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.
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.
Original String: "12312123;www.qwerty.com"
With this Model.getList().get(0).split(";")[1]
I get: "www.qwerty.com"
I tried doing this: Model.getList().get(0).split(";")[1].split(".")[1]
But it didnt work I get exception. How can I solve this?
I want only "qwerty"
Try this, to achieve "qwerty":
Model.getList().get(0).split(";")[1].split("\\.")[1]
You need escape dot symbol
Try to use split(";|\\.") like this:
for (String string : "12312123;www.qwerty.com".split(";|\\.")) {
System.out.println(string);
}
Output:
12312123
www
qwerty
com
You can split a string which has multiple delimiters. Example below:
String abc = "11;xyz.test.com";
String[] tokens = abc.split(";|\\.");
System.out.println(tokens[tokens.length-2]);
The array index 1 part doesn't make sense here. It will throw an ArrayIndexOutOfBounds Exception or something of the sort.
This is because splitting based on "." doesn't work the way you want it to. You would need to escape the period by putting "\." instead. You will find here that "." means something completely different.
You'd need to escape the ., i.e. "\\.". Period is a special character in regular expressions, meaning "any character".
What your current split means is "split on any character"; this means that it splits the string into a number of empty strings, since there is nothing between consecutive occurrences of " any character".
There is a subtle gotcha in the behaviour of the String.split method, which is that it discards trailing empty strings from the token array (unless you pass a negative number as the second parameter).
Since your entire token array consists of empty strings, all of these are discarded, so the result of the split is a zero-length array - hence the exception when you try to access one of its element.
Don't use split, use a regular expression (directly). It's safer, and faster.
String input = "12312123;www.qwerty.com";
String regex = "([^.;]+)\\.[^.;]+$";
Matcher m = Pattern.compile(regex).matcher(input);
if (m.find()) {
System.out.println(m.group(1)); // prints: qwerty
}
I am trying to split the string using Split function in java
String empName="employee name | employee Email";
String[] empDetails=empName.split("|");
it gives me result as
empDetails[0]="e";
empDetails[1]="m";
empDetails[2]="p";
empDetails[3]="l";
empDetails[4]="o";
empDetails[5]="y";
empDetails[6]="e";
empDetails[7]="e";
.
.
.
but when i try following code
String empName="employee name - employee Email";
String[] empDetails=empName.split("-");
it gives me
empDetails[0]="employee name ";
empDetails[1]=" employee Email";
why java split function can not split the string seperated by "|"
String#split() method accepts a regex and not a String.
Since | is a meta character, and it's have a special meaning in regex.
It works when you escape that.
String[] empDetails=empName.split("\\|");
Update:
Handling special characters in java:OFFICIAL DOCS.
As a side note:
In java method names starts with small letters.it should be split() not Split() ..not the capital and small s
but my question is why we have to use escape in case of "|" and not for "-"
Because "|" is a regex meta-character. It means "alternation"; e.g. "A|B" means match "A" or "B". If you have problems understanding Java regexes, the javadocs for Pattern describe the complete Java regex syntax.
So when you split on "|" (without the escaping!), you are specifying that the separator is "nothing or nothing", and that matches between each character of the target string.
(For the record, "-" is also a meta-character, but only in a "[..]" character group. In other contexts it doesn't require escaping.)
You should use .split("\\|"); instead of .split("|");
Try
String[] empDetails=empName.split("\\|");
I need to cut certain strings for an algorithm I am making. I am using substring() but it gets too complicated with it and actually doesn't work correctly. I found this topic how to cut string with two regular expression "_" and "."
and decided to try with split() but it always gives me
java.util.regex.PatternSyntaxException: Dangling meta character '+' near index 0
+
^
So this is the code I have:
String[] result = "234*(4-5)+56".split("+");
/*for(int i=0; i<result.length; i++)
{
System.out.println(result[i]);
}*/
Arrays.toString(result);
Any ideas why I get this irritating exception ?
P.S. If I fix this I will post you the algorithm for cutting and then the algorithm for the whole calculator (because I am building a calculator). It is gonna be a really badass calculator, I promise :P
+ in regex has a special meaning. to be treated as a normal character, you should escape it with backslash.
String[] result = "234*(4-5)+56".split("\\+");
Below are the metacharaters in regex. to treat any of them as normal characters you should escape them with backslash
<([{\^-=$!|]})?*+.>
refer here about how characters work in regex.
The plus + symbol has meaning in regular expression, which is how split parses it's parameter. You'll need to regex-escape the plus character.
.split("\\+");
You should split your string like this: -
String[] result = "234*(4-5)+56".split("[+]");
Since, String.split takes a regex as delimiter, and + is a meta-character in regex, which means match 1 or more repetition, so it's an error to use it bare in regex.
You can use it in character class to match + literal. Because in character class, meta-characters and all other characters loose their special meaning. Only hiephen(-) has a special meaning in it, which means a range.
+ is a regex quantifier (meaning one or more of) so needs to be escaped in the split method:
String[] result = "234*(4-5)+56".split("\\+");
I have a question about strings in Java. Let's say, I have a string like so:
String str = "The . startup trace ?state is info?";
As the string contains the special character like "?" I need the string to be replaced with "\?" as per my requirement. How do I replace special characters with "\"? I tried the following way.
str.replace("?","\?");
But it gives a compilation error. Then I tried the following:
str.replace("?","\\?");
When I do this it replaces the special characters with "\\". But when I print the string, it prints with single slash. I thought it is taking single slash only but when I debugged I found that the variable is taking "\\".
Can anyone suggest how to replace the special characters with single slash ("\")?
On escape sequences
A declaration like:
String s = "\\";
defines a string containing a single backslash. That is, s.length() == 1.
This is because \ is a Java escape character for String and char literals. Here are some other examples:
"\n" is a String of length 1 containing the newline character
"\t" is a String of length 1 containing the tab character
"\"" is a String of length 1 containing the double quote character
"\/" contains an invalid escape sequence, and therefore is not a valid String literal
it causes compilation error
Naturally you can combine escape sequences with normal unescaped characters in a String literal:
System.out.println("\"Hey\\\nHow\tare you?");
The above prints (tab spacing may vary):
"Hey\
How are you?
References
JLS 3.10.6 Escape Sequences for Character and String Literals
See also
Is the char literal '\"' the same as '"' ?(backslash-doublequote vs only-doublequote)
Back to the problem
Your problem definition is very vague, but the following snippet works as it should:
System.out.println("How are you? Really??? Awesome!".replace("?", "\\?"));
The above snippet replaces ? with \?, and thus prints:
How are you\? Really\?\?\? Awesome!
If instead you want to replace a char with another char, then there's also an overload for that:
System.out.println("How are you? Really??? Awesome!".replace('?', '\\'));
The above snippet replaces ? with \, and thus prints:
How are you\ Really\\\ Awesome!
String API links
replace(CharSequence target, CharSequence replacement)
Replaces each substring of this string that matches the literal target sequence with the specified literal replacement sequence.
replace(char oldChar, char newChar)
Returns a new string resulting from replacing all occurrences of oldChar in this string with newChar.
On how regex complicates things
If you're using replaceAll or any other regex-based methods, then things becomes somewhat more complicated. It can be greatly simplified if you understand some basic rules.
Regex patterns in Java is given as String values
Metacharacters (such as ? and .) have special meanings, and may need to be escaped by preceding with a backslash to be matched literally
The backslash is also a special character in replacement String values
The above factors can lead to the need for numerous backslashes in patterns and replacement strings in a Java source code.
It doesn't look like you need regex for this problem, but here's a simple example to show what it can do:
System.out.println(
"Who you gonna call? GHOSTBUSTERS!!!"
.replaceAll("[?!]+", "<$0>")
);
The above prints:
Who you gonna call<?> GHOSTBUSTERS<!!!>
The pattern [?!]+ matches one-or-more (+) of any characters in the character class [...] definition (which contains a ? and ! in this case). The replacement string <$0> essentially puts the entire match $0 within angled brackets.
Related questions
Having trouble with Splitting text. - discusses common mistakes like split(".") and split("|")
Regular expressions references
regular-expressions.info
Character class and Repetition with Star and Plus
java.util.regex.Pattern and Matcher
In case you want to replace ? with \?, there are 2 possibilities: replace and replaceAll (for regular expressions):
str.replace("?", "\\?")
str.replaceAll("\\?","\\\\?");
The result is "The . startup trace \?state is info\?"
If you want to replace ? with \, just remove the ? character from the second argument.
But when I print the string, it prints
with single slash.
Good. That's exactly what you want, isn't it?
There are two simple rules:
A backslash inside a String literal has to be specified as two to satisfy the compiler, i.e. "\". Otherwise it is taken as a special-character escape.
A backslash in a regular expresion has to be specified as two to satisfy regex, otherwise it is taken as a regex escape. Because of (1) this means you have to write 2x2=4 of them:"\\\\" (and because of the forum software I actually had to write 8!).
String str="\\";
str=str.replace(str,"\\\\");
System.out.println("New String="+str);
Out put:- New String=\
In java "\\" treat as "\". So, the above code replace a "\" single slash into "\\".