Hiho everyone! :)
I have an application, in which the user can insert a string into a textbox, which will be used for a String.Format output later. So the user's input must have a certain format:
I would like to replace exactly one placeholder, so the string should be of a form like this: "Text{0}Text". So it has to contain at least one '{0}', but no other statement between curly braces, for example no {1}.
For the text before and after the '{0}', I would allow any characters.
So I think, I have to respect the following restrictions: { must be written as {{, } must be written as }}, " must be written as \" and \ must be written as \.
Can somebody tell me, how I can write such a RegEx? In particular, can I do something like 'any character WITHOUT' to exclude the four characters ( {, }, " and \ ) above instead of listing every allowed character?
Many thanks!!
Nikki:)
I hate to be the guy who doesn't answer the question, but really it's poor usability to ask your user to format input to work with String.Format. Provide them with two input requests, so they enter the part before the {0} and the part after the {0}. Then you'll want to just concatenate the strings instead of use String.Format- using String.Format on user-supplied text is just a bad idea.
[^(){}\r\n]+\{0}[^(){}\r\n]+
will match any text except (, ), {, } and linebreaks, then match {0}, then the same as before. There needs to be at least one character before and after the {0}; if you don't want that, replace + with *.
You might also want to anchor the regex to beginning and end of your input string:
^[^(){}\r\n]+\{0}[^(){}\r\n]+$
(Similar to Tim's answer)
Something like:
^[^{}()]*(\{0})[^{}()]*$
Tested at http://www.regular-expressions.info/javascriptexample.html
It sounds like you're looking for the [^CHARS_GO_HERE] construct. The exact regex you'd need depends on your regex engine, but it would resemble [^({})].
Check out the "Negated Character Classes" section of the Character Class page at Regular-Expressions.info.
I think your question can be answered by the regexp:
^(((\{\{|\}\}|\\"|\\\\|[^\{\}\"\\])*(\{0\}))+(\{\{|\}\}|\\"|\\\\|[^\{\}\"\\])*$
Explanation:
The expression is built up as follows:
^(allowed chars {0})+(allowed chars)*$
one or more sequences of allowed chars followed by a {0} with optional allowed chars at the end.
allowed chars is built of the 4 sequences you mentioned (I assumed the \ escape is \\ instead of \.) plus all chars that do not contain the escapes chars:
(\{\{|\}\}|\\"|\\\\|[^\{\}\"\\])
combined they make up the regexp I started with.
Related
I want to select some string combination (with dots(.)) from a very long string (sql). The full string could be a single line or multiple line with new line separator, and this combination could be in start (at first line) or a next line (new line) or at both place.
I need help in writing a regex for it.
Examples:
String s = I am testing something like test.test.test in sentence.
Expected output: test.test.test
Example2 (real usecase):
UPDATE test.table
SET access = 01
WHERE access IN (
SELECT name FROM project.dataset.tablename WHERE name = 'test' GROUP BY 1 )
Expected output: test.table and project.dataset.tablename
, can I also add some prefix or suffix words or space which should be present where ever this logic gets checked. In above case if its update regex should pick test.table, but if the statement is like select test.table regex should not pick it up this combinations and same applies for suffix.
Example3: This is to illustrate the above theory.
INS INTO test.table
SEL 'abcscsc', wu_id.Item_Nbr ,1
FROM test.table as_t
WHERE as_t.old <> 0 AND as_t.date = 11
AND (as_t.numb IN ('11') )
Expected Output: test.table, test.table (Key words are INTO and FROM)
Things Not Needed in selection:as_t.numb, as_t.old, as_t.date
If I get the regex I can use in program to extract this word.
Note: Before and after string words to the combination could be anything like update, select { or(, so we have to find the occurrence of words which are joined together with .(dot) and all the number of such occurrence.
I tried something like this:
(?<=.)(.?)(?=.)(.?) -: This only selected the word between two .dot and not all.
.(?<=.)(.?)(?=.)(.?). - This everything before and after.
To solve your initial problem, we can just use some negation. Here's the pattern I came up with:
[^\s]+\.[^\s]+
[^ ... ] Means to make a character class including everything except for what's between the brackets. In this case, I put \s in there, which matches any whitespace. So [^\s] matches anything that isn't whitespace.
+ Is a quantifier. It means to find as many of the preceding construct as you can without breaking the match. This would happily match everything that's not whitespace, but I follow it with a \., which matches a literal .. The \ is necessary because . means to match any character in regex, so we need to escape it so it only has its literal meaning. This means there has to be a . in this group of non-whitespace characters.
I end the pattern with another [^\s]+, which matches everything after the . until the next whitespace.
Now, to solve your secondary problem, you want to make this match only work if it is preceded by a given keyword. Luckily, regex has a construct almost specifically for this case. It's called a lookbehind. The syntax is (?<= ... ) where the ... is the pattern you want to look for. Using your example, this will only match after the keywords INTO and FROM:
(?<=(?:INTO|FROM)\s)[^\s]+\.[^\s]+
Here (?:INTO|FROM) means to match either the text INTO or the text FROM. I then specify that it should be followed by a whitespace character with \s. One possible problem here is that it will only match if the keywords are written in all upper case. You can change this behavior by specifying the case insensitive flag i to your regex parser. If your regex parser doesn't have a way to specify flags, you can usually still specify it inline by putting (?i) in front of the pattern, like so:
(?i)(?<=(?:INTO|FROM)\s)[^\s]+\.[^\s]+
If you are new to regex, I highly recommend using the www.regex101.com website to generate regex and learn how it works. Don't forget to check out the code generator part for getting the regex code based on the programming language you are using, that's a cool feature.
For your question, you need a regex that understands any word character \w that matches between 0 and unlimited times, followed by a dot, followed by another series of word character that repeats between 0 and unlimited times.
So here is my solution to your question:
Your regex in JavaScript:
const regex = /([\w][.][\w])+/gm;
in Java:
final String regex = "([\w][.][\w])+";
in Python:
regex = r"([\w][.][\w])+"
in PHP:
$re = '/([\w][.][\w])+/m';
Note that: this solution is written for your use case (to be used for SQL strings), because now if you have something like '.word' or 'word..word', it will still catch it which I assume you don't have a string like that.
See this screenshot for more details
In Java RegEx, how to find out the difference between .(dot) the meta character and the normal dot as we using in any sentence. How to handle this kind of situation for other meta characters too like (*,+,\d,...)
If you want the dot or other characters with a special meaning in regexes to be a normal character, you have to escape it with a backslash. Since regexes in Java are normal Java strings, you need to escape the backslash itself, so you need two backslashes e.g. \\.
Solutions proposed by the other members don't work for me.
But I found this :
to escape a dot in java regexp write [.]
Perl-style regular expressions (which the Java regex engine is more or less based upon) treat the following characters as special characters:
.^$|*+?()[{\ have special meaning outside of character classes,
]^-\ have special meaning inside of character classes ([...]).
So you need to escape those (and only those) symbols depending on context (or, in the case of character classes, place them in positions where they can't be misinterpreted).
Needlessly escaping other characters may work, but some regex engines will treat this as syntax errors, for example \_ will cause an error in .NET.
Some others will lead to false results, for example \< is interpreted as a literal < in Perl, but in egrep it means "word boundary".
So write -?\d+\.\d+\$ to match 1.50$, -2.00$ etc. and [(){}[\]] for a character class that matches all kinds of brackets/braces/parentheses.
If you need to transform a user input string into a regex-safe form, use java.util.regex.Pattern.quote.
Further reading: Jan Goyvaert's blog RegexGuru on escaping metacharacters
Escape special characters with a backslash. \., \*, \+, \\d, and so on. If you are unsure, you may escape any non-alphabetical character whether it is special or not. See the javadoc for java.util.regex.Pattern for further information.
Here is code you can directly copy paste :
String imageName = "picture1.jpg";
String [] imageNameArray = imageName.split("\\.");
for(int i =0; i< imageNameArray.length ; i++)
{
system.out.println(imageNameArray[i]);
}
And what if mistakenly there are spaces left before or after "." in such cases? It's always best practice to consider those spaces also.
String imageName = "picture1 . jpg";
String [] imageNameArray = imageName.split("\\s*.\\s*");
for(int i =0; i< imageNameArray.length ; i++)
{
system.out.println(imageNameArray[i]);
}
Here, \\s* is there to consider the spaces and give you only required splitted strings.
I wanted to match a string that ends with ".*"
For this I had to use the following:
"^.*\\.\\*$"
Kinda silly if you think about it :D
Heres what it means. At the start of the string there can be any character zero or more times followed by a dot "." followed by a star (*) at the end of the string.
I hope this comes in handy for someone. Thanks for the backslash thing to Fabian.
If you want to end check whether your sentence ends with "." then you have to add [\.\]$ to the end of your pattern.
I am doing some basic array in JGrasp and found that with an accessor method for a char[][] array to use ('.') to place a single dot.
I was trying to split using .folder. For this use case, the solution to use \\.folder and [.]folder didn't work.
The following code worked for me
String[] pathSplited = Pattern.compile("([.])(folder)").split(completeFilePath);
This question already has answers here:
Regular expression to match a line that doesn't contain a word
(34 answers)
Closed 2 years ago.
I know that I can negate group of chars as in [^bar] but I need a regular expression where negation applies to the specific word - so in my example how do I negate an actual bar, and not "any chars in bar"?
A great way to do this is to use negative lookahead:
^(?!.*bar).*$
The negative lookahead construct is the pair of parentheses, with the opening parenthesis followed by a question mark and an exclamation point. Inside the lookahead [is any regex pattern].
Unless performance is of utmost concern, it's often easier just to run your results through a second pass, skipping those that match the words you want to negate.
Regular expressions usually mean you're doing scripting or some sort of low-performance task anyway, so find a solution that is easy to read, easy to understand and easy to maintain.
Solution:
^(?!.*STRING1|.*STRING2|.*STRING3).*$
xxxxxx OK
xxxSTRING1xxx KO (is whether it is desired)
xxxSTRING2xxx KO (is whether it is desired)
xxxSTRING3xxx KO (is whether it is desired)
You could either use a negative look-ahead or look-behind:
^(?!.*?bar).*
^(.(?<!bar))*?$
Or use just basics:
^(?:[^b]+|b(?:$|[^a]|a(?:$|[^r])))*$
These all match anything that does not contain bar.
The following regex will do what you want (as long as negative lookbehinds and lookaheads are supported), matching things properly; the only problem is that it matches individual characters (i.e. each match is a single character rather than all characters between two consecutive "bar"s), possibly resulting in a potential for high overhead if you're working with very long strings.
b(?!ar)|(?<!b)a|a(?!r)|(?<!ba)r|[^bar]
I came across this forum thread while trying to identify a regex for the following English statement:
Given an input string, match everything unless this input string is exactly 'bar'; for example I want to match 'barrier' and 'disbar' as well as 'foo'.
Here's the regex I came up with
^(bar.+|(?!bar).*)$
My English translation of the regex is "match the string if it starts with 'bar' and it has at least one other character, or if the string does not start with 'bar'.
The accepted answer is nice but is really a work-around for the lack of a simple sub-expression negation operator in regexes. This is why grep --invert-match exits. So in *nixes, you can accomplish the desired result using pipes and a second regex.
grep 'something I want' | grep --invert-match 'but not these ones'
Still a workaround, but maybe easier to remember.
If it's truly a word, bar that you don't want to match, then:
^(?!.*\bbar\b).*$
The above will match any string that does not contain bar that is on a word boundary, that is to say, separated from non-word characters. However, the period/dot (.) used in the above pattern will not match newline characters unless the correct regex flag is used:
^(?s)(?!.*\bbar\b).*$
Alternatively:
^(?!.*\bbar\b)[\s\S]*$
Instead of using any special flag, we are looking for any character that is either white space or non-white space. That should cover every character.
But what if we would like to match words that might contain bar, but just not the specific word bar?
(?!\bbar\b)\b\[A-Za-z-]*bar[a-z-]*\b
(?!\bbar\b) Assert that the next input is not bar on a word boundary.
\b\[A-Za-z-]*bar[a-z-]*\b Matches any word on a word boundary that contains bar.
See Regex Demo
Extracted from this comment by bkDJ:
^(?!bar$).*
The nice property of this solution is that it's possible to clearly negate (exclude) multiple words:
^(?!bar$|foo$|banana$).*
I wish to complement the accepted answer and contribute to the discussion with my late answer.
#ChrisVanOpstal shared this regex tutorial which is a great resource for learning regex.
However, it was really time consuming to read through.
I made a cheatsheet for mnemonic convenience.
This reference is based on the braces [], (), and {} leading each class, and I find it easy to recall.
Regex = {
'single_character': ['[]', '.', {'negate':'^'}],
'capturing_group' : ['()', '|', '\\', 'backreferences and named group'],
'repetition' : ['{}', '*', '+', '?', 'greedy v.s. lazy'],
'anchor' : ['^', '\b', '$'],
'non_printable' : ['\n', '\t', '\r', '\f', '\v'],
'shorthand' : ['\d', '\w', '\s'],
}
Just thought of something else that could be done. It's very different from my first answer, as it doesn't use regular expressions, so I decided to make a second answer post.
Use your language of choice's split() method equivalent on the string with the word to negate as the argument for what to split on. An example using Python:
>>> text = 'barbarasdbarbar 1234egb ar bar32 sdfbaraadf'
>>> text.split('bar')
['', '', 'asd', '', ' 1234egb ar ', '32 sdf', 'aadf']
The nice thing about doing it this way, in Python at least (I don't remember if the functionality would be the same in, say, Visual Basic or Java), is that it lets you know indirectly when "bar" was repeated in the string due to the fact that the empty strings between "bar"s are included in the list of results (though the empty string at the beginning is due to there being a "bar" at the beginning of the string). If you don't want that, you can simply remove the empty strings from the list.
I had a list of file names, and I wanted to exclude certain ones, with this sort of behavior (Ruby):
files = [
'mydir/states.rb', # don't match these
'countries.rb',
'mydir/states_bkp.rb', # match these
'mydir/city_states.rb'
]
excluded = ['states', 'countries']
# set my_rgx here
result = WankyAPI.filter(files, my_rgx) # I didn't write WankyAPI...
assert result == ['mydir/city_states.rb', 'mydir/states_bkp.rb']
Here's my solution:
excluded_rgx = excluded.map{|e| e+'\.'}.join('|')
my_rgx = /(^|\/)((?!#{excluded_rgx})[^\.\/]*)\.rb$/
My assumptions for this application:
The string to be excluded is at the beginning of the input, or immediately following a slash.
The permitted strings end with .rb.
Permitted filenames don't have a . character before the .rb.
Is there a simple way to ignore the white space in a target string when searching for matches using a regular expression pattern? For example, if my search is for "cats", I would want "c ats" or "ca ts" to match. I can't strip out the whitespace beforehand because I need to find the begin and end index of the match (including any whitespace) in order to highlight that match and any whitespace needs to be there for formatting purposes.
You can stick optional whitespace characters \s* in between every other character in your regex. Although granted, it will get a bit lengthy.
/cats/ -> /c\s*a\s*t\s*s/
While the accepted answer is technically correct, a more practical approach, if possible, is to just strip whitespace out of both the regular expression and the search string.
If you want to search for "my cats", instead of:
myString.match(/m\s*y\s*c\s*a\*st\s*s\s*/g)
Just do:
myString.replace(/\s*/g,"").match(/mycats/g)
Warning: You can't automate this on the regular expression by just replacing all spaces with empty strings because they may occur in a negation or otherwise make your regular expression invalid.
Addressing Steven's comment to Sam Dufel's answer
Thanks, sounds like that's the way to go. But I just realized that I only want the optional whitespace characters if they follow a newline. So for example, "c\n ats" or "ca\n ts" should match. But wouldn't want "c ats" to match if there is no newline. Any ideas on how that might be done?
This should do the trick:
/c(?:\n\s*)?a(?:\n\s*)?t(?:\n\s*)?s/
See this page for all the different variations of 'cats' that this matches.
You can also solve this using conditionals, but they are not supported in the javascript flavor of regex.
You could put \s* inbetween every character in your search string so if you were looking for cat you would use c\s*a\s*t\s*s\s*s
It's long but you could build the string dynamically of course.
You can see it working here: http://www.rubular.com/r/zzWwvppSpE
If you only want to allow spaces, then
\bc *a *t *s\b
should do it. To also allow tabs, use
\bc[ \t]*a[ \t]*t[ \t]*s\b
Remove the \b anchors if you also want to find cats within words like bobcats or catsup.
This approach can be used to automate this
(the following exemplary solution is in python, although obviously it can be ported to any language):
you can strip the whitespace beforehand AND save the positions of non-whitespace characters so you can use them later to find out the matched string boundary positions in the original string like the following:
def regex_search_ignore_space(regex, string):
no_spaces = ''
char_positions = []
for pos, char in enumerate(string):
if re.match(r'\S', char): # upper \S matches non-whitespace chars
no_spaces += char
char_positions.append(pos)
match = re.search(regex, no_spaces)
if not match:
return match
# match.start() and match.end() are indices of start and end
# of the found string in the spaceless string
# (as we have searched in it).
start = char_positions[match.start()] # in the original string
end = char_positions[match.end()] # in the original string
matched_string = string[start:end] # see
# the match WITH spaces is returned.
return matched_string
with_spaces = 'a li on and a cat'
print(regex_search_ignore_space('lion', with_spaces))
# prints 'li on'
If you want to go further you can construct the match object and return it instead, so the use of this helper will be more handy.
And the performance of this function can of course also be optimized, this example is just to show the path to a solution.
The accepted answer will not work if and when you are passing a dynamic value (such as "current value" in an array loop) as the regex test value. You would not be able to input the optional white spaces without getting some really ugly regex.
Konrad Hoffner's solution is therefore better in such cases as it will strip both the regest and test string of whitespace. The test will be conducted as though both have no whitespace.
What's wrong with this expression?
^[a-zA-Z]+(([\''\-][a-zA-Z])?[a-zA-Z]*)*$
I want to allow alpha characters with space,-, and ' characters
for example O'neal;Jackson-Peter, Mary Jane
The following is all you need:
^[a-zA-Z' -]+$
The important thing is that the "-" is the last character in the group, otherwise it'd be interpreted as a range (unless you escaped it with "\")
How you actually input that expression as a string in your target language is different depending on the language. For C#, I usually use "#" strings, like so:
var regex = new Regex(#"^[a-zA-Z' -]+$");
This will match any string made up of at least one character, which can be alpha characters, hyphen or the single quote mark:
^[a-zA-Z-\']+$
This will also include empty strings:
^[a-zA-Z-\']*$
If it needs to begin and end with alpha characters (as names do):
^[a-zA-Z][a-zA-Z-\']*[a-zA-Z]$
Something like this?
^[a-zA-Z '\-,]*$