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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]", ""));

Ignoring ? in a regex in String replaceFirst method

I have below program:
String one = "Hello I am IN (?)";
String two = one.replaceFirst(" IN (?)", " IN ('one','two')");
System.out.println(two);
The output is:
Hello I am IN ('one','two')(?)
The output I am expecting is:
Hello I am IN ('one','two')
What is the issue? It has to do with replaceFirst method having first argument as a regex.

replaceFirst uses a regular expression as its first argument. ( and ) are meta characters used to denote capturing groups, ? being the non-greedy quantifier
You can either escape meta characters
String two = one.replaceFirst(" IN \\(\\?\\)", " IN ('one','two')");
or (better) using the replace method as mentioned by #RJ
String two = one.replace(" IN (?)", " IN ('one','two')");

Since the first parameter of replaceFirst() is a regex, you need to escape the (?) as (?) has a special meaning in regex(read meta character).
String two = one.replaceFirst(" IN \\(\\?\\)", " IN ('one','two')");
To avoid escaping, you can use the replace() method, which doesn't take a regex as the replace string.
String two = one.replace(" IN (?)", " IN ('one','two')");

Try this one
String one = "Hello I am IN (?)";
String two = one.replaceFirst(" IN \\(\\?\\)", " IN ('one','two')");
System.out.println(two);

Regular expression for string with apostrophes

I'm trying to build regex which will filter form string all non-alphabetical characters, and if any string contains single quotes then I want to keep it as an exception to the rule.
So for example when I enter
car's34
as a result I want to get
car's
when I enter
*&* Lisa's car 0)*
I want to get
Lisa's
at the moment I use this:
string.replaceAll("[^A-Za-z]", "")
however, it gives me only alphabets, and removed the desired single quotas.

This will also remove apostrophes that are not "part if words":
string = string.replaceAll("[^A-Za-z' ]+|(?<=^|\\W)'|'(?=\\W|$)", "")
.replaceAll(" +", " ").trim();
This first simply adds an apostrophe to the list of chars you want to keep, but uses look arounds to find apostrophes not within words, so
I'm a ' 123 & 'test'
would become
I'm a test
Note how the solitary apostrophe was removed, as well as the apostrophes wrapping test, but I'm was preserved.
The subsequent replaceAll() is to replace multiple spaces with a single space, which will result if there's a solitary apostrophe in the input. A further call to trim() was added in case it occurs at the end of the input.
Here's a test:
String string = "I'm a ' 123 & 'test'";
string = string.replaceAll("[^A-Za-z' ]+|(?<=^|\\W)'|'(?=\\W|$)", "").replaceAll(" +", " ").trim();
System.out.println(string);
Output:
I'm a test

Isn't this working ?
[^A-Za-z']

The obvious solution would be:
string.replaceAll("[^A-Za-z']", "")
I suspect you want something more.

You can try the regular expression:
[^\p{L}' ]
\p{L} denote the category of Unicode letters.
In ahother hand, you need to use a constant of Pattern for avoid recompiled the expression every time, something like that:
private static final Pattern REGEX_PATTERN =
Pattern.compile("[^\\p{L}' ]");
public static void main(String[] args) {
String input = "*&* Lisa's car 0)*";
System.out.println(
REGEX_PATTERN.matcher(input).replaceAll("")
); // prints " Lisa's car "
}

#Bohemian has a good idea but word boundaries are called for instead of lookaround:
string.replaceAll("([^A-Za-z']|\B'|'\B)+", " ");

Parse and remove special characters in java regex

So we were looking at some of the other regex posts and we are having trouble removing a special case in one instance; the special character is in the beginning of the word.
We have the following line in our code:
String k = s.replaceAll("([a-z]+)[()?:!.,;]*", "$1");
where s is a singular word. For example, when parsing the sentence "(hi hi hi)" by tokenizing it, and then performing the replaceAll function on each token, we get an output of:
(hi
hi
hi
What are we missing in our regex?

You can use an easier approach - replace the characters that you do not want with spaces:
String k = s.replaceAll("[()?:!.,;]+", " ");

Position matters so you would need to match the excluded charcters before the capturing group also:
String k = s.replaceAll("[()?:!.,;]*([a-z]+)[()?:!.,;]*", "$1");

your replace just removed the "special chars" after the [a-z]+, that's why the ( before hi is left there.
If you know s is a single word
you could either:
String k = s.replaceAll("\\W*(\\w+)\\W*", "$1");
or
String k = s.replaceAll("\\W*", "");

This can be more simple
try this :
String oldString = "Hi There ##$ What is %#your name?##$##$ 0123$$";
System.out.println(oldString.replaceAll("[\\p{Punct}\\s\\d]+", " ");
output :
Hi There What is your name 0123
So it also accepts numeric.
.replaceAll("[\p{Punct}\s\d]+", " ");
will replace alll the Punctuations used which includes almost all the special characters.

Match only first and last character of a string

I had a look at other stackoverflow questions and couldn't find one that asked the same question, so here it is:
How do you match the first and last characters of a string (can be multi-line or empty).
So for example:
String = "this is a simple sentence"
Note that the string includes the beginning and ending quotation marks.
How do I get match the first and last characters where the string begins and ends with a quotation mark (").
I tried:
^"|$" and \A"\Z"
but these do not produce the desired result.
Thanks for your help in advance :)

Is this what you are looking for?
String input = "\"this is a simple sentence\"";
String result = input.replaceFirst("(?s)^\"(.*)\"$", " $1 ");
This will replace the first and last character of the input string with spaces if it starts and ends with ". It will also work across multiple lines since the DOTALL flag is specified by (?s).

The regex that matches the whole input ".*". In java, it looks like this:
String regex = "\".*\"";
System.out.println("\"this is a simple sentence\"".matches(regex)); // true
System.out.println("this is a simple sentence".matches(regex)); // false
System.out.println("this is a simple sentence\"".matches(regex)); // false
If you want to remove the quotes, use this:
String input = "\"this is a simple sentence\"";
input = input.replaceAll("(^\"|\"$)", "")); // this is a simple sentence (without any quotes)
If you want this to work over multiple lines, use this:
String input = "\"this is a simple sentence\"\n\"and another sentence\"";
System.out.println(input + "\n");
input = input.replaceAll("(?m)(^\"|\"$)", "");
System.out.println(input);
which produces output:
"this is a simple sentence"
"and another sentence"
this is a simple sentence
and another sentence
Explanation of regex (?m)(^"|"$):
(?m) means "Caret and dollar match after and before newlines for the remainder of the regular expression"
(^"|"$) means ^" OR "$, which means "start of line then a double quote" OR "double quote then end of line"

Why not use the simple logic of getting the first and last characters based on charAt method of String? Place a few checks for empty/incomplete strings and you should be done.

String regexp = "(?s)\".*\"";
String data = "\"This is some\n\ndata\"";
Matcher m = Pattern.compile(regexp).matcher(data);
if (m.find()) {
System.out.println("Match starts at " + m.start() + " and ends at " + m.end());
}