Here what the program is expectiong as the output:
if originalString = "CATCATICATAMCATCATGREATCATCAT";
Output should be "I AM GREAT".
The code must find the sequence of characters (CAT in this case), and remove them. Plus, the resulting String must have spaces in between words.
String origString = remixString.replace("CAT", "");
I figured out I have to use String.replace, But what could be the logic for finding out if its not cat and producing the resulting string with spaces in between the words.
First off, you probably want to use the replaceAll method instead, to make sure you replace all occurrences of "CAT" within the String. Then, you want to introduce spaces, so instead of an empty String, replace "CAT" with " " (space).
As pointed out by the comment below, there might be multiple spaces between words - so we use a regular expression to replace multiple instances of "CAT" with a single space. The '+' symbol means "one or more",.
Finally, trim the String to get rid of leading and trailing white space.
remixString.replaceAll("(CAT)+", " ").trim()
You can use replaceAll which accepts a regular expression:
String remixString = "CATCATICATAMCATCATGREATCATCAT";
String origString = remixString.replaceAll("(CAT)+", " ").trim();
Note: the naming of replace and replaceAll is very confusing. They both replace all instances of the matching string; the difference is that replace takes a literal text as an argument, while replaceAll takes a regular expression.
Maybe this will help
String result = remixString.replaceAll("(CAT){1,}", " ");
Related
I'm trying to learn regex in Java.
So far, I've been trying some little mini challenges and I'm wondering if there is a way to define a nth character.
For instance, let's say I have this string: todayiwasnotagoodday
If I want to replace the third (fourth or seventh) character, how I can define a regex in order to change an specific "index", for this example the 'd' for an empty space "".
I've been searching about it, but so far my implementations match from the first element to the third: ^[a-z]{3}
¿Is it possible to define this regex?
Thanks in advance.
If you want to replace the third character with a space via regex, you could try a regex replace all:
String input = "todayiwasnotagoodday";
String output = input.replaceAll("^(.{2}).(.*)$", "$1 $2");
System.out.println(output); // to ayiwasnotagoodday
Note that you could also avoid regex here, and just use substring operations:
String output = input.substring(0, 2) + " " + input.substring(3);
System.out.println(output); // to ayiwasnotagoodday
I am wondering if I am going about splitting a string on a . the right way? My code is:
String[] fn = filename.split(".");
return fn[0];
I only need the first part of the string, that's why I return the first item. I ask because I noticed in the API that . means any character, so now I'm stuck.
split() accepts a regular expression, so you need to escape . to not consider it as a regex meta character. Here's an example :
String[] fn = filename.split("\\.");
return fn[0];
I see only solutions here but no full explanation of the problem so I decided to post this answer
Problem
You need to know few things about text.split(delim). split method:
accepts as argument regular expression (regex) which describes delimiter on which we want to split,
if delim exists at end of text like in a,b,c,, (where delimiter is ,) split at first will create array like ["a" "b" "c" "" ""] but since in most cases we don't really need these trailing empty strings it also removes them automatically for us. So it creates another array without these trailing empty strings and returns it.
You also need to know that dot . is special character in regex. It represents any character (except line separators but this can be changed with Pattern.DOTALL flag).
So for string like "abc" if we split on "." split method will
create array like ["" "" "" ""],
but since this array contains only empty strings and they all are trailing they will be removed (like shown in previous second point)
which means we will get as result empty array [] (with no elements, not even empty string), so we can't use fn[0] because there is no index 0.
Solution
To solve this problem you simply need to create regex which will represents dot. To do so we need to escape that .. There are few ways to do it, but simplest is probably by using \ (which in String needs to be written as "\\" because \ is also special there and requires another \ to be escaped).
So solution to your problem may look like
String[] fn = filename.split("\\.");
Bonus
You can also use other ways to escape that dot like
using character class split("[.]")
wrapping it in quote split("\\Q.\\E")
using proper Pattern instance with Pattern.LITERAL flag
or simply use split(Pattern.quote(".")) and let regex do escaping for you.
Split uses regular expressions, where '.' is a special character meaning anything. You need to escape it if you actually want it to match the '.' character:
String[] fn = filename.split("\\.");
(one '\' to escape the '.' in the regular expression, and the other to escape the first one in the Java string)
Also I wouldn't suggest returning fn[0] since if you have a file named something.blabla.txt, which is a valid name you won't be returning the actual file name. Instead I think it's better if you use:
int idx = filename.lastIndexOf('.');
return filename.subString(0, idx);
the String#split(String) method uses regular expressions.
In regular expressions, the "." character means "any character".
You can avoid this behavior by either escaping the "."
filename.split("\\.");
or telling the split method to split at at a character class:
filename.split("[.]");
Character classes are collections of characters. You could write
filename.split("[-.;ld7]");
and filename would be split at every "-", ".", ";", "l", "d" or "7". Inside character classes, the "." is not a special character ("metacharacter").
As DOT( . ) is considered as a special character and split method of String expects a regular expression you need to do like this -
String[] fn = filename.split("\\.");
return fn[0];
In java the special characters need to be escaped with a "\" but since "\" is also a special character in Java, you need to escape it again with another "\" !
String str="1.2.3";
String[] cats = str.split(Pattern.quote("."));
Wouldn't it be more efficient to use
filename.substring(0, filename.indexOf("."))
if you only want what's up to the first dot?
Usually its NOT a good idea to unmask it by hand. There is a method in the Pattern class for this task:
java.util.regex
static String quote(String s)
The split must be taking regex as a an argument... Simply change "." to "\\."
The solution that worked for me is the following
String[] fn = filename.split("[.]");
Note: Further care should be taken with this snippet, even after the dot is escaped!
If filename is just the string ".", then fn will still end up to be of 0 length and fn[0] will still throw an exception!
This is, because if the pattern matches at least once, then split will discard all trailing empty strings (thus also the one before the dot!) from the array, leaving an empty array to be returned.
Using ApacheCommons it's simplest:
File file = ...
FilenameUtils.getBaseName(file.getName());
Note, it also extracts a filename from full path.
split takes a regex as argument. So you should pass "\." instead of "." because "." is a metacharacter in regex.
I have string which should be split on "." (point) and " " (space). I have tried:
s.split("[\\s\\.]")
but it doesn't work, because it hasn't split this string normally - "123 456 . 11323 1".
How should I change my regular expression?
I think, what you want is this:
s.split("[\\s\\.]+");
Note the +. You don't seem to want to split on every single (!) occurrence of whitespace or dots. You want to match all lengths of combinations of whitespace or dots. That's why you have to greedy match as many as possible of those characters
Simply use "[\\s.]+" as the regex.
You will get a lot of blank spaces if you only split on a single character.
s.split("[\\s\\.]+")
will produce "123", "456", "11323", "1".
The + causes it to treat any run of spaces and dots as a single break instead of returning a string between adjacent spaces and dots.
You might still get blank strings at either end of your results since given " 123" it will split between the start of the string and "123".
I have databasefield called TelephoneName. In this field, I got different formats of telephone number.
What I need now is to seperate them into countrycode and subscribernumber.
For example, I saw a telephone number +49 (0)711 / 61947-xx.
I want to remove all the slash,brackets,minus,space. The result could be +49 (countrycode) and 071161947**(subsribernumber).
How can I do that with replaceAll method?
replaceAll("//()-","") is that correct?
The thing is I got a lot of unformatted telephone number such as:
+49 04261 85120
+32027400050
It is different to apply every telephone number with same algorithms
The replaceAll method takes a regular expression as argument. To remove everything except digits and +, you could thus do
str = str.replaceAll("[^0-9+]", "")
Here's a more complete example that also figures out the country code (based on the index of the ( symbol):
String str = "+49 (0)711 / 61947-12";
int lpar = str.indexOf('(');
String countryCode = str.substring(0, lpar).trim();
String subscriber = str.substring(lpar).trim();
subscriber = subscriber.replaceAll("[^0-9]", "");
System.out.println(countryCode); // prints +49
System.out.println(subscriber); // prints 07116194712
replaceAll("//()-","") is that correct?
No, not quite. That will remove all //- substrings. To remove those characters you need to put them in [...], like this: replaceAll("[/()-]", "") (and / does not need to be escaped).
The first argument of replaceAll() is a regex pattern, so what you want to do is make it match all non digits (and +). You can do this using the "[^...]" (not one of...) construct :
mystring.replaceAll("[^0-9+]", "")
No, that doesn't work.
ReplaceAll() Replaces each substring of this string that matches the given regular expression with the given replacement.
So your expression would replace all instances in the number that look like /()' with an empty space.
You need to do something like
String output = "+49 (0)711 / 61947-xx".replaceAll("[//()-]","");
The square brackets make it a regex character class ('Either slash or open bracket or close bracket or hypen'), rather than a literal ('slash followed by open bracket followed by close bracket followed by hypen.').
This can be done simply by using :
s=s.replace("/","");
s=s.replace("(","");
s=s.replace(")","");
Then substring it to get country code.
I am wondering if I am going about splitting a string on a . the right way? My code is:
String[] fn = filename.split(".");
return fn[0];
I only need the first part of the string, that's why I return the first item. I ask because I noticed in the API that . means any character, so now I'm stuck.
split() accepts a regular expression, so you need to escape . to not consider it as a regex meta character. Here's an example :
String[] fn = filename.split("\\.");
return fn[0];
I see only solutions here but no full explanation of the problem so I decided to post this answer
Problem
You need to know few things about text.split(delim). split method:
accepts as argument regular expression (regex) which describes delimiter on which we want to split,
if delim exists at end of text like in a,b,c,, (where delimiter is ,) split at first will create array like ["a" "b" "c" "" ""] but since in most cases we don't really need these trailing empty strings it also removes them automatically for us. So it creates another array without these trailing empty strings and returns it.
You also need to know that dot . is special character in regex. It represents any character (except line separators but this can be changed with Pattern.DOTALL flag).
So for string like "abc" if we split on "." split method will
create array like ["" "" "" ""],
but since this array contains only empty strings and they all are trailing they will be removed (like shown in previous second point)
which means we will get as result empty array [] (with no elements, not even empty string), so we can't use fn[0] because there is no index 0.
Solution
To solve this problem you simply need to create regex which will represents dot. To do so we need to escape that .. There are few ways to do it, but simplest is probably by using \ (which in String needs to be written as "\\" because \ is also special there and requires another \ to be escaped).
So solution to your problem may look like
String[] fn = filename.split("\\.");
Bonus
You can also use other ways to escape that dot like
using character class split("[.]")
wrapping it in quote split("\\Q.\\E")
using proper Pattern instance with Pattern.LITERAL flag
or simply use split(Pattern.quote(".")) and let regex do escaping for you.
Split uses regular expressions, where '.' is a special character meaning anything. You need to escape it if you actually want it to match the '.' character:
String[] fn = filename.split("\\.");
(one '\' to escape the '.' in the regular expression, and the other to escape the first one in the Java string)
Also I wouldn't suggest returning fn[0] since if you have a file named something.blabla.txt, which is a valid name you won't be returning the actual file name. Instead I think it's better if you use:
int idx = filename.lastIndexOf('.');
return filename.subString(0, idx);
the String#split(String) method uses regular expressions.
In regular expressions, the "." character means "any character".
You can avoid this behavior by either escaping the "."
filename.split("\\.");
or telling the split method to split at at a character class:
filename.split("[.]");
Character classes are collections of characters. You could write
filename.split("[-.;ld7]");
and filename would be split at every "-", ".", ";", "l", "d" or "7". Inside character classes, the "." is not a special character ("metacharacter").
As DOT( . ) is considered as a special character and split method of String expects a regular expression you need to do like this -
String[] fn = filename.split("\\.");
return fn[0];
In java the special characters need to be escaped with a "\" but since "\" is also a special character in Java, you need to escape it again with another "\" !
String str="1.2.3";
String[] cats = str.split(Pattern.quote("."));
Wouldn't it be more efficient to use
filename.substring(0, filename.indexOf("."))
if you only want what's up to the first dot?
Usually its NOT a good idea to unmask it by hand. There is a method in the Pattern class for this task:
java.util.regex
static String quote(String s)
The split must be taking regex as a an argument... Simply change "." to "\\."
The solution that worked for me is the following
String[] fn = filename.split("[.]");
Note: Further care should be taken with this snippet, even after the dot is escaped!
If filename is just the string ".", then fn will still end up to be of 0 length and fn[0] will still throw an exception!
This is, because if the pattern matches at least once, then split will discard all trailing empty strings (thus also the one before the dot!) from the array, leaving an empty array to be returned.
Using ApacheCommons it's simplest:
File file = ...
FilenameUtils.getBaseName(file.getName());
Note, it also extracts a filename from full path.
split takes a regex as argument. So you should pass "\." instead of "." because "." is a metacharacter in regex.