I have been sitting for hours to figure out a regExp for a preg_match_all function in php.
My problem is that i whant two different things from the string.
Say you have the string "Code is fun [and good for the brain.] But the [brain is] tired."
What i need from this an array of all the word outside of the brackets and the text in the brackets together as one string.
Something like this
[0] => Code
[1] => is
[2] => fun
[3] => and good for the brain.
[4] => But
[5] => the
[6] => brain is
[7] => tired.
Help much appreciated.
You could try the below regex also,
(?<=\[)[^\]]*|[.\w]+
DEMO
Code:
<?php
$data = "Code is fun [and good for the brain.] But the [brain is] tired.";
$regex = '~(?<=\[)[^\]]*|[.\w]+~';
preg_match_all($regex, $data, $matches);
print_r($matches);
?>
Output:
Array
(
[0] => Array
(
[0] => Code
[1] => is
[2] => fun
[3] => and good for the brain.
[4] => But
[5] => the
[6] => brain is
[7] => tired.
)
)
The first lookbind (?<=\[)[^\]]* matches all the characters which are present inside the braces [] and the second [.\w]+ matches one or more word characters or dot from the remaining string.
You can use the following regex:
(?:\[([\w .!?]+)\]+|(\w+))
The regex contains two alternations: one to match everything inside the two square brackets, and one to capture every other word.
This assumes that the part inside the square brackets doesn't contain any characters other than alphabets, digits, _, !, ., and ?. In case you need to add more punctuation, it should be easy enough to add them to the character class.
If you don't want to be that specific about what should be captured, then you can use a negated character class instead — specify what not to match instead of specifying what to match. The expression then becomes: (?:\[([^\[\]]+)\]|(\w+))
Visualization:
Explanation:
(?: # Begin non-capturing group
\[ # Match a literal '['
( # Start capturing group 1
[\w .!?]+ # Match everything in between '[' and ']'
) # End capturing group 1
\] # Match literal ']'
| # OR
( # Begin capturing group 2
\w+ # Match rest of the words
) # End capturing group 2
) # End non-capturing group
Demo
Related
So, say I have string like:
abcd[efg]hi[jkl]m
What I want is to split the string in such a way that each character is placed in its own index in an array, except for the characters within the brackets. They should remain grouped together. In other words, I want to create an array from the input string as such:
[0] => a
[1] => b
[2] => c
[3] => d
[4] => efg
[5] => h
[6] => i
[7] => jkl
[8] => m
I know I can split the input string in the way shown below using preg_split('/[\[]*[\][]/U', $value, -1):
[0] => abcd
[1] => efg
[2] => hi
[3] => jkl
[4] => m
However, I do not know the regular expression I'd need to use to get me exactly what I want. What regular expression will give me my desired solution?
Effectively, you split on every "optionally occurring" square-braced expression. IOW, if a square braced expression is encountered, split on the whole expression (without consuming it) and if there is no expression encountered, then split on that position in the string.
The pattern can look like either of these: (Regex Demo)
~\[([^\]]+)]|~ # (shorter pattern, 44 steps on the sample input string)
or
~(?:\[([^\]]+)])?~ # (longer pattern, 64 steps on the sample input string)
Both preg flags are essential to ensure that no substrings are lost and no empty elements are generated during the process.
Code: (Demo)
$string = 'abcd[efg]hi[jkl]m';
var_export(
preg_split(
'~\[([^\]]+)]|~',
$string,
0,
PREG_SPLIT_NO_EMPTY|PREG_SPLIT_DELIM_CAPTURE
)
);
Output:
array (
0 => 'a',
1 => 'b',
2 => 'c',
3 => 'd',
4 => 'efg',
5 => 'h',
6 => 'i',
7 => 'jkl',
8 => 'm',
)
Pattern Breakdown:
~ #starting pattern delimiter
\[ #literal opening square brace
( #open capturing group
[^\]]+ #match one or more non-closing square braces
) #close capturing group
] #literal closing square brace
| #OR operator, to allow a match on current position (not a brace expression)
~
The literal braces are written outside of the inner capture group so that they are discarded/consumed during the splitting. Normally, the inner capture group would be consumed as well, but the PREG_SPLIT_DELIM_CAPTURE flag demands that that substring should remain after the splitting.
I'm not good at Regex and I've been trying for hours now so I hope you can help me. I have this text:
✝his is *✝he* *in✝erne✝*
I need to capture (using PREG_OFFSET_CAPTURE) only the ✝ in a word surrounded with *, so I only need to capture the last three ✝ in this example. The output array should look something like this:
[0] => Array
(
[0] => ✝
[1] => 17
)
[1] => Array
(
[0] => ✝
[1] => 32
)
[2] => Array
(
[0] => ✝
[1] => 44
)
I've tried using (✝) but ofcourse this will select all instances including the words without asterisks. Then I've tried \*[^ ]*(✝)[^ ]*\* but this only gives me the last instance in one word. I've tried many other variations but all were wrong.
To clarify: The asterisk can be at all places in the string, but always at the beginning and end of a word. The opening asterisk always precedes a space except at the beginning of the string and the closing asterisk always ends with a space except at the end of the string. I must add that punctuation marks can be inside these asterisks. ✝ is exactly (and only) what I need to capture and can be at any position in a word.
You could make use of the \G anchor to get iterative matches between the *. The anchor matches either at the start of the string, or at the end of the previous match.
(?:\*|\G(?!^))[^&*]*(?>&(?!#)[^&*]*)*\K✝(?=[^*]*\*)
Explanation
(?: Non capture group
\* Match *
| Or
\G(?!^) Assert the end of the previous match, not at the start
) Close non capture group
[^&*]* Match 0+ times any char except & and *
(?> Atomic group
&(?!#) Match & only when not directly followed by #
[^&*]* Match 0+ times any char except & and *
)* Close atomic group and repeat 0+ times
\K Clear the match buffer (forget what is matched until now)
✝ Match literally
(?=[^*]*\*) Positive lookahead, assert a * at the right
Regex demo | Php demo
For example
$re = '/(?:\*|\G(?!^))[^&*]*(?>&(?!#)[^&*]*)*\K✝(?=[^*]*\*)/m';
$str = '✝his is *✝he* *in✝erne✝*';
preg_match_all($re, $str, $matches, PREG_OFFSET_CAPTURE);
print_r($matches[0]);
Output
Array
(
[0] => Array
(
[0] => ✝
[1] => 16
)
[1] => Array
(
[0] => ✝
[1] => 31
)
[2] => Array
(
[0] => ✝
[1] => 43
)
)
Note The the offset is 1 less than the expected as the string starts counting at 0. See PREG_OFFSET_CAPTURE
If you want to match more variations, you could use a non capturing group and list the ones that you would accept to match. If you don't want to cross newline boundaries you can exclude matching those in the negated character class.
(?:\*|\G(?!^))[^&*\r\n]*(?>&(?!#)[^&*\\rn]*)*\K&#(?:x271D|169);(?=[^*\r\n]*\*)
Regex demo
Let's say I'm having the following string:
$string = 'cats[Garfield,Tom,Azrael]';
I need to capture the following strings:
cats
Garfield
Tom
Azrael
That string can be any word-like text, followed by brackets with the list of comma-separated word-like entries. I tried the following:
preg_match('#^(\w+)\[(\w+)(?:,(\w+))*\]$#', $string, $matches);
The problem is that $matches ignores Tom, matching only the first and the last cat.
Now, I know how to do that with more calls, perhaps combining preg_match() and explode(), so the question is not how to do it in general.
The question is: can that be done in single preg_match(), so I could validate and match on one go?
The underlying question seems to be: is it possible to extract each occurrence of a repeated capture group?
The answer is no.
However, several workarounds exists:
The most understandable uses two steps: you capture the full list and then you split it. Something like:
$str = 'cats[Garfield,Tom,Azrael,Supermatou]';
if ( preg_match('~(?<item>\w+)\[(?<list>\w+(?:,\w+)*)]~', $str, $m) )
$result = [ $m['item'], explode(',', $m['list']) ];
(or any structure you want)
An other workaround uses preg_match_all in conjunction with the \G anchor that matches either the start of the string or the position after a successful match:
$pattern = '~(?:\G(?!\A),|(?<item>\w+)\[(?=[\w,]+]))(?<elt>\w+)~';
if ( preg_match_all($pattern, $str, $matches) )
print_r($matches);
This design ensures that all elements are between the brackets.
To obtain a more flat result, you can also write it like this:
$pattern = '~\G(?!\A)[[,]\K\w+|\w+(?=\[[\w,]+])~';
details of this last pattern:
~
# first alternative (can't be the first match)
\G (?!\A) # position after the last successful match
# (the negative lookahead discards the start of the string)
[[,] # an opening bracket or a comma
\K # return the whole match from this position
\w+ # an element
| # OR
# second alternative (the first match)
\w+ # the item
(?= # lookahead to check forward if the format is correct
\[ # opening bracket
[\w,]+ # word characters and comma (feel free to be more descriptive
# like \w+(?:,\w+)* or anything you want)
] # closing bracket
)
~
Why not a simple preg_match_all:
$string = 'cats[Garfield,Tom,Azrael], entity1[child11,child12,child13], entity2:child21&child22&child23';
preg_match_all('#\w+#', $string, $matches);
print_r($matches);
Output:
Array
(
[0] => Array
(
[0] => cats
[1] => Garfield
[2] => Tom
[3] => Azrael
[4] => entity1
[5] => child11
[6] => child12
[7] => child13
[8] => entity2
[9] => child21
[10] => child22
[11] => child23
)
)
having a string like this:
$str = "dateto:'2015-10-07 15:05' xxxx datefrom:'2015-10-09 15:05' yyyy asdf"
the desired result is:
[0] => Array (
[0] => dateto:'2015-10-07 15:05'
[1] => xxxx
[2] => datefrom:'2015-10-09 15:05'
[3] => yyyy
[4] => asdf
)
what I get with:
preg_match_all("/\'(?:[^()]|(?R))+\'|'[^']*'|[^(),\s]+/", $str, $m);
is:
[0] => Array (
[0] => dateto:'2015-10-07
[1] => 15:05'
[2] => xxxx
[3] => datefrom:'2015-10-09
[4] => 15:05'
[5] => yyyy
[6] => asdf
)
Also tried with preg_split("/[\s]+/", $str) but no clue how to escape if value is between quotes. Can anyone show me how and also please explain the regex. Thank you!
I would use PCRE verb (*SKIP)(*F),
preg_split("~'[^']*'(*SKIP)(*F)|\s+~", $str);
DEMO
Often, when you are looking to split a string, using preg_split isn't the best approach (that seems a little counter intuitive, but that's true most of the time). A more efficient way consists to find all items (with preg_match_all) using a pattern that describes all that is not the delimiter (white-spaces here):
$pattern = <<<'EOD'
~(?=\S)[^'"\s]*(?:'[^']*'[^'"\s]*|"[^"]*"[^'"\s]*)*~
EOD;
if (preg_match_all($pattern, $str, $m))
$result = $m[0];
pattern details:
~ # pattern delimiter
(?=\S) # the lookahead assertion only succeeds if there is a non-
# white-space character at the current position.
# (This lookahead is useful for two reasons:
# - it allows the regex engine to quickly find the start of
# the next item without to have to test each branch of the
# following alternation at each position in the strings
# until one succeeds.
# - it ensures that there's at least one non-white-space.
# Without it, the pattern may match an empty string.
# )
[^'"\s]* #"'# all that is not a quote or a white-space
(?: # eventual quoted parts
'[^']*' [^'"\s]* #"# single quotes
|
"[^"]*" [^'"\s]* # double quotes
)*
~
demo
Note that with this a little long pattern, the five items of your example string are found in only 60 steps. You can use this shorter/more simple pattern too:
~(?:[^'"\s]+|'[^']*'|"[^"]*")+~
but it's a little less efficient.
For your example, you can use preg_split with negative lookbehind (?<!\d), i.e.:
<?php
$str = "dateto:'2015-10-07 15:05' xxxx datefrom:'2015-10-09 15:05' yyyy asdf";
$matches = preg_split('/(?<!\d)(\s)/', $str);
print_r($matches);
Output:
Array
(
[0] => dateto:'2015-10-07 15:05'
[1] => xxxx
[2] => datefrom:'2015-10-09 15:05'
[3] => yyyy
[4] => asdf
)
Demo:
http://ideone.com/EP06Nt
Regex Explanation:
(?<!\d)(\s)
Assert that it is impossible to match the regex below with the match ending at this position (negative lookbehind) «(?<!\d)»
Match a single character that is a “digit” «\d»
Match the regex below and capture its match into backreference number 1 «(\s)»
Match a single character that is a “whitespace character” «\s»
I need to split my text into an array at every period, exclamation and question mark.
Example with a full-width period and exclamation mark:
$string = "日本語を勉強しているみんなを応援したいです。一緒に頑張りましょう！";
I am looking for the following output:
Array (
[0] => 日本語を勉強しているみんなを応援したいです。
[1] => 一緒に頑張りましょう！ )
I need the same code to work with half-width.
Example with a mix of full-width and half-width:
$string = "Hi. I am Bob! Nice to meet you. 日本語を勉強しています。Do you understand me?";
Output:
Array (
[0] => Hi.
[1] => I am Bob!
[2] => Nice to meet you.
[3] => 日本語を勉強しています。
[4] => Do you understand me? )
I suck at regular expressions and can't figure out a solution nor find one.
I tried:
$string = preg_split('(.*?[。？！])', $string);
First of all, you forgot your delimiters (most commonly a slash).
You can split on \pP (a unicode punctuation - remember the u modifier meaning unicode):
You can see the rest of the special unicode characters here.
<?php
$str = 'Hi. I am Bob! Nice to meet you. 日本語を勉強しています。Do you understand me?';
$array = preg_split('/(?<=\pP)\s*/u', $str, null, PREG_SPLIT_NO_EMPTY);
print_r($array);
The PREG_SPLIT_NO_EMPTY is there to make sure that we don't include an empty match if your last character is punctuation.
Output:
Array
(
[0] => Hi.
[1] => I am Bob!
[2] => Nice to meet you.
[3] => 日本語を勉強しています。
[4] => Do you understand me?
)
Regex autopsy:
/ - the start delimiter - this must also come at the end before our modifiers
(?<=\pP) - a positive lookbehind matching \pP (a unicode punctuation - we could just use \pP, but then the punctuation would not be included in our final string - a positive lookbehind includes it)
\s* - a white space character matched 0 to infinity times - this is to make sure that we don't include the white space after the punctuation
/u - the end delimiter (/) and our modifier (u meaning "unicode")
DEMO
Your first sentence would result in the following array:
Array
(
[0] => 日本語を勉強しているみんなを応援したいです。
[1] => 一緒に頑張りましょう！
)
Please note that this includes all punctuation including commas.
Array
(
[0] => This is my sentence,
[1] => and it is very nice.
)
This can be fixed by using a negative lookbehind in front of our positive lookbehind:
/(?<![,、;；"”\'’｀`])(?<=\pP)\s*/u