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How do I make multi replace() java

I have a string with \r\n, \r, \n or \" characters in it. How can I replace them faster?
What I already have is:
String s = "Kerner\\r\\n kyky\\r hihi\\n \\\"";
System.out.println(s.replace("\\r\\n", "\n").replace("\\r", "").replace("\\n", "").replace("\\", ""));
But my code does not look beautiful enough.
I found on the Internet something like:
replace("\\r\\n|\\r|\\n|\\", "")
I tried that, but it didn't work.

You can wrap it in a method, put /r/n, /n and /r in a list. iterate the list and replace all such characters and return the modified string.
public String replaceMultipleSubstrings(String original, List<String> mylist){
String tmp = original;
for(String str: mylist){
tmp = tmp.replace(str, "");
}
return tmp;
}
Test:
mylist.add("\\r");
mylist.add("\\r\\n");
mylist.add("\\n");
mylist.add("\\"); // add back slash
System.out.println("original:" + s);
String x = new Main().replaceMultipleSubstrings(s, mylist);
System.out.println("modified:" + x);
Output:
original:Kerner\r\n kyky\r hihi\n \"
modified:Kerner kyky hihi "

I don't know if your current replacement logic be correct, but it says now that either \n, \r, or \r\n gets replaced with empty string, and backslash also gets replaced with empty string. If so, then you can try the following regex replace all:
String s = "Kerner\\r\\n kyky\\r hihi\\n \\\"";
System.out.println(s.replaceAll("\\r|\\n|\\r\\n|\\\\", ""));
One problem I saw with your attempt is that you are calling replace(), not replaceAll(), so it would only do a single replacement and then stop.

String.replaceAll() can be used, in your question you tried to use String.replace() which does not interpret regular expressions, only plain replacement strings...
You also need to escape the \\ again, i.e. \\\\ instead of \\
String s = "Kerner\\r\\n kyky\\r hihi\\n \\\"";
System.out.println(s.replaceAll("\\\\r|\\\\n|\\\\\"", ""));
Output
Kerner kyky hihi
Note the differences between String.replaceAll() and String.replace()
String.replaceAll()
Replaces each substring of this string that matches the given regular
expression with the given replacement.
String.replace()
Replaces each substring of this string that matches the literal target
sequence with the specified literal replacement sequence.

Use a regular expression if you want to do all the replaces in one go.
http://www.javamex.com/tutorials/regular_expressions/search_replace.shtml

Replace character with a particular String in Java

I have a program which should replace a alternate characters in the string with a new string. Lets say I have...
String s1 = "JAVAJAVA";
String s2 = "VA";
Output:
VAAVAAVAAVAA
Character in the each alternate index of s1 should be replaced with s2. I've tried using StringBulider but I'm not able to proceed further with it. Can someone help me out on this please. thanks

Try this:
s1 = s1.replaceAll(".(.)", s2+"$1");
Explanation: Regular Expression ".(.)" matches every 2 characters. The second char is "remembered" (brackets), so you can re-use it in the replacement part ($1):

If you want to go other way than REGEX other simple solution can be, though regex one should be preferred one
1) Split String to char array with String class toCharArray() function
2) Replace the new character at alternate position by running loop
3) Convert back array to string with new String(charArray)

have you tried string replace function?
here are some examples: http://javarevisited.blogspot.ch/2011/12/java-string-replace-example-tutorial.html
You can use it like this:
String newString = s1.replace("J", s2);

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]", ""));

having trouble with arrays and maybe split

String realstring = "&&&.&&&&";
Double value = 555.55555;
String[] arraystring = realstring.split(".");
String stringvalue = String.valueof(value);
String [] valuearrayed = stringvalue.split(".");
System.out.println(arraystring[0]);
Sorry if it looks bad. Rewrote on my phone. I keep getting ArrayIndexOutOfBoundsException: 0 at the System.out.println. I have looked and can't figure it out. Thanks for the help.

split() takes a regexp as argument, not a literal string. You have to escape the dot:
string.split("\\.");
or
string.split(Pattern.quote("."));
Or you could also simply use indexOf('.') and substring() to get the two parts of your string.
And if the goal is to get the integer part of a double, you could also simply use
long truncated = (long) doubleValue;

split uses regex as parameter and in regex . means "any character except line separators", so you could expect that "a.bc".split(".") would create array of empty strings like ["","","","",""]. Only reason it is not happening is because (from split javadoc)
This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.
so because all strings are empty you get empty array (and that is because you see ArrayIndexOutOfBoundsException).
To turn off removal mechanism you would have to use split(regex, limit) version with negative limit.
To split on . literal you need to escape it with \. (which in Java needs to be written as "\\." because \ is also Strings metacharacter) or [.] or other regex mechanism.

Dot (.) is a special character so you need to escape it.
String realstring = "&&&.&&&&";
String[] partsOfString = realstring.split("\\.");
String part1 = partsOfString[0];
String part2 = partsOfString[1];
System.out.println(part1);
this will print expected result of
&&&
Its also handy to test if given string contains this character. You can do this by doing :
if (string.contains(".")) {
// Split it.
} else {
throw new IllegalArgumentException("String " + string + " does not contain .");
}

Removing backslashes in the numbers sequence

What regular expression can get a number sequence from the input string, contains backslashes and not a numbers, for example -
"12\34a56ss7890"
I need to -
1234567890

If we assume you have this in a String. You could do something like:
string = string.replaceAll("\\D", "");
This will replace all non digit Characters from your String.

str.replaceAll("[^\d]", "");
bootnote: im not a java developer, but the regex itself should be correct

Sorry for adding another Answer but this is needed because this won't fit to an Comment.
I think this is because of the \34. If I do call System.out.print("12\34a56ss7890"); I will get the following output 12a56ss7890. This is because the \34 will be escaped. This is an Issue in Java. You can fix this by first calling this Method on your InputStream:
private InputStreamReader replaceBackSlashes() throws Exception {
FileInputStream fis = new FileInputStream(new File("PATH TO A FILE");
Scanner in = new Scanner(fis, "UTF-8");
ByteArrayOutputStream out = new ByteArrayOutputStream();
while (in.hasNext()) {
String nextLine = in.nextLine().replace("\", "");
out.write(nextLine.getBytes());
out.write("\n".getBytes());
}
return new InputStreamReader(new ByteArrayInputStream(out.toByteArray()));
}
BTW: Sorry for my Edit, but there was a little Mistake in the Code.
After calling this Method you will convert your InputStream to a String and the call this on the String:
string = string.replaceAll("\\D", "");
This should hopefully work now :)

String num;
String str =" 12\34a56ss7890";
str= str.replace("\34", "34");
String regex = "[\\d]+";
Matcher matcher = Pattern.compile( regex ).matcher( str);
while (matcher.find( ))
{
num = matcher.group();
System.out.print(num);
}
replace \34 by 34 and match the rest using regular expression.

User a regular exxpression.
String numvber;
String str =" 12\34a56ss7890";
str= str.replace("\34", "34");
String regex = "[\\d]+";//match only digits.
Matcher matcher = Pattern.compile( regex ).matcher( str);
while (matcher.find( ))
{
num = matcher.group();
System.out.print(num);
}

The following example:
String a ="1\2sas";
String b ="1\\2sas";
System.out.println(a.replaceAll("[a-zA-Z\\\\]",""));
System.out.println(b.replaceAll("[a-zA-Z\\\\]",""));
gives output:
1X
12
where X is not a X but a little rectangle - a symbol which is shown when the text showing control does not know how to draw it, a so called non printable character.
It is because in String a the "\2" part obviously tries to be interpreted as a single escaped sign "\u0002"- similar to "\n" "\t" - you can see this in debugger (i tried it using NetBeans)
Since the first argument of a replaceAll method is passed to [Pattern.compile](http://docs.oracle.com/javase/6/docs/api/java/lang/String.html#replaceAll(java.lang.String, java.lang.String)) it needs to be escaped twice as opposed to String literal (like b).
So if the String "12\34a56ss7890" looks like this on screen you have printed it out like this:
System.out.println("12\\34a56ss7890");
which is solved in the second example.
However if the literal is given as "12\34a56ss7890" then I think you can't handle it with a single regexp, because if the backslash is followed by a number it gets interpreted as as \u0000 -\u0009 so the best I can think of is a very ugly solution:
str.replaceAll("\u0000","0").replaceAll("\u0001","1") ... .replaceAll("\u0009","9").replaceAll("[^\\d]")
the first then replacements (\u0000-\u0009) might be rewritten as a for loop to make it look elegant.
+1 for an EXCELLENT question :)
EDIT:
actually if a backslash is followed by more than one number they all get interpreted as a single sign - up to three numbers after a backslash, the fourth number will be treated as a single number.
Therefore, my solution is not generally correct, but could be extended to be. I would recommend Robin's solution below as it is far more efficient.

The character \34 is an octal number in the string 12\34a56ss7890, so you could use:
str.replaceAll("\034", "34").replaceAll("\\D", "")