In java is there a way to replace specific special characters with another special characters within entire text without using if else .
Eg:
String s = abcd&c!&%^ .
Replace & with ~
Replace ! with ¬ etc on the above example string.
String has a replace function, so you can do s = s.replace('&','~');
public String replace(char oldChar, char newChar)
Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.
If the character oldChar does not occur in the character sequence represented by this String object, then a reference to this String object is returned. Otherwise, a new String object is created that represents a character sequence identical to the character sequence represented by this String object, except that every occurrence of oldChar is replaced by an occurrence of newChar.
String.replace(char oldChar, char newChar);
All things you can do with String: https://docs.oracle.com/en/java/javase/11/docs/api/java.base/java/lang/String.html
I recommend reading these docs when you're working with anything Java you don't know. Like Arrays, or Lists, and so on.
Related
I have a string "EAD\rgonzalez" which is passed to me.
I need to pull out "rgonzalez" from it.
I am running into problems with the "\" character.
I cannot find the index of it, I cannot replace it, etc.
Any help on pulling the data after the "\" would be appreciated.
The string that i receive is in the format of domain\username; the data can vary.
Another example would be US\ngross where \n would be interpreted as a newline character.
To clarify, I am not adding a '\', i am trying to split a string on a '\'
This string contains '\r' which in itself is a character, a special one.
I need a way to make \r contained within my string two separate characters, a '\' and an 'r'.
You haven't provided any code, but I'm assuming what you're doing is something like this:
String user = request.getParameter("user"); // user = "EAD\rgonzalez"
If you were to declare a static string in your application, you would have to escape the backslash because it is a special character for Java strings:
String user = "EAD\\rgonzalez";
To split that string on the backslash you must escape it twice in the regex that you pass to the split method. Once because backslash is a special character for Java strings and again because backslash is a special character for regex strings. So instead of one backlash you have four. The one is escaped so then you have two, and then both of them are escaped again.
String[] parts = user.split("\\\\");
Now you have split the string:
System.out.println(parts[0]); // "EAD"
System.out.println(parts[1]); // "rgonzalez"
The string that i receive is in the format of domain\username... the data can vary
The data shouldn't vary if that is the input your program expects.
where \n would be interpreted as a newline character
I'm not sure how you'd get newlines from a single line input form. If you are, then your input is invalid because it does not follow the format you're specified and are expecting. In the case where you did interpret newlines and other whitespace characters, you would either treat the whole thing as the domain, or the username, thus potentially breaking your program logic... Since you have stated the requirement of domain\username, and I don't think that requires you to handle any other form of input.
I am collecting this string from the header data from the request object in a webapp.
In that case, the raw value should not contain an escape character and is actually represented as the form "domain\\username" as a Java string. When you print the value, the escape characters aren't shown
I cannot find the index of it,
With the correct representation, indexOf("\\") will work...
pulling the data after the "\"
Since you would have the value as domain\\username, you need to escape both of the backslashes within the method of split(String pattern) since that is a regular expression.
For example,
public static void main (String[] args) throws java.lang.Exception
{
String in = "EAD\\rgonzalez";
System.out.println(in.indexOf("\\")); // find the index of '\'
String[] parts = in.split("\\\\"); // split on '\\'
System.out.println(Arrays.toString(parts));
}
Again, the string "EAD\rgonzalez" is not in the form of domain\username, as demonstrated here
System.out.print("EAD\rgonzalez".matches("[A-Z]+\\[a-z]+")); // false
The magic you need is in org.apache.commons.lang.StringEscapeUtils
Here is a demo:
package ignoreescapeseq2;
import org.apache.commons.lang.StringEscapeUtils;
/*
* #author Charles Knell
*/
public class IgnoreEscapeSeq2 {
public static void main(String[] args) {
String string = "EAD\rgonzalez"; // REQUIRED INPUT STRING
String eString = StringEscapeUtils.escapeJava(string);
String [] sArray = eString.split("\\\\");
System.out.println("domain: " + sArray[0]);
System.out.println("username: " + sArray[1]);
}
}
Here is the output:
Although this MAY answer the question, there does still seem to be a problem
if you must define the string in java. As you said, "EAD\xgonzalez" isn't a
valid java string because \x isn't a valid escape character. The solution above only works if the input string never has to be explictly defined, as in the demo.
This question already has answers here:
Replace with empty string replaces newChar around all the characters in original string
(4 answers)
Closed 6 years ago.
I'm confused with a code
public class StringReplaceWithEmptyString
{
public static void main(String[] args)
{
String s1 = "asdfgh";
System.out.println(s1);
s1 = s1.replace("", "1");
System.out.println(s1);
}
}
And the output is:
asdfgh
1a1s1d1f1g1h1
So my first opinion was every character in a String is having an empty String "" at both sides. But if that's the case after 'a' (in the String) there should be two '1' coming in the second line of output (one for end of 'a' and second for starting of 's').
Now I checked whether the String is represented as a char[] in these links In Java, is a String an array of chars? and String representation in Java I got answer as YES.
So I tried to assign an empty character '' to a char variable, but its giving me a compiler error,
Invalid character constant
The same process gives a compiler error when I tried in char[]
char[] c = {'','a','','s'}; // CTE
So I'm confused about three things.
How an empty String is represented by char[] ?
Why I'm getting that output for the above code?
How the String s1 is represented in char[] when it is initialized first time?
Sorry if I'm wrong at any part of my question.
Just adding some more explanation to Tim Biegeleisen answer.
As of Java 8, The code of replace method in java.lang.String class is
public String replace(CharSequence target, CharSequence replacement) {
return Pattern.compile(target.toString(), Pattern.LITERAL).matcher(
this).replaceAll(Matcher.quoteReplacement(replacement.toString()));
}
Here You can clearly see that the string is replaced by Regex Pattern matcher and in regex "" is identified by Zero-Length character and it is present around any Non-Zero length character.
So, behind the scene your code is executed as following
Pattern.compile("".toString(), Pattern.LITERAL).matcher("asdfgh").replaceAll(Matcher.quoteReplacement("1".toString()));
The the output becomes
1a1s1d1f1g1h1
Going with Andy Turner's great comment, your call to String#replace() is actually implemented using String#replaceAll(). As such, there is a regex replacement happening here. The matches occurs before the first character, in between each character in the string, and after the last character.
^|a|s|d|f|g|h|$
^ this and every pipe matches to empty string ""
The match you are making is a zero length match. In Java's regex implementation used in String.replaceAll(), this behaves as the example above shows, namely matching each inter-character position and the positions before the first and after the last characters.
Here is a reference which discusses zero length matches in more detail: http://www.regexguru.com/2008/04/watch-out-for-zero-length-matches/
A zero-width or zero-length match is a regular expression match that does not match any characters. It matches only a position in the string. E.g. the regex \b matches between the 1 and , in 1,2.
This is because it does a regex match of the pattern/replacement you pass to the replace().
public String replace(CharSequence target, CharSequence replacement) {
return Pattern.compile(target.toString(), Pattern.LITERAL).matcher(
this).replaceAll(Matcher.quoteReplacement(replacement.toString()));
}
Replaces each substring of this string that matches the literal target
sequence with the specified literal replacement sequence. The
replacement proceeds from the beginning of the string to the end, for
example, replacing "aa" with "b" in the string "aaa" will result in
"ba" rather than "ab".
Parameters:
target The sequence of char values
to be replaced
replacement The replacement sequence of char values
Returns: The resulting string
Throws: NullPointerException if target
or replacement is null.
Since:
1.5
Please read more at the link below ... (Also browse through the source code).
http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/lang/String.java#String.replace%28java.lang.CharSequence%2Cjava.lang.CharSequence%29
A regex such as "" would match every possible empty string in a string. In this case it happens to be every empty space at the start and end and after every character in the string.
The Below java code replaces all the character in string variable BusDetails with blank even though i don't see a (. dot) for the method to replace it. Why ?
Output = _BusDetails
String BusDetails = " BUS_12_UFV_BOURQUIN_COMMUTER_TO_UFV";
String table_UniqueBusNameTimings = BusDetails.replaceAll(".", "")+"_BusTimings";
System.out.println("TableName: "+table_UniqueBusNameDetails);
replaceAll treats its first argument as a regular expression, and in regular expressions a dot matches any single character except a newline.
To replace one fixed string with another you should use the replace method that takes two CharSequence parameters instead - despite its name, this method does in fact replace all occurrences of the first CharSequence with the second one.
String table_UniqueBusNameTimings = BusDetails.replace(".", "")+"_BusTimings";
You need to escape that meta character
String table_UniqueBusNameTimings = BusDetails.replaceAll("\\.", "")+"_BusTimings";
See how escapes works in java
http://docs.oracle.com/javase/jndi/tutorial/beyond/names/syntax.html
you need to escape the period, like this:
String BusDetails = " BUS_12_UFV_BOURQUIN_COMMUTER_TO_UFV";
String table_UniqueBusNameTimings = BusDetails.replaceAll("\\.", "")+"_BusTimings";
System.out.println("TableName: "+table_UniqueBusNameTimings);
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 .");
}
String aStr="TEST-1-TV_50";
System.out.println(aStr.matches("^[A-Z0-9\\-\\_]+")); //TRUE.
But why this is not working..?
String aStr1= "$local:TEST12-1-TV_50 as xs:boolean";
int strtIndex=aStr.indexOf(":");
int endIndex=aStr.indexOf("as");
String extractedStr=aStr1.substring(strtIndex+1,endIndex); //TEST12-1-TV_50
System.out.println(extractedStr.matches("^[A-Z0-9\\-\\_]+")); //FALSE.
Why its giving result as false.???
There's a trailing space in extractedStr.
So it contains "TEST12-1-TV_50 " (not that there's a space after the final 0).
You can either replace endIndex with aStr.indexOf(" as") (starting space) or simply call trim() on extractedStr:
String extractedStr=aStr1.substring(strtIndex+1,endIndex).trim();
You need to include space also in character class:
extractedStr.matches("^[A-Z0-9 _-]+"); // true
OR else call trim() before matches
extractedStr.trim().matches("^[A-Z0-9_-]+"); // true
PS: You don't need to escape _ in character class and hyphen as well (if used at start or end)