I have a text file where every line is a random combination of any of the following groups
Numbers - English Letters - Arabic Letters - Punctuation
\w which is composed of a-zA-Z0-9_ for the first 2 groups
\p{InArabic} for the third group
\p{Punct} which is composed of !"#$%&'()*+,-./:;<=>?#[]^_`{|}~ for the fifth group
I got this info from here
i read a line. The ONLY time I do something to this line is if the line contains Arabic letters AND (English letters OR Unicode Symbols)
After reading this post and this post I came up with the following expression. Obviously it's wrong as my output is all wrong >.<
pattern = Pattern.compile("(?=\\p{InArabic})(?=[a-zA-Z])");
Here's the input
1
1a
a!
aش
شa
ششa
aشش
شaش
aشa
!aش
The first three shouldn't be matched but my output shows that NONE are a match.
Edit: sorry I just realized that I forgot to change my title. But if any of you feel that searching is better performance wise then please suggest a search algorithm. Using search algo instead of regex looks ugly but I'd go with it if it performed better. Thanks to the posts I read, I learned that I can make regex faster if I put this in the constructor so that it'd be executed once only instead of including them in my loop thereby being executed everytime
pattern = Pattern.compile("(?=\\p{InArabic})(?=[a-zA-Z])");
matcher = pattern.matcher("");
To follow your idea, the correct pattern is:
pattern = Pattern.compile("(?=.*\\p{InArabic})(?=.*[a-zA-Z\\p{Punct}])");
The same position in a string can not be followed by an arabic letter and a punctuation character or a latin letter at the same time. In other words, you have written an always false condition. Adding .* allows characters to be anywhere in the string.
If you want a more optimised pattern, you can use Jason C idea but with negative character classes to reduce the backtracking:
pattern = Pattern.compile("\\p{inArabic}[^a-zA-Z\\p{Punct}]*[a-zA-Z\\p{Punct}]|[a-zA-Z\\p{Punct}]\\P{inArabic}*\\p{inArabic}");
If you want to find a line with a mix, all you really need are 2 boundry condition checks.
A sucessfull match indicates a mix.
# "\\p{InArabic}(?=[\\w\\p{Punct}])|(?<=[\\w\\p{Punct}])\\p{InArabic}"
\p{InArabic}
(?= [\w\p{Punct}] )
|
(?<= [\w\p{Punct}] )
\p{InArabic}
Related
Background
Writing a straight quote to curly quote converter and am looking to separate substitution into a few different steps. The first step is to replace contractions in text using a lexicon of known contractions. This won't solve ambiguities, but should convert straight quote usages in common contractions.
Problem
In Java, \b and \w don't include apostrophes as part of a word, which makes this problem a bit finicky. The issue is in matching words that:
contain one or more apostrophes, but do not start or end with one (inner);
begin with an apostrophe, may contain one or more, but do not end with one (began);
end with an apostrophe, may contain one or more, but do not start with one (ended); and
begin and end with an apostrophe, but may not contain one (outer).
Code
Given some nonsensical text:
'Twas---Wha'? Wouldn'tcha'? 'Twas, or 'twasn't, 'tis what's 'tween dawn 'n' dusk 'n stuff. Cookin'? 'Sams' place, 'yo''
the regexes should capture the following words:
inner: what's
began: 'Twas, 'Twas, 'twasn't, 'tis, 'tween, 'n
ended: Wha', Wouldn'tcha', Cookin'
outer: 'n', 'Sams', 'yo'
Here are non-working expressions, a mix-mash of maladroit thoughts:
inner: \p{L}+'\p{L}*\p{L}
began: ((?<=[^\p{L}])|^)'\p{L}+('\p{L}|\p{L})?
ended: (\p{L}|\p{L}')+'(?=[^\p{L}]|$)
This one appears to work:
outer: ((?<=[^\p{L}])|^)'\p{L}+'(?!\p{L})
Question
What regular expressions would categorize this quartet of contractions correctly?
This regex should do what you want. It uses named capture groups to categorise the words with appropriate lookarounds to ensure that we match the whole words with the required outer quotes:
(?<inner>(?<![\p{L}'])(?:\p{L}+')+\p{L}+(?![\p{L}']))|
(?<began>(?<!\p{L})(?:'\p{L}+)+(?![\p{L}']))|
(?<ended>(?<![\p{L}'])(?:\p{L}+')+(?!\p{L}))|
(?<outer>(?<!\p{L})'\p{L}+'(?!\p{L}))
Group inner looks for a string with some number of groups of letters followed by a quote (?:\p{L}+')+ followed by some number of letters \p{L}+.
Group began looks for a string with some number of groups of a quote followed by some number of letters (?:'\p{L}+)+.
Group ended looks for a string with some number of groups of letters followed by a quote (?:\p{L}+')+.
Group outer looks for a string with quotes on either end and some number of letters in the middle '\p{L}+'.
Demo on regex101
I try to have a regex validating an input field.
What i call "joker" chars are '?' and '*'.
Here is my java regex :
"^$|[^\\*\\s]{2,}|[^\\*\\s]{2,}[\\*\\?]|[^\\*\\s]{2,}[\\?]{1,}[^\\s\\*]*[\\*]{0,1}"
What I'm tying to match is :
Minimum 2 alpha-numeric characters (other than '?' and '*')
The '*' can only appears one time and at the end of the string
The '?' can appears multiple time
No WhiteSpace at all
So for example :
abcd = OK
?bcd = OK
ab?? = OK
ab*= OK
ab?* = OK
??cd = OK
*ab = NOT OK
??? = NOT OK
ab cd = NOT OK
abcd = Not OK (space at the begining)
I've made the regex a bit complicated and I'm lost can you help me?
^(?:\?*[a-zA-Z\d]\?*){2,}\*?$
Explanation:
The regex asserts that this pattern must appear twice or more:
\?*[a-zA-Z\d]\?*
which asserts that there must be one character in the class [a-zA-Z\d] with 0 to infinity questions marks on the left or right of it.
Then, the regex matches \*?, which means an 0 or 1 asterisk character, at the end of the string.
Demo
Here is an alternative regex that is faster, as revo suggested in the comments:
^(?:\?*[a-zA-Z\d]){2}[a-zA-Z\d?]*\*?$
Demo
Here you go:
^\?*\w{2,}\?*\*?(?<!\s)$
Both described at demonstrated at Regex101.
^ is a start of the String
\?* indicates any number of initial ? characters (must be escaped)
\w{2,} at least 2 alphanumeric characters
\?* continues with any number of and ? characters
\*? and optionally one last * character
(?<!\s) and the whole String must have not \s white character (using negative look-behind)
$ is an end of the String
Other way to solve this problem could be with look-ahead mechanism (?=subregex). It is zero-length (it resets regex cursor to position it was before executing subregex) so it lets regex engine do multiple tests on same text via construct
(?=condition1)
(?=condition2)
(?=...)
conditionN
Note: last condition (conditionN) is not placed in (?=...) to let regex engine move cursor after tested part (to "consume" it) and move on to testing other things after it. But to make it possible conditionN must match precisely that section which we want to "consume" (earlier conditions didn't have that limitation, they could match substrings of any length, like lets say few first characters).
So now we need to think about what are our conditions.
We want to match only alphanumeric characters, ?, * but * can appear (optionally) only at end. We can write it as ^[a-zA-Z0-9?]*[*]?$. This also handles non-whitespace characters because we didn't include them as potentially accepted characters.
Second requirement is to have "Minimum 2 alpha-numeric characters". It can be written as .*?[a-zA-Z0-9].*?[a-zA-Z0-9] or (?:.*?[a-zA-Z0-9]){2,} (if we like shorter regexes). Since that condition doesn't actually test whole text but only some part of it, we can place it in look-ahead mechanism.
Above conditions seem to cover all we wanted so we can combine them into regex which can look like:
^(?=(?:.*?[a-zA-Z0-9]){2,})[a-zA-Z0-9?]*[*]?$
I was trying to made regex for extracting word at the place of Delhi in text
sending to: GK Delhi, where the sending to: is fixed and i don't want to capture whatever at the place of GK. Actually GK will be one word in my case, what i made which should work is: (?<=sending to: \w )Delhi, means if word starts with sending to: and ends with Delhi then return Delhi.
Please help me to fix this.
Three points,
\w matches a single word character. Use \w+ to match one or more or \w* to match zero or more word characters.
Don't forget about space between DK and Delhi: \s+.
Just a note: The (?<= construct is the positive lookbehind, not negative one.
So the regex could look like this:
(?<=sending to:\s*\w+\s+)Delhi
Please also note that arbitrary-length lookbehind is only supported by very few regex engines, but you didn't say anything about the tool you are using.
Update:
Java doesn't support arbitrary-length lookbehind expressions.
The possibilities you have are:
The matched text will always be Delhi (on successful match). So if you are only checking for a match, then you could just use the regex: sending to:\s*\w+\s+Delhi.
If you want to extend the regex to other towns in future, then you could use a capturing group. The regex would be, for example, sending to:\s*\w+\s+(Delhi|Mumbai) and in Java code you would get the city name via matcher.group(1).
Please post your actual Java code of how you are using the regex if you want a more detailed advice.
I need to replace the duplicate characters in a string. I tried using
outputString = str.replaceAll("(.)(?=.*\\1)", "");
This replaces the duplicate characters but the position of the characters changes as shown below.
input
haih
output
aih
But I need to get an output hai. That is the order of the characters that appear in the string should not change. Given below are the expected outputs for some inputs.
input
aaaassssddddd
output
asd
input
cdddddggggeeccc
output
cdge
How can this be achieved?
It seems like your code is leaving the last character, so how about this?
outputString = new StringBuilder(str).reverse().toString();
// outputString is now hiah
outputString = outputString.replaceAll("(.)(?=.*\\1)", "");
// outputString is now iah
outputString = new StringBuilder(outputString).reverse().toString();
// outputString is now hai
Overview
It's possible with Oracle's implementation, but I wouldn't recommend this answer for many reasons:
It relies on a bug in the implementation, which interprets *, + or {n,} as {0, 0x7FFFFFFF}, {1, 0x7FFFFFFF}, {n, 0x7FFFFFFF} respectively, which allows the look-behind to contains such quantifiers. Since it relies on a bug, there is no guarantee that it will work similarly in the future.
It is unmaintainable mess. Writing normal code and any people who have some basic Java knowledge can read it, but using the regex in this answer limits the number of people who can understand the code at a glance to people who understand the in and out of regex implementation.
Therefore, this answer is for educational purpose, rather than something to be used in production code.
Solution
Here is the one-liner replaceAll regex solution:
String output = input.replaceAll("(.)(?=(.*))(?<=(?=\\1.*?\\1\\2$).+)","")
Printing out the regex:
(.)(?=(.*))(?<=(?=\1.*?\1\2$).+)
What we want to do is to look-behind to see whether the same character has appeared before or not. The capturing group (.) at the beginning captures the current character, and the look-behind group is there to check whether the character has appeared before. So far, so good.
However, since backreferences \1 doesn't have obvious length, it can't appear in the look-behind directly.
This is where we make use of the bug to look-behind up to the beginning of the string, then use a look-ahead inside the look-behind to include the backreference, as you can see (?<=(?=...).+).
This is not the end of the problem, though. While the non-assertion pattern inside look-behind .+ can't advance past the position after the character in (.), the look-ahead inside can. As a simple test:
"haaaaaaaaa".replaceAll("h(?<=(?=(.*)).*)","$1")
> "aaaaaaaaaaaaaaaaaa"
To make sure that the search doesn't spill beyond the current character, I capture the rest of the string in a look-ahead (?=(.*)) and use it to "mark" the current position (?=\\1.*?\\1\\2$).
Can this be done in one replacement without using look-behind?
I think it is impossible. We need to differentiate the first appearance of a character with subsequent appearance of the same character. While we can do this for one fixed character (e.g. a), the problem requires us to do so for all characters in the string.
For your information, this is for removing all subsequent appearance of a fixed character (h is used here):
.replaceAll("^([^h]*h[^h]*)|(?!^)\\Gh+([^h]*)","$1$2")
To do this for multiple characters, we must keep track of whether the character has appeared before or not, across matches and for all characters. The regex above shows the across matches part, but the other condition kinda makes this impossible.
We obviously can't do this in a single match, since subsequent occurrences can be non-contiguous and arbitrary in number.
I'm no expert in regex but I need to parse some input I have no control over, and make sure I filter away any strings that don't have A-z and/or 0-9.
When I run this,
Pattern p = Pattern.compile("^[a-zA-Z0-9]*$"); //fixed typo
if(!p.matcher(gottenData).matches())
System.out.println(someData); //someData contains gottenData
certain spaces + an unknown symbol somehow slip through the filter (gottenData is the red rectangle):
In case you're wondering, it DOES also display Text, it's not all like that.
For now, I don't mind the [?] as long as it also contains some string along with it.
Please help.
[EDIT] as far as I can tell from the (very large) input, the [?]'s are either white spaces either nothing at all; maybe there's some sort of encoding issue, also perhaps something to do with #text nodes (input is xml)
The * quantifier matches "zero or more", which means it will match a string that does not contain any of the characters in your class. Try the + quantifier, which means "One or more": ^[a-zA-Z0-9]+$ will match strings made up of alphanumeric characters only. ^.*[a-zA-Z0-9]+.*$ will match any string containing one or more alphanumeric characters, although the leading .* will make it much slower. If you use Matcher.lookingAt() instead of Matcher.matches, it will not require a full string match and you can use the regex [a-zA-Z0-9]+.
You have an error in your regex: instead of [a-zA-z0-9]* it should be [a-zA-Z0-9]*.
You don't need ^ and $ around the regex.
Matcher.matches() always matches the complete string.
String gottenData = "a ";
Pattern p = Pattern.compile("[a-zA-z0-9]*");
if (!p.matcher(gottenData).matches())
System.out.println("doesn't match.");
this prints "doesn't match."
The correct answer is a combination of the above answers. First I imagine your intended character match is [a-zA-Z0-9]. Note that A-z isn't as bad as you might think it include all characters in the ASCII range between A and z, which is the letters plus a few extra (specifically [,\,],^,_,`).
A second potential problem as Martin mentioned is you may need to put in the start and end qualifiers, if you want the string to only consists of letters and numbers.
Finally you use the * operator which means 0 or more, therefore you can match 0 characters and matches will return true, so effectively your pattern will match any input. What you need is the + quantifier. So I will submit the pattern you are most likely looking for is:
^[a-zA-Z0-9]+$
You have to change the regexp to "^[a-zA-Z0-9]*$" to ensure that you are matching the entire string
Looks like it should be "a-zA-Z0-9", not "a-zA-z0-9", try correcting that...
Did anyone consider adding space to the regex [a-zA-Z0-9 ]*. this should match any normal text with chars, number and spaces. If you want quotes and other special chars add them to the regex too.
You can quickly test your regex at http://www.regexplanet.com/simple/
You can check input value is contained string and numbers? by using regex ^[a-zA-Z0-9]*$
if your value just contained numberString than its show match i.e, riz99, riz99z
else it will show not match i.e, 99z., riz99.z, riz99.9
Example code:
if(e.target.value.match('^[a-zA-Z0-9]*$')){
console.log('match')
}
else{
console.log('not match')
}
}
online working example