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Regex to convert this String

How can I convert this String AB23-01-0001 to AB23010001( replacing the "-" with "") and AB230001 (removing the middle part) using regex in Java, right row I'm using replace for the first case and substring and appending them into a SB for the second case. Just wanted to know how to achieve it using REGEX.

Why not use the method built into the String class?
String newString = "AB23-01-0001".replaceAll("[-]", "");
Note the use of [] - a regex string, since you are just replacing a -, you can omit them.

str = "AB23-01-0001"
happy = str.replaceAll("[^a-zA-Z0-9]", "");
from https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replaceAll(java.lang.String,%20java.lang.String)

Try this:
Pattern p1 = Pattern.compile("-[^-]*-");
Matcher m1 = p1.matcher("AB23-01-0001");
System.out.println(m1.replaceAll(""));
To test it
A replacement to the patter is Pattern.compile("-[\\d\\w]+-")

You can use build-in method to remove both parts at once with RegEx:
String value = "AB23-01-0001";
value = value.replaceAll("-[\\d\\w]+-", "");

String index out of range with replace all

How can I replace mapDir surrounded by <> to a certain string?
String mapDir = "D:\\mapping\\specialists\\ts_gpc\\";
String test = "foo: <mapDir> -bar";
println(test.replaceAll("<mapDir>", mapDir));
The above gives me a StringIndexOutOfBoundsException.
This code below for me, but I think pure java has to work as well.
static String replaceWord(String original, String find, String replacement) {
int i = original.indexOf(find);
if (i < 0) {
return original; // return original if 'find' is not in it.
}
String partBefore = original.substring(0, i);
String partAfter = original.substring(i + find.length());
return partBefore + replacement + partAfter;
}

You dont need replaceAll method as you are not using regex. Instead you could work with replace api like below:
String mapDir = "D:\\mapping\\specialists\\ts_gpc\\";
String test = "foo: <mapDir> -bar";
System.out.println(test.replace("<mapDir>", mapDir));

replaceAll in String uses a regex, as specified in the documentation:
Note that backslashes () and dollar signs ($) in the replacement string may cause the results to be different than if it were being treated as a literal replacement string; see Matcher.replaceAll. Use Matcher.quoteReplacement(java.lang.String) to suppress the special meaning of these characters, if desired.
Thus, you should escape your replacement string like this:
String mapDir = "D:\\mapping\\specialists\\ts_gpc\\";
String test = "foo: <mapDir> -bar";
System.out.println(test.replaceAll("<mapDir>", Matcher.quoteReplacement(mapDir)));
which gives the output:
foo: D:\mapping\specialists\ts_gpc\ -bar

Since replaceAll works with regex, you need to re-escape the backslashes:
String mapDir = "D:\\\\mapping\\\\specialists\\\\ts_gpc\\\\";
String test = "foo: <mapDir> -bar";
System.out.println(test.replaceAll("<mapDir>", mapDir));
Quoting this answer:
by the time the regex compiler sees the pattern you've given it, it
sees only a single backslash (since Java's lexer has turned the double
backwhack into a single one)

How to replace all numbers in java string

I have string like this String s="ram123",d="ram varma656887"
I want string like ram and ram varma so how to seperate string from combined string
I am trying using regex but it is not working
PersonName.setText(cursor.getString(cursor.getColumnIndex(cursor
.getColumnName(1))).replaceAll("[^0-9]+"));

The correct RegEx for selecting all numbers would be just [0-9], you can skip the +, since you use replaceAll.
However, your usage of replaceAll is wrong, it's defined as follows: replaceAll(String regex, String replacement). The correct code in your example would be: replaceAll("[0-9]", "").

You can use the following regex: \d for representing numbers. In the regex that you use, you have a ^ which will check for any characters other than the charset 0-9
String s="ram123";
System.out.println(s);
/* You don't need the + because you are using the replaceAll method */
s = s.replaceAll("\\d", ""); // or you can also use [0-9]
System.out.println(s);

To remove the numbers, following code will do the trick.
stringname.replaceAll("[0-9]","");

Please do as follows
String name = "ram varma656887";
name = name.replaceAll("[0-9]","");
System.out.println(name);//ram varma
alternatively you can do as
String name = "ram varma656887";
name = name.replaceAll("\\d","");
System.out.println(name);//ram varma
also something like given will work for you
String given = "ram varma656887";
String[] arr = given.split("\\d");
String data = new String();
for(String x : arr){
data = data+x;
}
System.out.println(data);//ram varma

i think you missed the second argument of replace all. You need to put a empty string as argument 2 instead of actually leaving it empty.
try
replaceAll(<your regexp>,"")

you can use Java - String replaceAll() Method.
This method replaces each substring of this string that matches the given regular expression with the given replacement.
Here is the syntax of this method:
public String replaceAll(String regex, String replacement)
Here is the detail of parameters:
regex -- the regular expression to which this string is to be matched.
replacement -- the string which would replace found expression.
Return Value:
This method returns the resulting String.
for your question use this
String s = "ram123", d = "ram varma656887";
System.out.println("s" + s.replaceAll("[0-9]", ""));
System.out.println("d" + d.replaceAll("[0-9]", ""));

Difference between String replace() and replaceAll()

What's the difference between java.lang.String 's replace() and replaceAll() methods,
other than later uses regex? For simple substitutions like, replace . with / ,
is there any difference?

In java.lang.String, the replace method either takes a pair of char's or a pair of CharSequence's (of which String is a subclass, so it'll happily take a pair of String's). The replace method will replace all occurrences of a char or CharSequence. On the other hand, the first String arguments of replaceFirst and replaceAll are regular expressions (regex). Using the wrong function can lead to subtle bugs.

Q: What's the difference between the java.lang.String methods replace() and replaceAll(), other than that the latter uses regex.
A: Just the regex. They both replace all :)
http://docs.oracle.com/javase/8/docs/api/java/lang/String.html
PS:
There's also a replaceFirst() (which takes a regex)

Both replace() and replaceAll() replace all occurrences in the String.
Examples
I always find examples helpful to understand the differences.
replace()
Use replace() if you just want to replace some char with another char or some String with another String (actually CharSequence).
Example 1
Replace all occurrences of the character x with o.
String myString = "__x___x___x_x____xx_";
char oldChar = 'x';
char newChar = 'o';
String newString = myString.replace(oldChar, newChar);
// __o___o___o_o____oo_
Example 2
Replace all occurrences of the string fish with sheep.
String myString = "one fish, two fish, three fish";
String target = "fish";
String replacement = "sheep";
String newString = myString.replace(target, replacement);
// one sheep, two sheep, three sheep
replaceAll()
Use replaceAll() if you want to use a regular expression pattern.
Example 3
Replace any number with an x.
String myString = "__1_6____3__6_345____0";
String regex = "\\d";
String replacement = "x";
String newString = myString.replaceAll(regex, replacement);
// __x_x____x__x_xxx____x
Example 4
Remove all whitespace.
String myString = " Horse Cow\n\n \r Camel \t\t Sheep \n Goat ";
String regex = "\\s";
String replacement = "";
String newString = myString.replaceAll(regex, replacement);
// HorseCowCamelSheepGoat
See also
Documentation
replace(char oldChar, char newChar)
replace(CharSequence target, CharSequence replacement)
replaceAll(String regex, String replacement)
replaceFirst(String regex, String replacement)
Regular Expressions
Tutorial
List of patterns

The replace() method is overloaded to accept both a primitive char and a CharSequence as arguments.
Now as far as the performance is concerned, the replace() method is a bit faster than replaceAll() because the latter first compiles the regex pattern and then matches before finally replacing whereas the former simply matches for the provided argument and replaces.
Since we know the regex pattern matching is a bit more complex and consequently slower, then preferring replace() over replaceAll() is suggested whenever possible.
For example, for simple substitutions like you mentioned, it is better to use:
replace('.', '\\');
instead of:
replaceAll("\\.", "\\\\");
Note: the above conversion method arguments are system-dependent.

Both replace() and replaceAll() accepts two arguments and replaces all occurrences of the first substring(first argument) in a string with the second substring (second argument).
replace() accepts a pair of char or charsequence and replaceAll() accepts a pair of regex.
It is not true that replace() works faster than replaceAll() since both uses the same code in its implementation
Pattern.compile(regex).matcher(this).replaceAll(replacement);
Now the question is when to use replace and when to use replaceAll().
When you want to replace a substring with another substring regardless of its place of occurrence in the string use replace(). But if you have some particular preference or condition like replace only those substrings at the beginning or end of a string use replaceAll(). Here are some examples to prove my point:
String str = new String("==qwerty==").replaceAll("^==", "?"); \\str: "?qwerty=="
String str = new String("==qwerty==").replaceAll("==$", "?"); \\str: "==qwerty?"
String str = new String("===qwerty==").replaceAll("(=)+", "?"); \\str: "?qwerty?"

To throw more light with an example into how both are going to work for below code:
public static void main(String[] args)
{
String s = "My\\s aaab\\s is\\s aaab\\s name";
String s1 = s.replace("\\s", "c");
System.out.println(s1);
String s2 = s.replaceAll("\\s", "c");
System.out.println(s2);
}
Output:
Myc aaabc isc aaabc name
My\scaaab\scis\scaaab\scname
Explanation
s.replace replaces "\\s" sequence of characters with c. Hence, the output in first line.
s.replaceAll considers \\s as a regex rather(equivalent to space) and replaces spaces with c. \\s in String s is escaped with first \ encountered and becomes \s.
Intellij Idea is smart enough to notify you of the usage as well. If you take a closer look at below image, you will see the difference in interpretation by Intellij idea for replace and replaceAll usage.

String replace(char oldChar, char newChar)
Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.
String replaceAll(String regex, String replacement
Replaces each substring of this string that matches the given regular expression with the given replacement.

Old thread I know but I am sort of new to Java and discover one of it's strange things. I have used String.replaceAll() but get unpredictable results.
Something like this mess up the string:
sUrl = sUrl.replaceAll( "./", "//").replaceAll( "//", "/");
So I designed this function to get around the weird problem:
//String.replaceAll does not work OK, that's why this function is here
public String strReplace( String s1, String s2, String s )
{
if((( s == null ) || (s.length() == 0 )) || (( s1 == null ) || (s1.length() == 0 )))
{ return s; }
while( (s != null) && (s.indexOf( s1 ) >= 0) )
{ s = s.replace( s1, s2 ); }
return s;
}
Which make you able to do:
sUrl=this.strReplace("./", "//", sUrl );
sUrl=this.strReplace( "//", "/", sUrl );

As alluded to in wickeD's answer, with replaceAll the replacement string is handled differently between replace and replaceAll. I expected a[3] and a[4] to have the same value, but they are different.
public static void main(String[] args) {
String[] a = new String[5];
a[0] = "\\";
a[1] = "X";
a[2] = a[0] + a[1];
a[3] = a[1].replaceAll("X", a[0] + "X");
a[4] = a[1].replace("X", a[0] + "X");
for (String s : a) {
System.out.println(s + "\t" + s.length());
}
}
The output of this is:
\ 1
X 1
\X 2
X 1
\X 2
This is different from perl where the replacement does not require the extra level of escaping:
#!/bin/perl
$esc = "\\";
$s = "X";
$s =~ s/X/${esc}X/;
print "$s " . length($s) . "\n";
which prints
\X 2
This can be quite a nuisance, as when trying to use the value returned by java.sql.DatabaseMetaData.getSearchStringEscape() with replaceAll().

From Java 9 there is some optimizations in replace method.
In Java 8 it uses a regex.
public String replace(CharSequence target, CharSequence replacement) {
return Pattern.compile(target.toString(), Pattern.LITERAL).matcher(
this).replaceAll(Matcher.quoteReplacement(replacement.toString()));
}
From Java 9 and on.
And Stringlatin implementation.
Which perform way better.
https://medium.com/madhash/composite-pattern-in-a-nutshell-ad1bf78479cc?source=post_internal_links---------2------------------

replace() method doesn't uses regex pattern whereas replaceAll() method uses regex pattern. So replace() performs faster than replaceAll().

To add to the already selected "Best Answer" (and others that are just as good like Suragch's), String.replace() is constrained by replacing characters that are sequential (thus taking CharSequence). However, String.replaceAll() is not constrained by replacing sequential characters only. You could replace non-sequential characters as well as long as your regular expression is constructed in such a way.
Also (most importantly and painfully obvious), replace() can only replace literal values; whereas replaceAll can replace 'like' sequences (not necessarily identical).

replace works on char data type but replaceAll works on String datatype and both replace the all occurrences of first argument with second argument.

How to Replace dot (.) in a string in Java

I have a String called persons.name
I want to replace the DOT . with /*/ i.e my output will be persons/*/name
I tried this code:
String a="\\*\\";
str=xpath.replaceAll("\\.", a);
I am getting StringIndexOutOfBoundsException.
How do I replace the dot?

You need two backslashes before the dot, one to escape the slash so it gets through, and the other to escape the dot so it becomes literal. Forward slashes and asterisk are treated literal.
str=xpath.replaceAll("\\.", "/*/"); //replaces a literal . with /*/
http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replaceAll(java.lang.String,%20java.lang.String)

If you want to replace a simple string and you don't need the abilities of regular expressions, you can just use replace, not replaceAll.
replace replaces each matching substring but does not interpret its argument as a regular expression.
str = xpath.replace(".", "/*/");

Use Apache Commons Lang:
String a= "\\*\\";
str = StringUtils.replace(xpath, ".", a);
or with standalone JDK:
String a = "\\*\\"; // or: String a = "/*/";
String replacement = Matcher.quoteReplacement(a);
String searchString = Pattern.quote(".");
String str = xpath.replaceAll(searchString, replacement);

return sentence.replaceAll("\s",".");