So we were looking at some of the other regex posts and we are having trouble removing a special case in one instance; the special character is in the beginning of the word.
We have the following line in our code:
String k = s.replaceAll("([a-z]+)[()?:!.,;]*", "$1");
where s is a singular word. For example, when parsing the sentence "(hi hi hi)" by tokenizing it, and then performing the replaceAll function on each token, we get an output of:
(hi
hi
hi
What are we missing in our regex?
You can use an easier approach - replace the characters that you do not want with spaces:
String k = s.replaceAll("[()?:!.,;]+", " ");
Position matters so you would need to match the excluded charcters before the capturing group also:
String k = s.replaceAll("[()?:!.,;]*([a-z]+)[()?:!.,;]*", "$1");
your replace just removed the "special chars" after the [a-z]+, that's why the ( before hi is left there.
If you know s is a single word
you could either:
String k = s.replaceAll("\\W*(\\w+)\\W*", "$1");
or
String k = s.replaceAll("\\W*", "");
This can be more simple
try this :
String oldString = "Hi There ##$ What is %#your name?##$##$ 0123$$";
System.out.println(oldString.replaceAll("[\\p{Punct}\\s\\d]+", " ");
output :
Hi There What is your name 0123
So it also accepts numeric.
.replaceAll("[\p{Punct}\s\d]+", " ");
will replace alll the Punctuations used which includes almost all the special characters.
Related
I want to split a string based on text qualifier for example
"1","10411721","MikeTison","08/11/2009","21/11/2009","2800.00","002934538","051","New York","10411720-002",".\Images\b.jpg",".\RTF\b.rtf"
Qualifer="
Spliter = ,
I want to split string based on Spliter , but if Spliter comes inside qualifier " than ignore it and return string including Spliter .
Regular expression i am using is (?:|,)(\"(?:[^\"]+|\"\")*\"|[^,]*)
but this regular expression only returns commas,please help me in this perspective as i am new to regular expressions
please note that if we have newline characters in string ie \r\n than it should ignore newline character
"1","10411","Muis","a","21/11/2009","2800.06","0029683778","03005136851","Awan","10411720-001",".\Images\a.jpg",".\RTF\a.rtf"
"2","08/10/2009","07:32","Call","On-Net","030092343242342376543","Monk","00:00","1.500","0.000","10.000","0.200"
"2","08/10/2009","02:50","Call","Off-Net","030092343242342376543","Une","08:00","1.500","2.000","20.000","3.500"
"2","09/10/2009","03:55","SMS","On-Net","030092343242342376543","Mink","00:00","1.500","0.000","5.000","100.500"
"2","09/10/2009","12:30","Call","Off-Net","030092343242342376543","Zog","01:01","3.500","3.000","70.000","6.500"
"2","09/10/2009","09:11","Call","On-Net","030092343242342376543","Monk","02:30","2.00","2.000","90.000","4.000"
Probably easiest solution is not searching for place to split, but finding elements which you want to return. In your case these elements
starts "
ends with "
have no " inside.
So you try with something like
String data = "\"1\",\"10411721\",\"MikeTison\",\"08/11/2009\",\"21/11/2009\",\"2800.00\",\"002934538\",\"051\",\"New York\",\"10411720-002\",\".\\Images\\b.jpg\",\".\\RTF\\b.rtf\"";
Pattern p = Pattern.compile("\"([^\"]+)\"");
Matcher m = p.matcher(data);
while(m.find()){
System.out.println(m.group(1));
}
Output:
1
10411721
MikeTison
08/11/2009
21/11/2009
2800.00
002934538
051
New York
10411720-002
.\Images\b.jpg
.\RTF\b.rtf
You can split using this regex:
String[] arr = input.split( "(?=(([^\"]*\"){2})*[^\"]*$),+" );
This regex will split on commas if those are outside double quotes by using a lookahead to make sure there are even number of quotes after a comma.
Remove the first and the last character of the whole string. Then split with ","
String test = "\"1\",\"10411721\",\"MikeTison\",\"08/11/2009\",\"21/11/2009\",\"2800.00\",\"002934538\",\"051\",\"New York\",\"10411720-002\",\".\\Images\\b.jpg\",\".\\RTF\\b.rtf\"";
if (test.length() > 0)
test = test.substring(1, test.length()-1);
System.out.println(Arrays.toString(test.split("\",\"")));
This works even if you have new line character..try it out
String str="\"1\",\"10411721\",\"MikeTison\",\"08/11/2009\",\"21/11/2009\",\"2800.00\",\"002934538\",\"051\",\"New York\",\"10411720-002\",\".\\Images\\b.jpg\",\".\\RTF\\b.rtf\"";
System.out.println(Arrays.toString(str.split(",(?=([^\"]*\"[^\"]*\")*[^\"]*$)")));
i have seen to replace "," to "." by using ".$"|",$", but this logic is not working with alphabets.
i need to replace last letter of a word to another letter for all word in string containing EXAMPLE_TEST using java
this is my code
Pattern replace = Pattern.compile("n$");//here got the real problem
matcher2 = replace.matcher(EXAMPLE_TEST);
EXAMPLE_TEST=matcher2.replaceAll("k");
i also tried "//n$" ,"\n$" etc
Please help me to get the solution
input text=>njan ayman
output text=> njak aymak
Instead of the end of string $ anchor, use a word boundary \b
String s = "njan ayman";
s = s.replaceAll("n\\b", "k");
System.out.println(s); //=> "njak aymak"
You can use lookahead and group matching:
String EXAMPLE_TEST = "njan ayman";
s = EXAMPLE_TEST.replaceAll("(n)(?=\\s|$)", "k");
System.out.println("s = " + s); // prints: s = njak aymak
Explanation:
(n) - the matched word character
(?=\\s|$) - which is followed by a space or at the end of the line (lookahead)
The above is only an example! if you want to switch every comma with a period the middle line should be changed to:
s = s.replaceAll("(,)(?=\\s|$)", "\\.");
Here's how I would set it up:
(?=.\b)\w
Which in Java would need to be escaped as following:
(?=.\\b)\\w
It translates to something like "a character (\w) after (?=) any single character (.) at the end of a word (\b)".
String s = "njan ayman aowkdwo wdonwan. wadawd,.. wadwdawd;";
s = s.replaceAll("(?=.\\b)\\w", "");
System.out.println(s); //nja ayma aowkdw wdonwa. wadaw,.. wadwdaw;
This removes the last character of all words, but leaves following non-alphanumeric characters. You can specify only specific characters to remove/replace by changing the . to something else.
However, the other answers are perfectly good and might achieve exactly what you are looking for.
if (word.endsWith("char oldletter")) {
name = name.substring(0, name.length() - 1 "char newletter");
}
For the string value "ABCD_12" (including quotes), I would like to extract only the content and exclude out the double quotes i.e. ABCD_12 . My code is:
private static void checkRegex()
{
final Pattern stringPattern = Pattern.compile("\"([a-zA-Z_0-9])+\"");
Matcher findMatches = stringPattern.matcher("\"ABC_12\"");
if (findMatches.matches())
System.out.println("Match found" + findMatches.group(0));
}
Now I have tried doing findMatches.group(1);, but that only returns the last character in the string (I did not understand why !).
How can I extract only the content leaving out the double quotes?
Try this regex:
Pattern.compile("\"([a-zA-Z_0-9]+)\"");
OR
Pattern.compile("\"([^\"]+)\"");
Problem in your code is a misplaced + outside right parenthesis. Which is causing capturing group to capture only 1 character (since + is outside) and that's why you get only last character eventually.
A nice simple (read: non-regex) way to do this is:
String myString = "\"ABC_12\"";
String myFilteredString = myString.replaceAll("\"", "");
System.out.println(myFilteredString);
gets you
ABC_12
You should change your pattern to this:
final Pattern stringPattern = Pattern.compile("\"([a-zA-Z_0-9]+)\"");
Note that the + sign was moved inside the group, since you want the character repetition to be part of the group. In the code you posted, what you were actually searching for was a repetition of the group, which consisted in a single occurence of a single characters in [a-zA-Z_0-9].
If your pattern is strictly any text in between double quotes, then you may be better off using substring:
String str = "\"ABC_12\"";
System.out.println(str.substring(1, str.lastIndexOf('\"')));
Assuming it is a bit more complex (double quotes in between a larger string), you can use the split() function in the Pattern class and use \" as your regex - this will split the string around the \" so you can easily extract the content you want
Pattern p = Pattern.compile("\"");
// Split input with the pattern
String[] result =
p.split(str);
for (int i=0; i<result.length; i++)
System.out.println(result[i]);
}
http://docs.oracle.com/javase/1.4.2/docs/api/java/util/regex/Pattern.html#split%28java.lang.CharSequence%29
Hi please help me out in getting regular expression for the
following requirement
I have string type as
String vStr = "Every 1 nature(s) - Universe: (Air,Earth,Water sea,Fire)";
String sStr = "Every 1 form(s) - Earth: (Air,Fire) ";
from these strings after using regex I need to get values as "Air,Earth,Water sea,Fire" and "Air,Fire"
that means after
String vStrRegex ="Air,Earth,Water sea,Fire";
String sStrRegex ="Air,Fire";
All the strings that are input will be seperated by ":" and values needed are inside brackets always
Thanks
The regular expression would be something like this:
: \((.*?)\)
Spelt out:
Pattern p = Pattern.compile(": \\((.*?)\\)");
Matcher m = p.matcher(vStr);
// ...
String result = m.group(1);
This will capture the content of the parentheses as the first capture group.
Try the following:
\((.*)\)\s*$
The ending $ is important, otherwise you'll accidentally match the "(s)".
If you have each string separately, try this expression: \(([^\(]*)\)\s*$
This would get you the content of the last pair of brackets, as group 1.
If the strings are concatenated by : try to split them first.
Ask yourself if you really need a regex. Does the text you need always appear within the last two parentheses? If so, you can keep it simple and use substring instead:
String vStr = "Every 1 nature(s) - Universe: (Air,Earth,Water sea,Fire)";
int lastOpeningParens = vStr.lastIndexOf('(');
int lastClosingParens = vStr.lastIndexOf(')');
String text = vStr.substring(lastOpeningParens + 1, lastClosingParens);
This is much more readable than a regex.
I assume that there are only whitespace characters between : and the opening bracket (:
Pattern regex = Pattern.compile(":\\s+\\((.+)\\)");
You'll find your results in capturing group 1.
Try this regex:
.*\((.*)\)
$1 will contain the required string
I had a look at other stackoverflow questions and couldn't find one that asked the same question, so here it is:
How do you match the first and last characters of a string (can be multi-line or empty).
So for example:
String = "this is a simple sentence"
Note that the string includes the beginning and ending quotation marks.
How do I get match the first and last characters where the string begins and ends with a quotation mark (").
I tried:
^"|$" and \A"\Z"
but these do not produce the desired result.
Thanks for your help in advance :)
Is this what you are looking for?
String input = "\"this is a simple sentence\"";
String result = input.replaceFirst("(?s)^\"(.*)\"$", " $1 ");
This will replace the first and last character of the input string with spaces if it starts and ends with ". It will also work across multiple lines since the DOTALL flag is specified by (?s).
The regex that matches the whole input ".*". In java, it looks like this:
String regex = "\".*\"";
System.out.println("\"this is a simple sentence\"".matches(regex)); // true
System.out.println("this is a simple sentence".matches(regex)); // false
System.out.println("this is a simple sentence\"".matches(regex)); // false
If you want to remove the quotes, use this:
String input = "\"this is a simple sentence\"";
input = input.replaceAll("(^\"|\"$)", "")); // this is a simple sentence (without any quotes)
If you want this to work over multiple lines, use this:
String input = "\"this is a simple sentence\"\n\"and another sentence\"";
System.out.println(input + "\n");
input = input.replaceAll("(?m)(^\"|\"$)", "");
System.out.println(input);
which produces output:
"this is a simple sentence"
"and another sentence"
this is a simple sentence
and another sentence
Explanation of regex (?m)(^"|"$):
(?m) means "Caret and dollar match after and before newlines for the remainder of the regular expression"
(^"|"$) means ^" OR "$, which means "start of line then a double quote" OR "double quote then end of line"
Why not use the simple logic of getting the first and last characters based on charAt method of String? Place a few checks for empty/incomplete strings and you should be done.
String regexp = "(?s)\".*\"";
String data = "\"This is some\n\ndata\"";
Matcher m = Pattern.compile(regexp).matcher(data);
if (m.find()) {
System.out.println("Match starts at " + m.start() + " and ends at " + m.end());
}