I'm writing some PHP to convert BBcode to HTML.
I would like to convert this BBcode:
[quote]
Hello World
[/quote]
to the following:
<blockquote>Hello World</blockquote>
The preg_replace function that I'm using to perform this is:
preg_replace("/\[quote\](.+?)\[\/quote\]/s", "<blockquote>\\1</blockquote>", $bbCode);
This almost does everything I need it to, but my problem is that it carries through the \n's from before and after 'Hello World', and produces:
<blockquote>
Hello World
</blockquote>
Any ideas how I could fix this? All help very much appreciated.
Try this regular expression:
/\[quote\]\s*(.+?)\s*\[\/quote\]/s
You need to escape backslashes inside of double-quotes. Instead of "\[", you need "\\[".
A possibility would be to use the 'e' regex-modifier, to call, for instance, the trim function on the string.
Quoting that page of the manual :
e (PREG_REPLACE_EVAL) If this modifier is set, preg_replace()
does normal substitution of
backreferences in the replacement
string, evaluates it as PHP code, and
uses the result for replacing the
search string. Single quotes, double
quotes, backslashes (\) and NULL
chars will be escaped by backslashes
in substituted backreferences.
Only preg_replace() uses this
modifier; it is ignored by other PCRE
functions.
For instance, this code, only slightly different from yours :
$bbCode = <<<STR
[quote]
Hello World
[/quote]
STR;
$output = preg_replace("/\[quote\](.+?)\[\/quote\]/es", "'<blockquote>' . trim('\\1') . '</blockquote>'", $bbCode);
var_dump($output);
Would give you :
string '<blockquote>Hello World</blockquote>' (length=36)
ie, the trim function is called on what was matched -- note it will remove all white-spaces at the beginning and end of your string ; not only newlines, but also spaces and tabulations.
(For instance, you can take a look at Example #4 on the manual page of preg_replace)
(It's maybe a bit overkill in this case, should I add -- but it's nice to know anyway)
Related
I have a multi-line string shown below -
?a="text1
?bc="text23
I need to identify a pattern like using below regex
'/[?][a-z^A-Z]+[=]["]/'
and replace my string by just remove the double quote (") in it, expected output is shown below
?a=text1
?b=text23
Please help in solving the above issue using php.
Capture everything except the quote in a capture group () and replace:
$string = preg_replace('/([?][a-z^A-Z]+[=])["]/', '$1', $string);
But you really don't need all those character classes []:
/(\?[a-z^A-Z]+=)"/
I will give another solution because i see the php tag also. So let's say you have these:
$a='"text1';
$b='"text2';
if i echo them i get
"text1
"text2
in order to get rid of double quote there is a function trim in php that you can use like that:
echo trim($a,'"');
echo trim($b,'"');
the results will be
text1
text2
I dont think you need regex in this occasion as long as you use php. Php can take care of those small things without bother with complex regex expressions.
Lately I've been studying (more in practice to tell the truth) regex, and I'm noticing his power. This demand made by me (link), I am aware of 'backreference'. I think I understand how it works, it works in JavaScript, while in PHP not.
For example I have this string:
[b]Text B[/b]
[i]Text I[/i]
[u]Text U[/u]
[s]Text S[/s]
And use the following regex:
\[(b|i|u|s)\]\s*(.*?)\s*\[\/\1\]
This testing it on regex101.com works, the same for JavaScript, but does not work with PHP.
Example of preg_replace (not working):
echo preg_replace(
"/\[(b|i|u|s)\]\s*(.*?)\s*\[\/\1\]/i",
"<$1>$2</$1>",
"[b]Text[/b]"
);
While this way works:
echo preg_replace(
"/\[(b|i|u|s)\]\s*(.*?)\s*\[\/(b|i|u|s)\]/i",
"<$1>$2</$1>",
"[b]Text[/b]"
);
I can not understand where I'm wrong, thanks to everyone who helps me.
It is because you use a double quoted string, inside a double quoted string \1 is read as the octal notation of a character (the control character SOH = start of heading), not as an escaped 1.
So two ways:
use single quoted string:
'/\[(b|i|u|s)\]\s*(.*?)\s*\[\/\1\]/i'
or escape the backslash to obtain a literal backslash (for the string, not for the pattern):
"/\[(b|i|u|s)\]\s*(.*?)\s*\[\/\\1\]/i"
As an aside, you can write your pattern like this:
$pattern = '~\[([bius])]\s*(.*?)\s*\[/\1]~i';
// with oniguruma notation
$pattern = '~\[([bius])]\s*(.*?)\s*\[/\g{1}]~i';
// oniguruma too but relative:
// (the second group on the left from the current position)
$pattern = '~\[([bius])]\s*(.*?)\s*\[/\g{-2}]~i';
Lately I've been studying (more in practice to tell the truth) regex, and I'm noticing his power. This demand made by me (link), I am aware of 'backreference'. I think I understand how it works, it works in JavaScript, while in PHP not.
For example I have this string:
[b]Text B[/b]
[i]Text I[/i]
[u]Text U[/u]
[s]Text S[/s]
And use the following regex:
\[(b|i|u|s)\]\s*(.*?)\s*\[\/\1\]
This testing it on regex101.com works, the same for JavaScript, but does not work with PHP.
Example of preg_replace (not working):
echo preg_replace(
"/\[(b|i|u|s)\]\s*(.*?)\s*\[\/\1\]/i",
"<$1>$2</$1>",
"[b]Text[/b]"
);
While this way works:
echo preg_replace(
"/\[(b|i|u|s)\]\s*(.*?)\s*\[\/(b|i|u|s)\]/i",
"<$1>$2</$1>",
"[b]Text[/b]"
);
I can not understand where I'm wrong, thanks to everyone who helps me.
It is because you use a double quoted string, inside a double quoted string \1 is read as the octal notation of a character (the control character SOH = start of heading), not as an escaped 1.
So two ways:
use single quoted string:
'/\[(b|i|u|s)\]\s*(.*?)\s*\[\/\1\]/i'
or escape the backslash to obtain a literal backslash (for the string, not for the pattern):
"/\[(b|i|u|s)\]\s*(.*?)\s*\[\/\\1\]/i"
As an aside, you can write your pattern like this:
$pattern = '~\[([bius])]\s*(.*?)\s*\[/\1]~i';
// with oniguruma notation
$pattern = '~\[([bius])]\s*(.*?)\s*\[/\g{1}]~i';
// oniguruma too but relative:
// (the second group on the left from the current position)
$pattern = '~\[([bius])]\s*(.*?)\s*\[/\g{-2}]~i';
I would like to replace extra spaces (instances of consecutive whitespace characters) with one space, as long as those extra spaces are not in double or single quotes (or any other enclosures I may want to include).
I saw some similar questions, but I could not find a direct response to my needs above. Thank you!
Hope you're still looking, or come back to check! This seems to work for me:
'/\s+((["\']).*?(?=\2)\2)|\s\s+/'
...and replace with $1
EDIT
Also, if you need to allow for escaped quotes like \" or \', you could use this expression:
'/\s+((["\'])(\\\\\2|(?!\2).)*?(?=\2)\2)|\s\s+/'
It gets a bit stickier if you want to add support for "balanced" quotes like brackets (e.g. () or {})
END EDIT
Let me know if you find problems or would like some explanation!
HOPEFULLY FINAL EDIT AND WARNINGS
Potential problem: If a quoted string starts at the beginning of the string variable (or file), it will either not count as a quoted string (and have any whitespace reduced) or it will throw off the whole thing, making anything NOT in quotes get treated as though it was in quotes and vice versa -
A potential change that might remedy this is to use the following match expression
/(?:^|\s+)((["\'])(\\\\\2|(?!\2).)*?(?=\2)\2)|\s\s+/
this replaces \s+ with (?:^|\s+) at the beginning of the expression
this will add a space at the beginning of the variable if the string starts with a quote - just trim() or remove that whitespace to continue
I seem to have used the "line by line" approach (like sed, if I'm not mistaken) to reach my original results - if you use the "whole file" or "whole string" setting or approach, carriage-return-line-feed seems to count as two whitespace characters (can't imagine why...), thus turning any newlines into single spaces (unless they are inside quotes and "dot-matches-newline" is used, of course)
this could be resolved by replacing the . and \s shorthand character classes with the specific characters you want to match, like the following:
/(?:^|[ \t]+)((["\'])(\\\\\2|(?!\2)[\s\S])*?(?=\2)\2)|[ \t]{2,}/
this does not require the dot-matches-newline switch and only replaces multiple spaces or tabs - not newlines - with a single space (and of course, only if they are not quoted)
EXAMPLE
This link shows an example of the first expression and last expression in use on sample text on http://codepad.viper-7.com
You could do it in several steps. Consider the following example:
$str = 'This is a string with "Bunch of extra spaces". Leave them "untouched !".';
$id = 0;
$buffer = array();
$str = preg_replace_callback('|".*?"|', function($m) use (&$id, &$buffer) {
$buffer[] = $m[0];
return '__' . $id++;
}, $str);
$str = preg_replace('|\s+|', ' ', $str);
$str = preg_replace_callback('|__(\d+)|', function($m) use ($buffer) {
return $buffer[$m[1]];
}, $str);
echo $str;
This will output the string:
This is a string with "Bunch of extra spaces". Leave them "untouched !".
Although this is is not the prettiest solution.
Given a literal string such as:
Hello\n\n\n\n\n\n\n\n\n\n\n\nWorld
I would like to reduce the repeated \n's to a single \n.
I'm using PHP, and been playing around with a bunch of different regex patterns. So here's a simple example of the code:
$testRegex = '/(\\n){2,}/';
$test = 'Hello\n\n\n\n\n\n\n\n\nWorld';
$test2 = preg_replace($testRegex ,'\n',$test);
echo "<hr/>test regex<hr/>".$test2;
I'm new to PHP, not that new to regex, but it seems '\n' conforms to special rules. I'm still trying to nail those down.
Edit: I've placed the literal code I have in my php file here, if I do str_replace() I can get good things to happen, but that's not a complete solution obviously.
To match a literal \n with regex, your string literal needs four backslashes to produce a string with two backlashes that’s interpreted by the regex engine as an escape for one backslash.
$testRegex = '/(\\\\n){2,}/';
$test = 'Hello\n\n\n\n\n\n\n\n\n\n\n\nWorld';
$test2 = preg_replace($testRegex, '\n', $test);
Perhaps you need to double up the escape in the regular expression?
$pattern = "/\\n+/"
$awesome_string = preg_replace($pattern, "\n", $string);
Edit: Just read your comment on the accepted answer. Doesn't apply, but is still useful.
If you're intending on expanding this logic to include other forms of white-space too:
$output = echo preg_replace('%(\s)*%', '$1', $input);
Reduces all repeated white-space characters to single instances of the matched white-space character.
it indeed conforms to special rules, and you need to add the "multiline"-modifier, m. So your pattern would look like
$pattern = '/(\n)+/m'
which should provide you with the matches. See the doc for all modifiers and their detailed meaning.
Since you're trying to reduce all newlines to one, the pattern above should work with the rest of your code. Good luck!
Try this regular expression:
/[\n]*/