I found this regex that works correctly but I didn't understand what is # (at the start) and at the end of the expression. Are not ^ and $ the start/end characters?
preg_match_all('#^/([^/]+)/([^/]+)/$#', $s, $matches);
Thanks
The matched pattern contains many /, thus the # is used as regex delimeter. These are identical
/^something$/
and
#^something$#
If you have multiple / in your pattern the 2nd example is better suited to avoid ugly masking with \/. This is how the RE would like like with using the standard // syntax:
/^\/([^\/]+)\/([^\/]+)\/$/
About #:
That's a delimiter of the regular expression itself. It's only meaning is to tell which delimiter is used for the expression. Commonly / is used, but others are possible. PCRE expressions need a delimiter with preg_match or preg_match_all.
About ^:
Inside character classes ([...]), the ^ has the meaning of not if it's the first character.
[abc] : matching a, b or c
[^abc] : NOT matching a, b or c, match every other character instead
Also # at the start and the end here are custom regex delimiters. Instead of the usual /.../ you have #...#. Just like perl.
These are delimiters. You can use any delimiter you want, but they must appear at the start and end of the regular expression.
Please see this documentation for a detail insight in to regular expressions:
http://www.php.net/manual/en/pcre.pattern.php
You can use pretty much anything as delimiters. The most common one is /.../, but if the pattern itself contains / and you don't want to escape any and all occurrences, you can use a different delimiter. My personal preference is (...) because it reminds me that $0 of the result is the entire pattern. But you can do anything, <...>, #...#, %...%, {...}... well, almost anything. I don't know exactly what the requirements are, but I think it's "any non-alphanumeric character".
Let me break it down:
# is the first character, so this is the character used as the delimiter of the regular expression - we know we've got to the end when we reach the next (unescaped) one of these
^ outside of a character class, this means the beginning of the string
/ is just a normal 'slash' character
([^/]+) This is a bracketed expression containing at least one (+) instance of any character that isn't a / (^ at the beginning of a character class inverts the character class - meaning it will only match characters that are not in this list)
/ again
([^/]+) again
/ again
$ this matches the end of the string
# this is the final delimeter, so we know that the regex is now finished.
Related
With perl like regular expression syntax, you are able to make inline comments using the /x modifier and the # character to annotate comments, but what if I'm using PHP and using # as delimiter for styling reasons, any way to make a comment then?
preg_replace("/foo # This is a comment\n/x", "bar","foobar")
works but
preg_replace("#foo # This is a comment\n#x", "bar","foobar")
doesnt work, neither does //, /**/ or any common comment sequence I tried.
In a PHP regex pattern, a delimiter has more "weight" than a pattern part. If you define a delimiter as # you cannot use it as a part of another special construct. So, "#foo # This is a comment\n#x" and "#foo (?# This is a comment\n)#x" won't work as the # signals the end of the pattern space inside the regex.
When you escape a #, it becomes a literal # symbol. The "#foo \\# This is a comment\n#x" will match "foo#Thisisacomment" as once it is escaped, it is matched as a literal symbol.
So, the best advice is available on the "Delimiters" page at php.net:
If the delimiter needs to be matched inside the pattern it must be escaped using a backslash. If the delimiter appears often inside the pattern, it is a good idea to choose another delimiter in order to increase readability.
I'm trying to write a simple regular expression that recognizes a sequence of characters that are not columns or are escaped columns.
I.e:
foo:bar //Does not match
but
foo\:bar //Does match
By my knowledge of Regular Languages, such language can be described by the regular expression
/([^:]|\\[:])*/
You can see a graphical representation of this expression in the wonderful tool Regexper
Using php's preg_match (that is based on the PCRE engine), such expression does not match "foo\:bar".
However, if substitute the class with the single char:
/([^:]|\\:)*/
the expression matches.
Do you have an explanation for this? Is this a sort of limitation of the PCRE engine on character classes?
PS: Testing the first expression on RegExr, that is based on AS3 Regexp engine, does not offer a match, while changing the alternation order:
/(\\[:]|[^:])*/
it does match, while the same expression does not match in PCRE.
preg_match() accepts a regular expression pattern as a string, so you need to double escape everything.
^(?:[^:\\\\]|\\\\:)+$
This matches one or more characters that are not colons or escape characters [^:\\\\], or an escaped colon \\\\:.
Why your first regular expression didn't work: /([^:]|\\[:])*/.
This matches a non-colon [^:], or it matches \\[:] which matches a literal [ followed by a literal : and then a literal ].
Why this works : /([^:]|\\:)*/ ?
This matches a non-colon [^:], or it matches a literal \\: so it effectively matches everything.
Edit: Why /([^:]|E[:])*/ won't match fooE:bar ?
This is what happens: [^:] matches the f then it matches o then the other o then it matches the E, now it finds a colon : and it can't match it, but since by default the PCRE engine doesn't look for the longest possible match it is satisfied with what is has matched so far and stops right there and returns fooE as a match without trying the other alternative E[:] (which is equal by the way to E:) at all.
If you want to match the entire sequence then you will to use an expression like this one:
/([^:E]|E[:])*/
This prevents [^:] from consuming that E.
You can try this. This allow the secuence \\: to have a chance before the negated character class [^:].
^(?:\\:|[^:])+$
If you use the values in the alternation bar inverted as in ^((?:[^:]|\\:)+$ it will not match escaped colon \: because the first alternative will consume the slash (\) before the second expression have a chance to try.
I am tired of being frightened of regular expressions. The topic of this post is limited to PHP implementation of regular expressions, however, any generic regular expression advice would obviously be appreciated (i.e. don't confuse me with scope that is not applicable to PHP).
The following (I believe) will remove any whitespace between numbers. Maybe there is a better way to do so, but I still want to understand what is going on.
$pat="/\b(\d+)\s+(?=\d+\b)/";
$sub="123 345";
$string=preg_replace($pat, "$1", $sub);
Going through the pattern, my interpretation is:
\b A word boundary
\d+ A subpattern of 1 or more digits
\s+ One or more whitespaces
(?=\d+\b) Lookahead assertion of one or more digit followed by a word boundary?
Putting it all together, search for any word boundary followed by one or more digits and then some whitespace, and then do some sort of lookahead assertion on it, and save the results in $1 so it can replace the pattern?
Questions:
Is my above interpretation correct?
What is that lookahead assertion all about?
What is the purpose of the leading / and trailing /?
Is my above interpretation correct?
Yes, your interpretation is correct.
What is that lookahead assertion all about?
That lookahead assertion is a way for you to match characters that have a certain pattern in front of them, without actually having to match the pattern.
So basically, using the regex abcd(?=e) to match the string abcde will give you the match: abcd.
The reason that this matches is that the string abcde does in fact contain:
An a
Followed by a b
Followed by a c
Followed by a d that has an e after it (this is a single character!)
It is important to note that after the 4th item it also contains an actual "e" character, which we didn't match.
On the other hand, trying to match the string against the regex abcd(?=f) will fail, since the sequence:
"a", followed by "b", followed by "c", followed by "d that has an f in front of it"
is not found.
What is the purpose of the leading / and trailing /
Those are delimiters, and are used in PHP to distinguish the pattern part of your string from the modifier part of your string. A delimiter can be any character, although I prefer # signs myself. Remember that the character you are using as a delimiter needs to be escaped if it is used in your pattern.
It would be a good idea to watch this video, and the 4 that follow this:
http://blog.themeforest.net/screencasts/regular-expressions-for-dummies/
The rest of the series is found here:
http://blog.themeforest.net/?s=regex+for+dummies
A colleague sent me the series and after watching them all I was much more comfortable using Regular Expressions.
Another good idea would be installing RegexBuddy or Regexr. Especially RegexBuddy is very useful for understanding the workings of a regular expression.
I'm trying to match all occurances of "string" in something like the following sequence except those inside ##
as87dio u8u u7o #string# ou os8 string os u
i.e. the second occurrence should be matched but not the first
Can anyone give me a solution?
You can use negative lookahead and lookbehind:
(?<!#)string(?!#)
EDIT
NOTE: As per Marks comments below, this would not match #string or string#.
You can try:
(?:[^#])string(?:[^#])
OK,
If you want to NOT match a character you put it in a character class (square brackets) and start it with the ^ character which negates it, for example [^a] means any character but a lowercase 'a'.
So if you want NOT at-sign, followed by string, followed by another NOT at-sign, you want
[^#]string[^#]
Now, the problem is that the character classes will each match a character, so in your example we'd get " string " which includes the leading and trailing whitespace. So, there's another construct that tells you not to match anything, and that is parens with a ?: in the beginning. (?: ). So you surround the ends with that.
(?:[^#])string(?:[^#])
OK, but now it doesn't match at the start of string (which, confusingly, is the ^ character doing double-duty outside a character class) or at the end of string $. So we have to use the OR character | to say "give me a non-at-sign OR start of string" and at the end "give me an non-at-sign OR end of string" like this:
(?:[^#]|^)string(?:[^#]|$)
EDIT: The negative backward and forward lookahead is a simpler (and clever) solution, but not available to all regular expression engines.
Now a follow-up question. If you had the word "astringent" would you still want to match the "string" inside? In other words, does "string" have to be a word by itself? (Despite my initial reaction, this can get pretty complicated :) )
What does the $/i mean in the following php code?
preg_match ('/^[A-Z \'.-]{2,20}$/i')
/ denotes the end of the pattern. The i is a modifier that makes the pattern case-insensitive, and the $ anchor matches the end of the string.
the $ is an anchor -- it means the end of the string should be there. the / is the end delimiter for the regular expression. The i means that the regular expressions should be case-insensitive (notice that [A-Z \'.-] only matches A-Z -- the i means it doesn't have to look for a-z as well).
Dollar sign is a common regex symbol meaning "end of line".
The slash at the end is the end of the expression itself.
Any letters after that slash are options you can turn on or off, called modifiers. In the case of i it means case-insensitive.
$ Matches at the end of the string the regex pattern is applied to. Matches a position rather than a character
/ is the ending delimiter of the regex pattern in PHP
i represents case insensitive regular expression search
you can also use this to understand things better, and can be used for testing/practice too.
http://gskinner.com/RegExr/