I'm diving into regex.
$subject = "abcdef";
$pattern = '/^def/';
preg_match($pattern, substr($subject,3), $matches, PREG_OFFSET_CAPTURE);
print_r($matches);
I see that forward slashes are used to mark the beginning and end of a regular expression. I'm trying to find out why that is the case. I can't find any documentation about these forward slashes.
The first and most important thing to know about the preg_...() functions is that they expect one character delimiter on each side of the pattern. For the delimiter, you can choose any character apart from backslashes and spaces. The forward slash is the most used delimiter.
And, everything is stated pretty clear in the documentation referencing delimiters.
For conventional/traditional reasons. Often we'd have three parts to the regex, the pattern, the substitution, and the modifiers. For example, the substitution command:
s/Hello/G'day/g
has three components: the Hello which is the pattern - we want to find "Hello", the G'day which is the substitution - we want to replace "Hello" with "G'day", and g which are the modifiers - in this case, g means to do it as many times as possible and not just once.
With some languages, they have the components as separate arguments rather than being delimited using slashes.
Often if you see it without any slashes, it means it's the pattern. If for some reason a language surrounds a pattern with slashes, you know it means it's the pattern.
Hope this helps. :)
Related
I'm diving into regex.
$subject = "abcdef";
$pattern = '/^def/';
preg_match($pattern, substr($subject,3), $matches, PREG_OFFSET_CAPTURE);
print_r($matches);
I see that forward slashes are used to mark the beginning and end of a regular expression. I'm trying to find out why that is the case. I can't find any documentation about these forward slashes.
The first and most important thing to know about the preg_...() functions is that they expect one character delimiter on each side of the pattern. For the delimiter, you can choose any character apart from backslashes and spaces. The forward slash is the most used delimiter.
And, everything is stated pretty clear in the documentation referencing delimiters.
For conventional/traditional reasons. Often we'd have three parts to the regex, the pattern, the substitution, and the modifiers. For example, the substitution command:
s/Hello/G'day/g
has three components: the Hello which is the pattern - we want to find "Hello", the G'day which is the substitution - we want to replace "Hello" with "G'day", and g which are the modifiers - in this case, g means to do it as many times as possible and not just once.
With some languages, they have the components as separate arguments rather than being delimited using slashes.
Often if you see it without any slashes, it means it's the pattern. If for some reason a language surrounds a pattern with slashes, you know it means it's the pattern.
Hope this helps. :)
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 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.
There are quite a few questions on removing multiple slashes using regex in PHP. However, I have a special case I would like to exclude.
I have a full URL as my input: http://localhost/path/to/whatever
I have written to regex to convert backslashes to forward slashes, and then remove multiple consecutive slashes:
$cleaned = preg_replace('/(\\\+)|(\/+)/', "/", trim($input));
This works fine for the most part, however I need to be able to exclude the :// case, otherwise using that expression will result in which is not the intended result:
http:/localhost/path/to/whatever
I have tried using /(\\\+)|^[:](\/+)/, but this doesn't seem to work.
How can I exclude the :// case in my expression?
$cleaned = preg_replace('~(?<!https:|http:)[/\\\\]+~', "/", trim($input));
The subexpression inside the lookbehind can't use quantifiers, so the obvious approach - (?<!https?:) - won't work. But it can be made up of two or more fixed-length alternatives with different lengths. For example:
(?<!https:|http:) # OK
Be aware that the alternation has to be at the top level of the lookbehind, so this won't work:
(?<!(https:|http:)) # error
There is something called "negative look behind" (also available in positive or look ahead)
http://www.phpro.org/tutorials/Introduction-to-PHP-Regex.html
With this you could add an exception by something like
(?<=^https?:)
Then your expression will only match in places NOT preceded by "http:"
Simply a negative look-behind for a colon, preceding two or more forward or backward slashes:
$cleaned = preg_replace('/(?<!:)(?:\\/|\\\\){2,}/', "/", trim($input));
I have tried to work this out myself (even bought a Kindle book!), but I am struggling with backreferences in php.
What I want is like the following example:
var $html = "hello %world|/worldlink/% again";
output:
hello world again
I tried stuff like:
preg_replace('/%([a-z]+)|([a-z]+)%/', '\1', $html);
but with no joy.
Any ideas please? I am sure someone will post the exact answer but I would like an explanation as well please - so that I don't have to keep asking these questions :)
The slashes "/" are not included in your allowed range [a-z]. Instead use
preg_replace('/%([a-z]+)\|([a-z\/]+)%/', '\1', $html);
Your expression:
'/%([a-z]+)|([a-z]+)%/'
Is only capturing one thing. The | in the middle means "OR". You're trying to capture both, so you don't need an OR in there. You want a literal | symbol so you need to escape it:
'/%([a-z]+)\|([a-z\/]+)%/'
The / character also needs to be included in your char set, and escaped as above.
Your regex (/%([a-z]+)|([a-z]+)%/) reads this way:
Match % followed by + (= one or
more) a-z characters (and store this
into backreference #1).
Or (the |):
Match + (= one or more) a-z
characters (and store this into
backreference #2) followed by a
%.
What you are looking for is:
preg_replace('~%([a-z]+)[|]([a-z/]+)%~', '$1', $html);
Basically I just escaped the | regex meta character (you can do this by either surrounding it with [] like I did or just prepending a backwards slash \, personally I find the former easier to read), and added a / to the second capture group.
I also changed your delimiters from / to ~ because tildes are much more unlikely to appear in strings, if you want to keep using / as your delimiter you also have to escape their occurrences in your regex.
It's also recommended that you use the $ syntax instead of \ in your replacement backreferences:
$replacement may contain references
of the form \\n or (since PHP 4.0.4)
$n, with the latter form being the
preferred one.
Here is a version that works according to the OPs data/information provided (using a non-slash delimiter to avoid escaping slashes):
preg_replace('#%([a-z]+)\|([a-z/]+)%#', '\1', $html);
Using a non slash delimiter, would alleviate the need to escape slashes.
Outputs:
hello world again
The Explanation
Why yours did not work. First up the | is an OR operator, and, in your example, should be escaped. Second up, since you are using /'s or expect slashes it is better to use a non-slash delimiter, such as #. Third up, the slash needed to be added to list of allowed matches. As stated before you may want to include a bit more options, as any type of word with numbers underscores periods hyphens will fail / break the script. Hopefully that is the explanation you were looking for.
Here's what works for me:
preg_replace('/%([a-z]+)\|([a-z\/]+)%/', '\1', $html);
Your regular expression doesn't escape the |, and doesn't include the proper characters for the URL.
Here's a basic live example supporting only a-z and slashes:
preg_replace('/%([a-z]+)\|([a-z\/]+)%/', '\1', $html);
In reality, you're going to want to change those [a-z]+ blocks to something more expressive. Do some searches for URL-matching regular expressions, and pick one that fits what you want.
$html = "hello %world|/worldlink/% again";
echo preg_replace('/([A-ZA-z_ ]*)%(.+)\|(.+)%([A-ZA-z_ ]*)/', '$1$2$4', $html);
output:
hello world again
here is a working code : http://www.ideone.com/0qhZ8