I want to delete every ! character from a string that is not immediately preceded by a word. To accomplish this task, I was thinking about preg_replace() to perform a Regex match.
That is, I'd like the following blasphemy of a text:
search! query ! !key!words that! acc!ept exclamation! marks!
... to become:
search! query keywords that! accept exclamation! marks!
There is no need to take double+ occurrences into account, since I filter those out using (![!]+) - although if someone knows of a solution that takes double+ occurrences into consideration, I'd be more than glad to welcome it, since it removes the need for an extra lookup.
So far I have (!\b)|(\s+!\s+)|(!\s+!) which - besides being a bit whacky in my opinion - works almost perfectly, but sometimes removes spacing between words, producing the result of
search! querykeywords that! accept exclamation! marks!
EDIT
I need to take accented and/or uppercase characters into consideration when parsing the string.
You want to remove an ! when
there's no word break before it (as in foo !)
or there is a word break after it (as in !foo)
That gives:
\B!|!\b
https://regex101.com/r/xF7bG6/1
([^a-z])\!+|\!+([a-z]), with a replacement of $1$2 should match multiple !'s that are not preceded by a letter (\W) or have a letter immediately after (\w).
If your regular expression language takes positive lookaheads/lookbehinds, then you can use (?<=[^a-z])\!+|\!+(?=[a-z]) with no replacement string.
Related
I have a string. An example might be "Contact /u/someone on reddit, or visit /r/subreddit or /r/subreddit2"
I want to replace any instance of "/r/x" and "/u/x" with "[/r/x](http://reddit.com/r/x)" and "[/u/x](http://reddit.com/u/x)" basically.
So I'm not sure how to 1) find "/r/" and then expand that to the rest of the word (until there's a space), then 2) take that full "/r/x" and replace with my pattern, and most importantly 3) do this for all "/r/" and "/u/" matches in a single go...
The only way I know to do this would be to write a function to walk the string, character by character, until I found "/", then look for "r" and "/" to follow; then keep going until I found a space. That would give me the beginning and ending characters, so I could do a string replacement; then calculate the new end point, and continue walking the string.
This feels... dumb. I have a feeling there's a relatively simple way to do this, and I just don't know how to google to get all the relevant parts.
A simple preg_replace will do what you want.
Try:
$string = preg_replace('#(/(?:u|r)/[a-zA-Z0-9_-]+)#', '[\1](http://reddit.com\1)', $string);
Here is an example: http://ideone.com/dvz2zB
You should see if you can discover what characters are valid in a Reddit name or in a Reddit username and modify the [a-zA-Z0-9_-] charset accordingly.
You are looking for a regular expression.
A basic pattern starts out as a fixed string. /u/ or /r/ which would match those exactly. This can be simplified to match one or another with /(?:u|r)/ which would match the same as those two patterns. Next you would want to match everything from that point up to a space. You would use a negative character group [^ ] which will match any character that is not a space, and apply a modifier, *, to match as many characters as possible that match that group. /(?:u|r)/[^ ]*
You can take that pattern further and add a lookbehind, (?<= ) to ensure your match is preceded by a space so you're not matching a partial which results in (?<= )/(?:u|r)/[^ ]*. You wrap all of that to make a capturing group ((?<= )/(?:u|r)/[^ ]*). This will capture the contents within the parenthesis to allow for a replacement pattern. You can express your chosen replacement using the \1 reference to the first captured group as [\1](http://reddit.com\1).
In php you would pass the matching pattern, replacement pattern, and subject string to the preg_replace function.
In my opinion regex would be an overkill for such a simple operation. If you just want to replace instance of "/r/x" with "[r/x](http://reddit.com/r/x)" and "/u/x" with "[/u/x](http://reddit.com/u/x)" you should use str_replace although with preg_replace it'll lessen the code.
str_replace("/r/x","[/r/x](http://reddit.com/r/x)","whatever_string");
use regex for intricate search string and replace. you can also use http://www.jslab.dk/tools.regex.php regular expression generator if you have something complex to capture in the string.
I'm trying to formulate a regular expression that will allow me to find a string within a piece of text, if the string exists on its own i.e. not within another word (but surrounded by special characters is ok).
/\bword\b/i
The above regex works fine, and finds "word" in the text. The problem comes when the word I want to find is something like "c++". In this case it matches on any occurrence of the "c" character on it's own. I've tried escaping the "+" characters but it doesn't make any difference. I'm assuming because "+" is a non-word character, I'm possibly going down the wrong route and using word boundaries is not what I should be doing.
So I guess the question is, how can I use a regular expression to find a string in a piece of text, on it's own, and regardless of whether the string is alphanumeric or contains special characters. So in the following piece of text it should match on the 3 occurences of "c++":
c++
(c++)
perl/c++/assembly
But it should not match on the following:
maniac++
c++abc
This is intended so that my script can tell if a specific skill exists within a user's CV/resume. I'm using this with PHP's preg_match_all() function.
I've done a lot of searching but can't come up with a solution, hopefully someone with good regex knowledge can help.
Try this:
/(?<!\w)(c\+\+)(?!\w)/
The (?<!\w) is a negative lookbehind clause, meaning that a word character should not immediately precede your pattern. The (?!\w) part is negative lookahead, meaning that a word character should not immediately follow.
Hope this helps!
I need a regex which can basically check for space, line break etc after string.
So conditions are,
Allow special characters ., _, -, + inside the string i.e.#hello.world, #hello_world, #helloworld, etc.
Discard anything including special characters where there is no alpha-numeric string after them i.e. #helloworld.<space>, #helloworld-<space>, #helloworld.?, etc. must be parsed as #helloworld
My existing RegEx is /#([A-Za-z0-9+_.-]+)/ which works perfectly Condition #1, but still there seems to be a problem Condition #2
I am using above RegEx in preg_replace()
Solution:
$str = preg_replace('##[\w+.\-]+\b#', '[[$0]]', $str);
This works perfectly.
Tested with
http://gskinner.com/RegExr/
You can use word boundaries to easily find the position between an alphanumeric letter and a non-alphanumeric letter:
$str = preg_replace('##[\w+.\-]+\b#', '[[$0]]', $str);
Working example: http://ideone.com/0ShCm
Here's an idea:
Use strrev to reverse the string
Use strcspn to find the longest prefix of the reversed string that does not contain any alphanumeric characters
Cut the prefix off with substr
Reverse the string again; this is your final result
See it in action.
I 'm not taking into account any requirement that restricts the legal characters in the string to some subset, but you can use your regular expression for that (or even strspn, which might be faster).
The reason is because it's reading the string as a whole. If you want it to parse out everything after the alphanumeric section you might have to do like and end(explode()); and run that through to make sure that it isn't valid and if it isn't valid then remove it from the equation, but then you'd have to check the end for every possible explode point i.e. .,-,~,etc.
Then again another trap that you might run into is that in the case of a item or anything w/ alphanumeric value it might just parse everything from after the last alphanumeric character on.
Sorry that this isn't much help, but I figured thinking aloud does help.
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 am trying to validate an input in PHP with REGEX. I want to check whether the input has the %s character group inside it and that it appears only once. Otherwise, the rule should fail.
Here's what I've tried:
preg_match('|^[0-9a-zA-Z_-\s:;,\.\?!\(\)\p{L}(%s){1}]*$|u', $value); (there are also some other rules besides this; I've tried the (%s){1} part and it doesn't work).
I believe it is a very easy solution to this, but I'm not really into REGEX's...Thank you for your help!
If I understand your question, you need a positive lookahead. The lookahead causes the expression to only match if it finds a single %s.
preg_match('|^(?=[^%s].*?[%s][^%s]*$)[0-9a-zA-Z_-\s:;,\.\?!\(\)\p{L}(%s){1}]*$|u', $value);
I'll explain how each part works
^(?=[^%s].*?[%s][^%s]*$) is a zero-width assertion -- (?=regex) a positive lookahead -- (meaning it must match, but does not "eat" any characters). It means that the whole line can have only 1 %s.
[0-9a-zA-Z_-\s:;,\.\?!\(\)\p{L}(%s){1}]*$ The remaining part of the regex also looks at the entire string and ensures that the whole string is composed only of the characters in the character class (like your original regex).
I managed to do this with PHP's substr_count() function, following Johnsyweb suggestion to use an alternate way to perform the validation and because the REGEX's suggested seem pretty complicated.
Thank you again!
Alternatively, you can use preg_match_all with your pattern and check the number of matches. If it's 1, then you're ok - something like this:
$result = (preg_match_all('|^[0-9a-zA-Z_-\s:;,\.\?!\(\)\p{L}(%s){1}]*$|u', $value) == 1)
Try this:
'|^(?=(?:(?!%s).)*%s(?:(?!%s).)*$)[0-9_\s:;,.?!()\p{L}-]+$|u'
The (%s){1} sequence inside the square brackets probably doesn't do what you think it does, but never mind, the solution is more complex. In fact, {1} should never appear anywhere in a regex. It doesn't ensure that there's only one of something, as many people assume. As a matter of fact, it doesn't do anything; it's pure clutter.
EDIT (in answer to the comment): To ensure that only one of a particular sequence is present in a string, you have to actively examine every single character, classifying it as either part-of-%s or not part-of-%s. To that end, (?:(?!%s).)* consumes one character at a time, after the negative lookahead has confirmed that the character is not the start of %s.
When that part of the lookahead expression quits matching, the next thing in the string has to be %s. Then the second (?:(?!%s).)*$ kicks in to confirm that there are no more %s sequences until the end of the string.
And don't forget that the lookahead expression must be anchored at both ends. Because the lookahead is the first thing after the main regex's start anchor you don't need to add another ^. But the lookahead must end with its own $ anchor.
If you're not "into" regular expressions, why not solve this with PHP?
One call to the builtin strpos() will tell you if the string has a match. A second call will tell you if it appears more than once.
This will be easier for you to read and for others to maintain.