I have rather hard time getting my head around regular expression, especially more complex formulas.
Currently I am writing my own markup language and am stumped by escaping. I want each special character to be "escapable", that is if *bold* would give me <b>bold</b>, then \*bold\* should leave it as-is, so I can do the stripping of backslashes later, but I can't think of a regular expression to convey this idea.
How can I select three groups:
Left asterisk if the number or BSes preceding it is even;
Content between asterisks;
Right asterisk if the number of BSes preceding it is even;
with one regular expression? I need it to be compliant with PHP's preg_replace.
This \\*(\*)\S(.)+?\S\\*(\*) would select both asterisks and content as three groups, but that doesn't check for 'evenity' and stuff.
UPDATE:
The second paragraph has been changed to better illustrate what I meant (please don't modify it anymore because the change that was made completely missed the point).
Plus, if that makes things easier, I can first parse any double backslash into some other character, so there is only need to check for ONE backslash before asterisk.
How about:
$rx = '/
([^\\]*|^) # no backslash or beginning of line
\\ # one backslash
\* # an asterisk
([^*\\]+) # one or more characters not being asterisks or BSs
\\ # one backslash
\* # one asterisk
# "mx" = multiline,extended regex
/mx';
preg_replace($rx, '\1\2', $content)
Well, I guess I found answer to my own question.
First I will have to replace each \\, and then use expression like this:
(?<!\\) #There is no backslash before...
\* #...Asterisk
( #Non-whitespace after first and before second asterisk
\S .*? \S
|
\S
)
(?<!\\) #There is no backslash before...
\* #...Asterisk
And from on here I can tweak it however I wish. Thanks for any input to anyone anyway :).
Related
I tried a few flavours of PHP markdown converter for converting *XYZ* into <em> tags, and **ABC** into <strong> tags. They were doing a bit too much for what I needed like adding paragraph tags, etc.
Note that I'm only using two markdown tags.
I wrote a RegExp which works okay, but I needed to escape the reserved characters incase the user wants a literal one of those characters, like I had to in my post.
This is what I have so far:
preg_replace("/(?<!\\\)\*\*([^\*\*]*)(?<!\\\)\*\*/", "<strong>$1</strong>", $line);
For those reading in the future that do not know RegEx too well, (?<!\\\) means don't match the following pattern if it is preceded by a backslash. ([^\*]*) is equivalent to .* but safer in that it says match everything until we get a double asterisk. The parens mean collect this answer so that I can use it as $1 in the next section
It breaks when I do 'My name is **Earle\***'. I would like it to output
My name is <strong>Earle*</strong>
But it outputs
My name is <em></em>Earle<em></em>*
What is wrong with my RegEx, and can you explain what the fixes are so that people in future know
You need to match escaped entities, you cannot use lookarounds for that.
\*\*([^*\\]*(?:\\.[^\\*]*)*)\*\*
See regex demo
Explanation:
\*\* - 2 leading asterisks
([^*\\]*(?:\\.[^\\*]*)*) - Group 1 matching
[^*\\]* - zero or more characters other than * and \
(?:\\.[^\\*]*)* - zero or more sequences of...
\\. - any escape sequence
[^\\*]* - zero or more characters other than * and \
\*\* - 2 trailing asterisks
The regex is based on the unroll-the-loop principle and should be efficient enough to work with any texts.
Also, you can use /s modifier to even support an escaped newline.
Can someone please help me improve this regex so that it captures everything that starts with http://, https://, or www and then continues until it reaches a ' or ". It includes punctuation and is case-insensitive.
Here is the regular expression right now:
(wwww|https?://)
/(?:https?:\/\/|www)[^'"]*/i
I escaped the slashes since they could conflict if you use /.../ notation. [^'"] is an inverted character class that allows everything but quotes.
Edit: I removed the caret to match any occurrence of the pattern, :? to make the group non-capturing.
#(www|https?://).*?(?=['"])#i
The .*? makes the quantifier reluctant so it will stop at the first quote rather than the last.
The following regex will work:
(?:https?:\/\/|www)[^'"]*
You can walk through the details of the match at www.debuggex.com.
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.
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','', $file);
will do this for you. However note it won't allow for any quoted double quotes (e.g. A "test \" quoted string" B will result in A quoted string" B with a leading space, not in A B as you might expect.
Question
What does it mean when a regular expression is surrounded by # symbols? Does that mean something different than being surround by slashes? What about when #x or #i are on the end? Now that I think about it, what do the surrounding slashes even mean?
Background
I saw this StackOverflow answer, posted by John Kugelman, in which he displays serious Regex skills.
Now, I'm used to seeing regexes surrounded by slashes as in
/^abc/
But he used a regex surrounded by # symbols:
'#
^%
(.{2}) # State, 2 chars
([^^]{0,12}.) # City, 13 chars, delimited by ^
([^^]{0,34}.) # Name, 35 chars, delimited by ^
([^^]{0,28}.) # Address, 29 chars, delimited by ^
\?$
#x'
In fact, it seems to be in the format:
#^abc#x
In the process of trying to google what that means (it's a tough question to google!), I also saw the format:
#^abc#i
It's clear the x and the i are not matched characters.
So what does it all mean???
Thanks in advance for any and all responses,
-gMale
The surrounding slashes are just the regex delimiters. You can use any character (afaik) to do that - the most commonly used is the /, other I've seen somewhat commonly used is #
So in other words, #whatever#i is essentially the same as /whatever/i (i is modifier for a case-insensitive match)
The reason you might want to use something else than the / is if your regex contains the character. You avoid having to escape it, similar to using '' for strings instead of "".
Found this from a "Related" link.
The delimiter can be any character that is not alphanumeric, whitespace or a backslash character.
/ is the most commonly used delimiter, since it is closely associated with regex literals, for instance in JavaScript where they are the only valid delimiter. However, any symbol can be used.
I have seen people use ~, #, #, even ! to delimit their regexes in a way that avoids using symbols that are also in the regex. Personally I find this ridiculous.
A lesser-known fact is that you can use a matching pair of brackets to delimit a regex in PHP. This has the tremendous advantage of having an obvious difference between the closing delimiter, and the symbol showing up in the pattern, and therefore don't need any escaping. My personal preference is this:
(^abc)i
By using parentheses, I remind myself that in a match, $m[0] is always the full match, and the subpatterns start at $m[1].