We Keep Coding

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

Java String.replace/replaceAll not working

So, I'm trying to parse a String input in Java that contains (opening) square brackets. I have str.replace("\\[", ""), but this does absolutely nothing. I've tried replaceAll also, with more than one different regex, but the output is always unchanged. Part of me wonders if this is possibly caused by the fact that all my back-slash characters appear as yen symbols (ever since I added Japanese to my languages), but it's been that way for over a year and hasn't caused me any issues like this before.
Any idea what I might be doing wrong here?

Strings are immutable in Java. Make sure you re-assign the return value to the same String variable:
str = str.replaceAll("\\[", "");
For the normal replace method, you don't need to escape the bracket:
str = str.replace("[", "");

public String replaceAll(String regex, String replacement)
As shown in the code above, replaceAll method expects first argument as regular expression and hence you need to escape characters like "(", ")" etc (with "\") if these exists in your replacement text which is to be replaced out of the string. For example :
String oldString = "This is (stringTobeReplaced) with brackets.";
String newString = oldString.replaceAll("\\(stringTobeReplaced\\)", "");
System.out.println(newString); // will output "This is with brackets."
Another way of doing this is to use Pattern.quote("str") :
String newString = oldString.replaceAll(Pattern.quote("(stringTobeReplaced)"), "");
This will consider the string as literal to be replaced.

As always, the problem is not that "xxx doesn't work", it is that you don't know how to use it.
First things first:
a String is immutable; if you read the javadoc of .replace() and .replaceAll(), you will see that both specify that a new String instance is returned;
replace() accepts a string literal as its first argument, not a regex literal.
Which means that you probably meant to do:
str = str.replace("[", "");
If you only ever do:
str.replace("[", "");
then the new instance will be created but you ignore it...
In addition, and this is a common trap with String (the other being that .matches() is misnamed), in spite of their respective names, .replace() does replace all occurrences of its first argument with its second argument; the only difference is that .replaceAll() accepts a regex as a first argument, and a "regex aware" expression as its second argument; for more details, see the javadoc of Matcher's .replaceAll().

For it to work it has to be inside a method.
for example:
public class AnyClass {
String str = "gtrg4\r\n" + "grtgy\r\n" + "grtht\r\n" + "htrjt\r\n" + "jtyjr\r\n" + "kytht";
public String getStringModified() {
str.replaceAll("\r\n", "");
return str;
}
}

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 .");
}

Why does reinserting line breaks after removing them not work?

I'm writing code that's supposed to remove actual line breaks from a block of text and replace them with the String "\n". Then, when the String is read at another time, it should replace the line breaks (in other words, search for all "\n" and insert \n. However, while the first conversion works fine, it's not doing the latter. It seems as though the second replace is doing nothing. Why?
The replace:
theString.replaceAll(Constants.LINE_BREAK, Constants.LINE_BREAK_DB_REPLACEMENT);
The re-replace:
theString.replaceAll(Constants.LINE_BREAK_DB_REPLACEMENT, Constants.LINE_BREAK);
The constants:
public static final String LINE_BREAK = "\n";
public static final String LINE_BREAK_DB_REPLACEMENT = "\\\\n";

In String.replaceAll(regex, replacement), both the regex string and replacement string treat backslash as an escape character:
regex represents a regular expression, which escapes a backslash as \\
replacement is a replacement string, which also escapes backslashes:
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.
This means backslashes must be escaped in both parameters. Further, string constants also use backslash as an escape character, so backslashes in string constants passed to the method must be double-escaped (see also this question).
This works fine for me:
// Replace newline with "\n"
theString.replaceAll("\\n", "\\\\n");
// Replace "\n" with newline
theString.replaceAll("\\\\n","\n");
You can also use the Matcher.quoteReplacement() method to treat the replacement string as a literal:
// Replace newline with "\n"
theString.replaceAll("\\n", Matcher.quoteReplacement("\\n"));
// Replace "\n" with newline
theString.replaceAll("\\\\n",Matcher.quoteReplacement("\n"));

You dont need four backslashes in the last replaceAll() method call.
This seems to work fine for me
String str = "abc\nefg\nhijklm";
String newStr = str.replaceAll("\n", "\\\\n");
String newnewStr = newStr.replaceAll("\\\\n", "\n");
The output is:
abc
efg
hijklm
abc\nefg\nhijklm
abc
efg
hijklm
Which I think is what you expected.