I'm trying to write a RegEx that would capture both instance of 'csv' delimited by commas in the string "csv,csv,csv,csv" and replace with "xml".
So my output would be "csv,xml,xml,csv".
The problem I'm running into is that my RegEx only captures the first instance of csv, and since the 'comma' at the end of the first match would also be the first delimiter of the second match, it doesn't get captured.
RegEx = ",csv," (pretty simple example).
Example 2:
RegEx = "(,dog,)"
Input "dog,dog,dog,dog"
Output "dog,cat,cat,dog"
Edit: I found what I was looking for, using lookahead and lookbehind.
RegEx = "(?<=,)dog(?=,)" would do the trick.
Use look arounds, which assert, but don't consume:
String replaced = str.replaceAll("(?<=,)csv(?=,)", "xml");
Make use of lookaround:
replace
(?<=,)xml(?=,)
with
csv
Demo https://regex101.com/r/ft7SQT/1
Related
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
I am trying to formulate a regex for the following scenario :
The String to match : mName87.com
So, the string may consist of any number of alpha numeric characters , but can contain only a single dot anywhere in the string .
I formulated this regex : [a-zA-Z0-9.], but it matches even multiple dots(.)
What am i doing wrong here ?
The regex you provided matches only a single character in the whole string you're trying to validate. There are a few things to take care of in your scenario
You want to match over the whole string, so your regex must start with ^ (beginning of the string) and end with $ (end of the string).
Then you want to accept any number of alpha-numeric characters, this is done with [a-zA-Z0-9]+, here the + means one or more characters.
Then match the point: \. (you must escape it here)
Finally accept more characters again.
All together the regex would then be:
^[a-zA-Z0-9]+\.[a-zA-Z0-9]+$
You can use this regex:
\\w*\\.\\w*
You can try here
Try with:
^([a-zA-Z0-9]+\.)+[a-zA-Z]$
use this regular expression ^[a-zA-Z0-9]*\.[a-zA-Z0-9.]*$
EDITED:
Try
([a-zA-Z0-9]+\.[a-zA-Z0-9]+)|(\.[a-zA-Z0-9]+)|([a-zA-Z0-9]+\.)
That is: [a word that ends with a dot] OR [two words and the dot in the middle] OR [a word that starts with a dot]
using only regular expression methods, the method String.replaceAll and ArrayList
how can i split a String into tokens, but ignore delimiters that exist inside quotes?
the delimiter is any character that is not alphanumeric or quoted text
for example:
The string :
hello^world'this*has two tokens'
should output:
hello
worldthis*has two tokens
I know there is a damn good and accepted answer already present but I would like to add another regex based (and may I say simpler) approach to split the given text using any non-alphanumeric delimiter which not inside the single quotes using
Regex:
/(?=(([^']+'){2})*[^']*$)[^a-zA-Z\\d]+/
Which basically means match a non-alphanumeric text if it is followed by even number of single quotes in other words match a non-alphanumeric text if it is outside single quotes.
Code:
String string = "hello^world'this*has two tokens'#2ndToken";
System.out.println(Arrays.toString(
string.split("(?=(([^']+'){2})*[^']*$)[^a-zA-Z\\d]+"))
);
Output:
[hello, world'this*has two tokens', 2ndToken]
Demo:
Here is a live working Demo of the above code.
Use a Matcher to identify the parts you want to keep, rather than the parts you want to split on:
String s = "hello^world'this*has two tokens'";
Pattern pattern = Pattern.compile("([a-zA-Z0-9]+|'[^']*')+");
Matcher matcher = pattern.matcher(s);
while (matcher.find()) {
System.out.println(matcher.group(0));
}
See it working online: ideone
You cannot in any reasonable way. You are posing a problem that regular expressions aren't good at.
Do not use a regular expression for this. It won't work. Use / write a parser instead.
You should use the right tool for the right task.
As a beginner with regex i believe im about to ask something too simple but ill ask anyway hope it won't bother you helping me..
Lets say i have a text like "hello 'cool1' word! 'cool2'"
and i want to get the first quote's text (which is 'cool1' without the ')
what should be my pattern? and when using matcher, how do i guarantee it will remain the first quote and not the second?
(please suggest a solution only with regex.. )
Use this regular expression:
'([^']*)'
Use as follows: (ideone)
Pattern pattern = Pattern.compile("'([^']*)'");
Matcher matcher = pattern.matcher(s);
if (matcher.find()) {
System.out.println(matcher.group(1));
}
Or this if you know that there are no new-line characters in your quoted string:
'(.*?)'
when using matcher, how do i guarantee it will remain the first quote and not the second?
It will find the first quoted string first because it starts seaching from left to right. If you ask it for the next match it will give you the second quoted string.
If you want to find first quote's text without the ' you can/should use Lookahead and Lookbehind mechanism like
(?<=').*?(?=')
for example
System.out.println("hello 'cool1' word! 'cool2'".replaceFirst("(?<=').*?(?=')", "ABC"));
//out -> hello 'ABC' word! 'cool2'
more info
You could just split the string on quotes and get the second piece (which will be between the first and second quotes).
If you insist on regex, try this:
/^.*?'(.*?)'/
Make sure it's set to multiline, unless you know you'll never have newlines in your input. Then, get the subpattern from the result and that will be your string.
To support double quotes too:
/^.*?(['"])(.*?)\1/
Then get subpattern 2.
I am running into this problem in Java.
I have data strings that contain entities enclosed between & and ; For e.g.
&Text.ABC;, &Links.InsertSomething;
These entities can be anything from the ini file we have.
I need to find these string in the input string and remove them. There can be none, one or more occurrences of these entities in the input string.
I am trying to use regex to pattern match and failing.
Can anyone suggest the regex for this problem?
Thanks!
Here is the regex:
"&[A-Za-z]+(\\.[A-Za-z]+)*;"
It starts by matching the character &, followed by one or more letters (both uppercase and lower case) ([A-Za-z]+). Then it matches a dot followed by one or more letters (\\.[A-Za-z]+). There can be any number of this, including zero. Finally, it matches the ; character.
You can use this regex in java like this:
Pattern p = Pattern.compile("&[A-Za-z]+(\\.[A-Za-z]+)*;"); // java.util.regex.Pattern
String subject = "foo &Bar; baz\n";
String result = p.matcher(subject).replaceAll("");
Or just
"foo &Bar; baz\n".replaceAll("&[A-Za-z]+(\\.[A-Za-z]+)*;", "");
If you want to remove whitespaces after the matched tokens, you can use this re:
"&[A-Za-z]+(\\.[A-Za-z]+)*;\\s*" // the "\\s*" matches any number of whitespace
And there is a nice online regular expression tester which uses the java regexp library.
http://www.regexplanet.com/simple/index.html
You can try:
input=input.replaceAll("&[^.]+\\.[^;]+;(,\\s*&[^.]+\\.[^;]+;)*","");
See it