET near index 32:
How can this be to achieved?. I have searched but could not find a perfect solution.
This worked for me:
String result = str.replaceAll("[^\\dA-Za-z ]", "").replaceAll("\\s+", "+");
For this input string:
/-+!##$%^&())";:[]{}\ |wetyk 678dfgh
It yielded this result:
+wetyk+678dfgh
replaceAll expects a regex:
public static final String specialChars2 = "[`~!##$%^&*()_+[\\]\\\\;\',./{}|:\"<>?]";
The problem with your first regex, is that "\W\S" means find a sequence of two characters, the first of which is not a letter or a number followed by a character which is not whitespace.
What you mean is "[^\w\s]". Which means: find a single character which is neither a letter nor a number nor whitespace. (we can't use "[\W\S]" as this means find a character which is not a letter or a number OR is not whitespace -- which is essentially all printable character).
The second regex is a problem because you are trying to use reserved characters without escaping them. You can enclose them in [] where most characters (not all) do not have special meanings, but the whole thing would look very messy and you have to check that you haven't missed out any punctuation.
Example:
String sequence = "qwe 123 :#~ ";
String withoutSpecialChars = sequence.replaceAll("[^\\w\\s]", "");
String spacesAsPluses = withoutSpecialChars.replaceAll("\\s", "+");
System.out.println("without special chars: '"+withoutSpecialChars+ '\'');
System.out.println("spaces as pluses: '"+spacesAsPluses+'\'');
This outputs:
without special chars: 'qwe 123 '
spaces as pluses: 'qwe+123++'
If you want to group multiple spaces into one + then use "\s+" as your regex instead (remember to escape the slash).
I had a similar problem to solve and I used following method:
text.replaceAll("\\p{Punct}+", "").replaceAll("\\s+", "+");
Code with time bench marking
public static String cleanPunctuations(String text) {
return text.replaceAll("\\p{Punct}+", "").replaceAll("\\s+", "+");
}
public static void test(String in){
long t1 = System.currentTimeMillis();
String out = cleanPunctuations(in);
long t2 = System.currentTimeMillis();
System.out.println("In=" + in + "\nOut="+ out + "\nTime=" + (t2 - t1)+ "ms");
}
public static void main(String[] args) {
String s1 = "My text with 212354 digits spaces and \n newline \t tab " +
"[`~!##$%^&*()_+[\\\\]\\\\\\\\;\\',./{}|:\\\"<>?] special chars";
test(s1);
String s2 = "\"Sample Text=\" with - minimal \t punctuation's";
test(s2);
}
Sample Output
In=My text with 212354 digits spaces and
newline tab [`~!##$%^&*()_+[\\]\\\\;\',./{}|:\"<>?] special chars
Out=My+text+with+212354+digits+spaces+and+newline+tab+special+chars
Time=4ms
In="Sample Text=" with - minimal punctuation's
Out=Sample+Text+with+minimal+punctuations
Time=0ms
you can use a regex like this:
[<#![CDATA[¢<(+|!$*);¬/¦,%_>?:#="~{#}\]]]#>]`
remove "#" at first and at end from expression
regards
#npinti
using "\w" is the same as "\dA-Za-z"
This worked for me:
String result = str.replaceAll("[^\\w ]", "").replaceAll("\\s+", "+");
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]", "");