I have user input where some tags are allowed inside square brackets. I've already wrote the regex pattern to find and validate what's inside the brackets.
In user input field opening-bracket could ([) be escaped with backslash, also backslash could be escaped with another backslash (\). I need look-behind sub-pattern to avoid odd number of consecutive backslashes before opening-bracket.
At the moment I must deal with something like this:
(?<!\\)(?:\\\\)*\[(?<inside brackets>.*?)]
It works fine, but problem is that this code still matches possible pairs of consecutive backslashes in front of brackets (even they are hidden) and look-behind just checks out if there's another single backslash appended to pairs (or directly to opening-bracket). I need to avoid them all inside look-behind group if possible.
Example:
my [test] string is ok
my \[test] string is wrong
my \\[test] string is ok
my \\\[test] string is wrong
my \\\\[test] string is ok
my \\\\\[test] string is wrong
...
etc
I work with PHP PCRE
Last time I checked, PHP did not support variable-length lookbehinds. That is why you cannot use the trivial solution (?<*\\).
The simplest workaround would be to simply match the entire thing, not just the brackets part:
(?<!\\)((?:\\\\)*)\[(?<inside_brackets>.*?)]
The difference is that now, if you're using that regex in a preg_replace, you gotta remember to prefix the replacement string by $1, to restore the backslashes being there.
You could do it without any look-behinds at all (the (\\\\|[^\\]) alternation eats anything but a single back-slash):
^(\\\\|[^\\])*\[(?<brackets>.*?)\]
Related
I have a document that's full of quotes, so like: "this is a quote". Some of those quotes have subclauses in two hyphens like: "this quote - this one right here - has em dashes", and some just have one hyphen like: "this quote has just one thing - a hyphen".
I'm trying to have some regex that matches all of the quotes with two hyphens, but not match any quotes with zero or one hyphen, and not match any of the text outside of the quotes. Also I should mention that there are some sentences with one or more hyphens that lie outside of quotes, I need to ignore them as well and not have them interfere with my matches in quotes. I want to change the properly matched quotes' double hyphens to proper em dash characters.
I've tried using lookaheads and negated characters, but can't seem to figure this one out.
Is this something regex can do, or do I need to come up with some kind of other approach (like splitting all of the text into an array and stepping through it, making my changes and then recombining it all at the end)? I can do that instead it just seems like a silly waste of time if there's a one-line regex statement that will do what I want.
Add a \b word boundary at the beginning of the quote, and check that the last character inside the quote is either a letter or number or some kind of punctuation.
("\b[^-"]*-[^-"]*-[^-"]*[\w.!?]")
"(?:[^-"]*-){2}[^-"]*" is about the best you can get with only regex, but it doesn't work if there are two hyphens outside of quotes. Splitting the text into an array is probably the best way to do what you want to.
I have a javascript regex
Value.match(/[A-Za-z0-9\-\,\.\(\)/]/)
This gives me 1 if a string contains alphabets, numbers, hyphen, comma, dot or braces; if any other character is found it gives 0.
When I apply same regex in PHP it is not working. Why?
You don't need to escape characters inside [] so you can try this /[A-Za-z0-9,.()]/ or even this one /[\w,.()]/ but if you want to check that the string contains only those characters that regex won't do, try:
/^[\w,.()]+$/
I noticed that you also have /. Is that intentional or a mistake, because you don't mention it in the question...
I am trying to validate a string of 3 numbers followed by / then 5 more numbers
I thought this would work
(/^([0-9]+[0-9]+[0-9]+/[0-9]+[0-9]+[0-9]+[0-9]+[0-9])/i)
but it doesn't, any ideas what i'm doing wrong
Try this
preg_match('#^\d{3}/\d{5}#', $string)
The reason yours is not working is due to the + symbols which match "one or more" of the nominated character or character class.
Also, when using forward-slash delimiters (the characters at the start and end of your expression), you need to escape any forward-slashes in the pattern by prefixing them with a backslash, eg
/foo\/bar/
PHP allows you to use alternate delimiters (as in my answer) which is handy if your expression contains many forward-slashes.
First of all, you're using / as the regexp delimiter, so you can't use it in the pattern without escaping it with a backslash. Otherwise, PHP will think that you're pattern ends at the / in the middle (you can see that even StackOverflow's syntax highlighting thinks so).
Second, the + is "greedy", and will match as many characters as it can, so the first [0-9]+ would match the first 3 numbers in one go, leaving nothing for the next two to match.
Third, there's no need to use i, since you're dealing with numbers which aren't upper- or lowercase, so case-sensitivity is a moot point.
Try this instead
/^\d{3}\/\d{5}$/
The \d is shorthand for writing [0-9], and the {3} and {5} means repeat 3 or 5 times, respectively.
(This pattern is anchored to the start and the end of the string. Your pattern was only anchored to the beginning, and if that was on purpose, the remove the $ from my pattern)
I recently found this site useful for debugging regexes:
http://www.regextester.com/index2.html
It assumes use of /.../ (meaning you should not include those slashes in the regex you paste in).
So, after I put your regex ^([0-9]+[0-9]+[0-9]+/[0-9]+[0-9]+[0-9]+[0-9]+[0-9]) in the Regex box and 123/45678 in the Test box I see no match. When I put a backslash in front of the forward slash in the middle, then it recognizes the match. You can then try matching 1234/567890 and discover it still matches. Then you go through and remove all the plus signs and then it correctly stops matching.
What I particularly like about this particular site is the way it shows the partial matches in red, allowing you to see where your regex is working up to.
I'm having a hard time removing text within double-quotes, especially those spread over multiple lines:
$file=file_get_contents('test.html');
$replaced = preg_replace('/"(\n.)+?"/m','', $file);
I want to remove ALL text within double-quotes (included). Some of the text within them will be spread over multiple lines.
I read that newlines can be \r\n and \n as well.
Try this expression:
"[^"]+"
Also make sure you replace globally (usually with a g flag - my PHP is rusty so check the docs).
Another edit: daalbert's solution is best: a quote followed by one or more non-quotes ending with a quote.
I would make one slight modification if you're parsing HTML: make it 0 or more non-quote characters...so the regex will be:
"[^"]*"
EDIT:
On second thought, here's a better one:
"[\S\s]*?"
This says: "a quote followed by either a non-whitespace character or white-space character any number of times, non-greedily, ending with a quote"
The one below uses capture groups when it isn't necessary...and the use of a wildcard here isn't explicit about showing that wildcard matches everything but the new-line char...so it's more clear to say: "either a non-whitespace char or whitespace char" :) -- not that it makes any difference in the result.
there are many regexes that can solve your problem but here's one:
"(.*?(\s)*?)*?"
this reads as:
find a quote optionally followed by: (any number of characters that are not new-line characters non-greedily, followed by any number of whitespace characters non-greedily), repeated any number of times non-greedily
greedy means it will go to the end of the string and try matching it. if it can't find the match, it goes one from the end and tries to match, and so on. so non-greedy means it will find as little characters as possible to try matching the criteria.
great link on regex: http://www.regular-expressions.info
great link to test regexes: http://regexpal.com/
Remember that your regex may have to change slightly based on what language you're using to search using regex.
You can use single line mode (also know as dotall) and the dot will match even newlines (whatever they are):
/".+?"/s
You are using multiline mode which simply changes the meaning of ^ and $ from beginning/end of string to beginning/end of text. You don't need it here.
"[^"]+"
Something like below. s is dotall mode where . will match even newline:
/".+?"/s
$replaced = preg_replace('/"[^"]*"/s','', $file);
will do this for you. However note it won't allow for any quoted double quotes (e.g. A "test \" quoted string" B will result in A quoted string" B with a leading space, not in A B as you might expect.
I got this issue figuring out how to build a regexp for verifying a netbios name. According to the ms standard these characters are illegal
\/:*?"<>|
So, thats what I'm trying to detect. My regex is looking like this
^[\\\/:\*\?"\<\>\|]$
But, that wont work.
Can anyone point me in the right direction? (not regexlib.com please...)
And if it matters, I'm using php with preg_match.
Thanks
Your regular expression has two problems:
you insist that the match should span the entire string. As Andrzej says, you are only matching strings of length 1.
you are quoting too many characters. In a character class (i.e. []), you only need to quote characters that are special within character classes, i.e. hyphen, square bracket, backslash.
The following call works for me:
preg_match('/[\\/:*?"<>|]/', "foo"); /* gives 0: does not include invalid characters */
preg_match('/[\\/:*?"<>|]/', "f<oo"); /* gives 1: does include invalid characters */
As it stands at the moment, your regex will match the start of the string (^), then exactly one of the characters in the square brackets (i.e. the illegal characters), then then end of the string ($).
So this likely isn't working because a string of length > 1 will trivially fail to match the regex, and thus be considered OK.
You likely don't need the start and end anchors (the ^ and $). If you remove these, then the regex should match one of the bracketed characters occurring anywhere on the input text, which is what you want.
(Depending on the exact regex dialect, you may canonically need less backslashes within the square brackets, but they are unlikely to do any harm in any case).