I have a problem trying to validate this string...
So, the user selects a template: q( ). Then, the user fills in the contents inside the brackets which can end up like this:
q(a,b,c)
I have tried different ways using regex to validate this String, but it keeps returning the answer "No". I believe the problem is "q(" and ")" in my regex as I am not sure how it should look like.
Here's a snippet of the code:
String data2 = "q(a,b,c)";
String regex2 = "q([a-zA-Z0-9,'])";
if(data2.matches(regex2)){
System.out.println("yes");
}
else{
System.out.println("No");
}
I do have an alternative way by removing "q(" and ")" in data2 string, but I rather have it in regex without the need of removing characters in a String.
Any suggestions?
You need to escape the parentheses (and escape the escape character so that it compiles) and add a + at the end to indicate one or more characters
String regex2 = "q\\([a-zA-Z0-9,']+\\)";
You can read the meaning of every character in a regular expression in the Pattern javadoc.
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.
Need help in getting the java regex to replace = sing between parenthesis with #, my input text is
8=FIX.4.&49=(550=0449)&35=RIO&76=(AB=4560)&
expected output string
8=FIX.4.&49=(550#0449)&35=RIO&76=(AB#4560)&
So would like to replace = char only within (550=0449) and (AB=4560) with # so the output should contain (550#0449) and (AB#4560).
I like anubhava's answer, but if you want to be more strict and assert there are non-blank terms and opening and closing brackets, capture the terms and write them back using back references:
str = str.replaceAll("(\\(\\w+)=(\\w+\\))", "$1#$2");
You can use:
String repl = str.replaceAll("=(?=[^()]*\\))", "#");
(?=[^()]*\)) is a lookahead that will make sure to match = only when there is a ) following it.
I have a String : testing<b>s<b>tringwit<b>h</b>nomean<b>s</b>ing
I want to replace the character s with some other character sequence suppose : <b>X</b> but i want the character sequence s to remain intact i.e. regex should not update the character s with a previous character as "<".
I used the JAVA code :
String str = testing<b>s<b>tringwit<b>h</b>nomean<b>s</b>ing;
str = str.replace("s[^<]", "<b>X</b>");
The problem is that the regex would match 2 characters, s and following character if it is not ">" and Sting.replace would replace both the characters. I want only s to be replaced and not the following character.
Any help would be appreciated. Since i have lots of such replacements i don't want to use a loop matching each character and updating it sequentially.
There are other ways, but you could, for example, capture the second character and put it back:
str = str.replaceAll("s([^<])", "<b>X\\1</b>");
Looks like you want a negative lookahead:
s(?!<)
String str = "testing<b>s<b>tringwit<b>h</b>nomean<b>s</b>ing;";
System.out.println(str.replaceAll("s(?!<)", "<b>X</b>"));
output:
te<b>X</b>ting<b>s<b>tringwit<b>h</b>nomean<b>s</b>ing;
Use look arounds to assert, but not capture, surrounding text:
str = str.replaceAll("s(?![^<]))", "whatever");
Or, capture and put back using a back reference $1:
str = str.replaceAll("s([^<])", "whatever$1");
Note that you need to use replaceAll() (which use regex), rather than replace() (which uses plain text).
Just could not get this one and googling did not help much either..
First something that I know: Given a string and a regex, how to replace all the occurrences of strings that matches this regular expression by a replacement string ? Use the replaceAll() method in the String class.
Now something that I am unable to do. The regex I have in my code now is [^a-zA-Z] and I know for sure that this regex is definitely going to have a range. Only some more characters might be added to the list. What I need as output in the code below is Worksheet+blah but what I get using replaceAll() is Worksheet++++blah
String homeworkTitle = "Worksheet%#5_blah";
String unwantedCharactersRegex = "[^a-zA-Z]";
String replacementString = "+";
homeworkTitle = homeworkTitle.replaceAll(unwantedCharactersRegex,replacementString);
System.out.println(homeworkTitle);
What is the way to achieve the output that I wish for? Are there any Java methods that I am missing here?
[^a-zA-Z]+
Will do it nicely.
You just need a greedy quantifier in order to match as many non-alphabetical characters you can, and replace the all match by one '+' (a - by default - greedy quantifier)
Note: [^a-zA-Z]+? would make the '+' quantifier lazy, and would have give you the same result than [^a-zA-Z], since it would only have matched only one non-alphabetical character at a time.
String unwantedCharactersRegex = "[^a-zA-Z]"
This matches a single non-letter. So each single non-letter is replaced by a +. You need to say "one or more", so try
String unwantedCharactersRegex = "[^a-zA-Z]+"