When I try to split a String around occurrences of "." the method split returns an array of strings with length 0.When I split around occurrences of "a" it works fine.Does anyone know why?Is split not supposed to work with punctuation marks?
split takes regex. Try split("\\.").
String a = "a.jpg";
String str = a.split(".")[0];
This will throw ArrayOutOfBoundException because split accepts regex arguments and "." is a reserved character in regular expression, representing any character.
Instead, we should use the following statement:
String str = a.split("\\.")[0]; //Yes, two backslashes
When the code is compiled, the regular expression is known as "\.", which is what we want it to be
Here is the link of my old blog post in case you are interested: http://junxian-huang.blogspot.com/2009/01/java-tip-how-to-split-string-with-dot.html
Related
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 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 a scenario where i need to break the below input string based on the keywords using regex.
Keywords are UPRCAS, REPLC, LOWCAS and TUPIL.
String input = "UPRCAS-0004-abcdREPLC-0003-123TUPIL-0005-adf2344LOWCAS-0003-ABCD";
The output should be as follows
UPRCAS-00040-abcd
REPLC-0003-123
TUPIL-0005-adf2344
LOWCAS-00030-ABCD
How can i achieve this using java regex.
I have tried using split by '-' and using regex but both the approach gives an array of strings and again i have to process each string and combine 3 strings together to form UPRCAS-00040-abcd. I felt this is not the efficient way to do as it takes an extra array and process them back.
String[] tokens = input.split("-");
String[] r = input.split("(?=\\p{Upper})");
Please let me know if we can split the string using regex based on the keyword. Basically i need to extract the string between the keyword boundary.
Edited question after understanding the limitation of existing problem
The regex should be generic to extract the string from input between the UPPERCASE characters
The regex should not contains keywords to split the string.
I understood that, it is a bad idea to add new keyword everytime in regex pattern for searching. My expectation is to be a generic as possible.
Thanks all for your time. Really appreciate it.
Split using the following regex:
(?=UPRCAS|REPLC|LOWCAS|TUPIL)
The (?=xxx) is a zero-width positive lookahead, meaning that it matches the empty space immediately preceding one of the 4 keywords.
See Regular-Expressions.info for more information: Lookahead and Lookbehind Zero-Length Assertions
Test
String input = "UPRCAS-0004-abcdREPLC-0003-123TUPIL-0005-adf2344LOWCAS-0003-ABCD";
String[] output = input.split("(?=UPRCAS|REPLC|LOWCAS|TUPIL)");
for (String value : output)
System.out.println(value);
Output
UPRCAS-0004-abcd
REPLC-0003-123
TUPIL-0005-adf2344
LOWCAS-0003-ABCD
You can try this regex:
\w+-\w+-(?:[a-z0-9]+|[A-Z]+)
Demo: https://regex101.com/r/etKBjI/3
This question already has answers here:
String.split() *not* on regular expression?
(8 answers)
Closed 2 years ago.
In my Java application I need to find indices and split strings using the same "target" for both occasions. The target is simply a dot.
Finding indices (by indexOf and lastIndexOf) does not use regex, so
String target = ".";
String someString = "123.456";
int index = someString.indexOf(target); // index == 3
gives me the index I need.
However, I also want to use this "target" to split some strings. But now the target string is interpreted as a regex string. So I can't use the same target string as before when I want to split a string...
String target = ".";
String someString = "123.456";
String[] someStringSplit = someString.split(target); // someStringSplit is an empty array
So I need either of the following:
A way to split into an array by a non-regex target
A way to "convert" a non-regex target string into a regex string
Can someone help? Would you agree that it seems a bit odd of the standard java platform to use regex for "split" while not using regex for "indexOf"?
You need to escape your "target" in order to use it as a regex.
Try
String[] someStringSplit = someString.split(Pattern.quote(target));
and let me know if that helps.
String::split do split without regex if the regex is:
a one-char String and this character is not one of the RegEx's meta characters .$|()[{^?*+\\
two-char String and the first char is the backslash and the second is
not the ascii digit or ascii letter.
Please see String::split() source code for details.
For escaped '.' target it is going to be split without regex.
You can try this one.
String target = ".";
String someString = "123.456";
StringTokenizer tokenValue = new StringTokenizer(someString, target);
while (tokenValue.hasMoreTokens()) {
System.out.println(tokenValue.nextToken());
}
So I'm using a String as a delimiter to use when I call the Split method.
String[] aExpr;
String strDelimiter = "[-+/=^//%//*//(//);:?]";
aExpr = expr.split(strDelimiter);
This fills aExpr with the strings broken accordingly with the strDelimiter.
The thing is that I also want the Split() method to compare not only the strDelimiter, but also this String:
String oprDelimiter = "[abcdefghijklmnopqrstuvwxyz0123456789]+"
Which is basically any characters followed by numbers. I could add all these characters to the First String, but the + in the end won't let me. The + means that any combination of the words will break the String. Any ideas of how could I do this?
Try using this regex:
(?<=[abcdefghijklmnopqrstuvwxyz0123456789])(?=[-+=^%*();:?])|(?<=[-+=^%*();:?])(?=[abcdefghijklmnopqrstuvwxyz0123456789])
as the delimiter. It will split on any location that is preceded by any of the characters abcdefghijklmnopqrstuvwxyz0123456789 and followed by any of the characters -+=^%*();:?, or vice versa. Explanation and demonstration here: http://regex101.com/r/mT3lL1.