Develop Reference

Java Regex from beginning to first char

How can I find any word from beginning of string to first char "~" using java?
Example:
Worddjjfdskfjsdkfjdsj ~ Word ~ Word
I want it to capture
Worddjjfdskfjsdkfjdsj

You can also do it without regex in a very simple way.
First of all use indexOf() String method to find the index of the "~" character. Then use the substring() method to extract the string you are lookin for.
Here is an example:
String stringToProcess = "hello~world";
int charIndex = stringToProcess.indexOf('~');
String finalString = stringToProcess.substring(0, charIndex);

You can use this regex to capture all character from start of string ^ to first occurrence of ~:
^[^~]*
[^~]* is negation based regex that matches 0 or more of anything but ~

Without regex it can be solved
Simply split your string by ~.
String str[] = "Worddjjfdskfjsdkfjdsj ~ Word ~ Word".split("~");
System.out.println(str[0]);

Here is regular expression that you can use: ^(.*?)~.
However in your simple case you do not need regular expressions at all. Use indexOf() and substring():
int tilda = str.indexOf('~');
if (tilda >= 0) {
word = str.substring(0, tilda);
}

How to replace \ with . in java String

I want to replace \ with . in String java.
Example src\main\java\com\myapp\AppJobExecutionListener
Here I want to get like src.main.java.com.myapp.AppJobExecutionListener
I tried str.replaceAll("\\","[.]") and str.replaceAll("\\","[.]") but it is not working.
I am still getting original string src\main\java\com\myapp\AppJobExecutionListener

String is immutable in Java, so whatever methods you invoke on the String object are not reflected on it unless you reassign it.
String s = "ABC";
s.replaceAll("B","D");
System.out.println(s); //still prints "ABC"
s = s.replaceAll("B","D");
System.out.println(s); //prints "ADC"

Currently you're using replaceAll, which takes regular expression patterns. That makes life much more complicated than it needs to be. Unless you're trying to use regular expressions, just use String.replace instead.
In fact, as you're only replacing one character with another, you can just use character literals:
String replaced = original.replace('\\', '.');
The \ is doubled as it's the escape character in Java character literals - but as the above doesn't use regular expressions, the period has no special meaning.

Assign it back to string str variable, .String#replaceAll doesn't changes the string itself, it returns a new String.
str = str.replaceAll("\\\\",".")

Can you try this:
String original = "Some text with \\ and rest of the text";
String replaced = original.replace("\\",".");
System.out.println(replaced);

'\' character is doubled in a string like '\\'. So '\\' character should be used to replace it with '.' character and also using replace instead of replaceAll would be enough to make it. Here is a sample;
public static void main(String[] args) {
String myString = "src\\main\\java\\com\\vxl\\appanalytix\\AppJobExecutionListener";
System.out.println("Before Replaced: " + myString);
myString = myString.replace("\\", ".");
System.out.println("After Replaced: " + myString);
}
This will give you:
Before Replaced: src\main\java\com\vxl\appanalytix\AppJobExecutionListener
After Replaced: src.main.java.com.vxl.appanalytix.AppJobExecutionListener

With String replaceAll(String regex, String replacement):
str = str.replaceAll("\\\\", ".");
With String replace(char oldChar, char newChar):
str = str.replace('\\', '.');
With String replace(CharSequence target, CharSequence replacement)
str = str.replace("\\", ".");

String replaced = original.replace('\', '.');
try this its works well

Use replace instead of replaceall
String my_str="src\\main\\java\\com\\vxl\\appanalytix\\AppJobExecutionListener";
String my_new_str = my_str.replace("\\", ".");
System.out.println(my_new_str);
DEMO AT IDEONE.COM

replaceAll takes a regex as the first parameter.
To replace the \ you need to double escape. You need an additional \ to escape the first . And as it is a regex input you need to escape those again. As other answers have said string is immutable so you will need to assign the result
String newStr = str.replaceAll("\\\\", ".");
The second parameter is not regex so you can just put . in there but note you need four slashes to replace one backslash if using replaceAll

i tried this:
String s="src\\main\\java\\com\\vxl\\appanalytix\\AppJobExecutionListener";
s = s.replace("\\", ".");
System.out.println("s: "+ s);
output: src.main.java.com.vxl.appanalytix.AppJobExecutionListener

Just change the line to
str = str.replaceAll("\\",".");
Edit : I didnt try it, because the problem here is not whether its a correct regex,but the problem here is that he is not assigning the str to new str value. Anyways regex corrected now.

replace all + with -

I am trying to replace a + character into a hyphen I have in my string.
String str = "word+word";
str.replaceAll('+ ', '-');
I tried using replace but it throwing an exception.Is there any other method to do this.

Use
str = str.replaceAll("\\+", "-");
A few errors in your code :
replaceAll takes strings, not chars
the + char must be escaped as the first argument is a regular expression (and \ itself must be escaped in java string literals)
you must take the return of the function : as String is immutable the function doesn't change it but returns another string

Just use replace:
str = str.replace('+', '-');
This one doesn't work on regex but take characters as they are.
Also as you see you have to reassing value again to your str variable because String in Java are immutable. In this case method replace doesn't change current String (str) but create new one with replaced + to '-'.

`replaceAll´ is for regular expressions and strings are immutable. Use:
str = str.replace("+", "-");
instead...

The replaceAll function takes a regular expression as its first argument. It so happens that + is a special character in regular expression language. Try replacing + with \\+. This will escape the plus sign, thus making the code to treat it like a normal character.
Also, the replaceAll method yields a string, so that will not work. Try doing:
String str = "word+word";
str = str.replaceAll("\\+ ", "-");

Use "" as opposed to '' in replaceAll.
String java.lang.String.replaceAll(String regex, String replacement)

If you are not sure about the escape sequence you need to use,
You could simply do this.
str = str.replaceAll(Pattern.quote("+"), "-");
This will automatically escape the regex predefined tokens to match in a literal way

Remove doubled letter from a string using java

I need to remove a doubled letter from a string using regex operations in java.
Eg: PRINCEE -> PRINCE
APPLE -> APLE

Simple Solution (remove duplicate characters)
Like this:
final String str = "APPLEE";
String replaced = str.replaceAll("(.)\\1", "$1");
System.out.println(replaced);
Output:
APLE
Not just any Chracters, Letters only
As #Jim comments correctly, the above matches any double character, not just letters. Here are a few variations that just match letters:
// the basics, ASCII letters. these two are equivalent:
str.replaceAll("([A-Za-z])\\1", "$1");
str.replaceAll("(\\p{Alpha})\\1", "$1");
// Unicode Letters
str.replaceAll("(\\p{L})\\1", "$1");
// anything where Character.isLetter(ch) returns true
str.replaceAll("(\\p{javaLetter})\\1", "$1");
References:
For additional reference:
Character.isLetter(ch) (javadocs)
any method in Character of
the form Character.isXyz(char)
enables a pattern named
\p{javaXyz} (mind the
capitalization). This mechanism is
described in the Pattern
javadocs
Unicode blocks and categories can
also be matched with the \p and
\P constructs as in Perl. \p{prop}
matches if the input has the
property prop, while \P{prop} does
not match if the input has that
property. This mechanism is also
described in the Pattern
javadocs

String s = "...";
String replaced = s.replaceAll( "([A-Z])\\1", "$1" );

If you want to replace just duplicate ("AA"->"A", "AAA" -> "AA") use
public String undup(String str) {
return str.replaceAll("(\\w)\\1", "$1");
}
To replace triplicates etc use: str.replaceAll("(\\w)\\1+", "$1");
To replace only a single dupe is a long string (AAAA->AAA, AAA->AA) use: str.replaceAll("(\\w)(\\1+)", "$2");

This can be done simply by iterating over the String instead of having to resort to regexes.
StringBuilder ret=new StringBuilder(text.length());
if (text.length()==0) return "";
ret.append(text.charAt(0));
for(int i=1;i<text.length();i++){
if (text.charAt(i)!=text.charAt(i-1))
ret.append(text.charAt(i));
}
return ret.toString();

Remove end of line characters from end of Java String

I have a string which I'd like to remove the end of line characters from the very end of the string only using Java
"foo\r\nbar\r\nhello\r\nworld\r\n"
which I'd like to become
"foo\r\nbar\r\nhello\r\nworld"
(This question is similar to, but not the same as question 593671)

You can use s = s.replaceAll("[\r\n]+$", "");. This trims the \r and \n characters at the end of the string
The regex is explained as follows:
[\r\n] is a character class containing \r and \n
+ is one-or-more repetition of
$ is the end-of-string anchor
References
regular-expressions.info/Anchors, Character Class, Repetition
Related topics
You can also use String.trim() to trim any whitespace characters from the beginning and end of the string:
s = s.trim();
If you need to check if a String contains nothing but whitespace characters, you can check if it isEmpty() after trim():
if (s.trim().isEmpty()) {
//...
}
Alternatively you can also see if it matches("\\s*"), i.e. zero-or-more of whitespace characters. Note that in Java, the regex matches tries to match the whole string. In flavors that can match a substring, you need to anchor the pattern, so it's ^\s*$.
Related questions
regex, check if a line is blank or not
how to replace 2 or more spaces with single space in string and delete leading spaces only

Wouldn't String.trim do the trick here?
i.e you'd call the method .trim() on your string and it should return a copy of that string minus any leading or trailing whitespace.

The Apache Commons Lang StringUtils.stripEnd(String str, String stripChars) will do the trick; e.g.
String trimmed = StringUtils.stripEnd(someString, "\n\r");
If you want to remove all whitespace at the end of the String:
String trimmed = StringUtils.stripEnd(someString, null);

Well, everyone gave some way to do it with regex, so I'll give a fastest way possible instead:
public String replace(String val) {
for (int i=val.length()-1;i>=0;i--) {
char c = val.charAt(i);
if (c != '\n' && c != '\r') {
return val.substring(0, i+1);
}
}
return "";
}
Benchmark says it operates ~45 times faster than regexp solutions.

If you have Google's guava-librariesin your project (if not, you arguably should!) you'd do this with a CharMatcher:
String result = CharMatcher.any("\r\n").trimTrailingFrom(input);

String text = "foo\r\nbar\r\nhello\r\nworld\r\n";
String result = text.replaceAll("[\r\n]+$", "");

"foo\r\nbar\r\nhello\r\nworld\r\n".replaceAll("\\s+$", "")
or
"foo\r\nbar\r\nhello\r\nworld\r\n".replaceAll("[\r\n]+$", "")