Understanding Regular Expressions - php

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.

Related

Using Multiple Regular Expressions in PHP

I need to write a regular expression that will evaluate the following conditions:
2 consecutive lower case characters
at least 1 digit
at least 1 upper case character
2 consecutive identical punctuation characters
For example, the string 'aa1A!!' should match, as should '!!A1aa'.
I have written the following regular expression:
'/(?=([a-z]){2,})(?=[0-9])(?=[A-Z])(?=(\W)\1)/'
I have found each individual expression works, but I am struggling to put it all together. What am I missing?
First, your pattern must be anchored to be sure that lookaheads are only tested from the position at the start of string. Then, since your characters can be everywhere in the string, you need to start the subpatterns inside lookahead with .*.
\W is a character class for non-word characters (all that is not [A-Za-z0-9_] that includes spaces, control characters, accented letters...). IMO, \pP or [[:punct:]] are more appropriate.
/^(?=.*[a-z]{2})(?=.*[0-9])(?=.*[A-Z])(?=.*(\pP)\1)/
About the idea to make 4 patterns instead of 1, it looks like a good idea, it tastes like a good idea, but it's useless and slower. However, it can be interesting if you want to know what particular rule fails.

(PHP) How to find words beginning with a pattern and replace all of them?

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.

what does the regular expression (?<!-) mean

I'm trying to understand a piece of code and came across this regular expression used in PHP's preg_replace function.
'/(?<!-)color[^{:]*:[^{#]*$/i'
This bit... (?<!-)
doesnt appear in any of my reg-exp manuals. Anyone know what this means please? (Google doesnt return anything - I dont think symbols work in google.)
The ?<! at the start of a parenthetical group is a negative lookbehind. It asserts that the word color (strictly, the c in the engine) was not preceded by a - character.
So, for a more concrete example, it would match color in the strings:
color
+color
someTextColor
But it will fail on something like -color or background-color. Also note that the engine will not technically "match" whatever precedes the c, it simply asserts that it is not a hyphen. This can be an important distinction depending on the context (illustrated on Rubular with a trivial example; note that only the b in the last string is matched, not the preceding letter).
PHP uses perl compatible regular expressions (PCRE) for the preg_* functions. From perldoc perlre:
"(?<!pattern)"
A zero-width negative look-behind assertion. For example
"/(?<!bar)foo/" matches any occurrence of "foo" that does
not follow "bar". Works only for fixed-width look-
behind.
I'm learning regular expressions using Python's re module!
http://docs.python.org/library/re.html
Matches if the current position in the string is not preceded by a match for .... This is called a negative lookbehind assertion. Similar to positive lookbehind assertions, the contained pattern must only match strings of some fixed length. Patterns which start with negative lookbehind assertions may match at the beginning of the string being searched.

What are those characters in a regular expression?

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.

Why does this regex not validate in the same way in PHP?

when I try preg_match with the following expression: /.{0,5}/, it still matches string longer than 5 characters.
It does, however, work properly when trying in online regexp matcher
The site you reference, myregexp.com, is focussed on Java.
Java has a specific function for matching an exact pattern, without needing to use anchor characters. This is the function which myregexp.com uses.
In most other languages, in order to match an exact pattern, you would need to add the anchoring characters ^ and $ at the start and end of the pattern respectively, otherwise the regex assumes it only needs to find the matched pattern somewhere within the string, rather than the whole string being the match.
This means that without the anchors, your pattern will match any string, of any length, because whatever the string, it will contain within it somewhere a match for "zero to five of any character".
So in PHP, and Perl, and virtually any other language, you need your pattern to look like this:
/^.{0,5}$/
Having explained all that, I would make one final observation though: this specific pattern really doesn't need to be a regular expression -- you could achieve the same thing with strlen(). In addition, the dot character in regex may not work exactly as you expect: it typically matches almost any character; some characters, including new line characters, are excluded by default, so if your string contains five characters, but one of them is a new line, it will fail your regex when you might have expected it to pass. With this in mind, strlen() would be a safer option (or mb_strlen() if you expect to have unicode characters).
If you need to match any character in regex, and the default behaviour of the dot isn't good enough, there are two options: One is to add the s modifier at the end of the expression (ie it becomes /^.{0,5}$/s). The s modifier tells regex to include new line characters in the dot "any character" match.
The other option (which is useful for languages that don't support the s modifier) is to use an expression and its negative together in a character class - eg [\s\S] - instead of the dot. \s matches any white space character, and \S is a negative of \s, so any character not matched by \s. So together in a character class they match any character. It's more long winded and less readable than a dot, but in some languages it's the only way to be sure.
You can find out more about this here: http://www.regular-expressions.info/dot.html
Hope that helps.
You need to anchor it with ^$. These symbols match the beginning and end of the string respectively, so it must be 0-5 characters between the beginning and end. Leaving out the anchors will match anywhere in the string so it could be longer.
/^.{0,5}$/
For better readability, I would probably also enclose the . in (), but that's kind of subjective.
/^(.){0,5}$/

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