I have this string :
$content = 'Hello <!--blapi[getinfoprix("prix_p1"=>"1-6048df6;image/purchase-small.png"]--> Hello<!--blapi[prix_p1->description]-->';
How can i get the two string <!--*--> in an array[2]?
I've made this :
$pattern = '/<!--blapi\[(.*)\]-->/sU';
preg_match($pattern, $content, $matches);
But I have this result :
array(2) {
[0]=>
string(74) "<!--blapi[getinfoprix("prix_p1"=>"1-6048df6;image/purchase-small.png")]-->"
[1]=>
string(60) "getinfoprix("prix_p1"=>"1-6048df6;image/purchase-small.png")"
}
I don't understand why it's ignoring the second string <!--blapi[prix_p1->description]-->...
I've used the flag "U". Maybe there is a better pattern for what I want to do?
EDITION :
I expect this result :
Array
(
[0] => getinfoprix("prix_p1"=>"1-6048df6;image/purchase-small.png"]
[1] => prix_p1->description
)
This preg_match_all should work:
$content = 'Hello <!--blapi[getinfoprix("prix_p1"=>"1-6048df6;image/purchase-small.png"]--> Hello<!--blapi[prix_p1->description]-->';
if ( preg_match_all('/<!--.*?\[(.*?)\]-->/', $content, $matches) )
print_r($matches[0]);
OUTPUT
Array
(
[0] => getinfoprix("prix_p1"=>"1-6048df6;image/purchase-small.png"
[1] => prix_p1->description
)
$pattern = '~<!--(blapi\[(?:.*?)\])-->~si';
Does this pattern produce the expected results? I understand you want to capture the blapi part too. But not sure...
Changed .*U to .*? and added i for case-insensitive at the end. The inner blapi is a non-capture group and the blapi[...] is now the capture group.
Also avoid wrapping a regex in / as it conflicts with URLs and HTML. Use ~ as it's seldom used and much safer. It's not nice to escape http:// to http:\/\/ just because of the wrap character.
You also need preg_match_all as preg_match capture only one match. It's mostly used for match testing, single-match search but not multiple match search.
Related
I have tried the non capturing group option ?:
Here is my data:
hello:"abcdefg"},"other stuff
Here is my regex:
/hello:"(.*?)"}/
Here is what it returns:
Array
(
[0] => Array
(
[0] => hello:"abcdefg"}
)
[1] => Array
(
[0] => abcdefg
)
)
I wonder, how can I make it so that [0] => abdefg and that [1] => doesnt exist?
Is there any way to do this? I feel like it would be much cleaner and improve my performance. I understand that regex is simply doing what I told it to do, that is showing me the whole string that it found, and the group inside the string. But how can I make it only return abcdefg, and nothing more? Is this possible to do?
Thanks.
EDIT: I am using the regex on a website that says it uses perl regex. I am not actually using the perl interpreter
EDIT Again: apparently I misread the website. It is indeed using PHP, and it is calling it with this function: preg_match_all('/hello:"(.*?)"}/', 'hello:"abcdefg"},"other stuff', $arr, PREG_PATTERN_ORDER);
I apologize for this error, I fixed the tags.
EDIT Again 2: This is the website http://www.solmetra.com/scripts/regex/index.php
preg_match_all
If you want a different captured string, you need to change your regex. Here I'm looking for anything not a double quote " between two quote " characters behind a : colon character.
<?php
$string = 'hello:"abcdefg"},"other stuff';
$pattern = '!(?<=:")[^"]+(?=")!';
preg_match_all($pattern,$string,$matches);
echo $matches[0][0];
?>
Output
abcdefg
If you were to print_r($matches) you would see that you have the default array and the matches in their own additional arrays. So to access the string you would need to use $matches[0][0] which provides the two keys to access the data. But you're always going to have to deal with arrays when you're using preg_match_all.
Array
(
[0] => Array
(
[0] => abcdefg
)
)
preg_replace
Alternatively, if you were to use preg_replace instead, you could replace all of the contents of the string except for your capture group, and then you wouldn't need to deal with arrays (but you need to know a little more about regex).
<?php
$string = 'hello:"abcdefg"},"other stuff';
$pattern = '!^[^:]+:"([^"]+)".+$!s';
$new_string = preg_replace($pattern,"$1",$string);
echo $new_string;
?>
Output
abcdefg
preg_match_all is returning exactly what is supposed to.
The first element is the entire string that matched the regex. Every other element are the capture groups.
If you just want the the capture group, then just ignore the 1st element.
preg_match_all('/hello:"(.*?)"}/', 'hello:"abcdefg"},"other stuff', $arr, PREG_PATTERN_ORDER);
$firstMatch = $arr[1];
I have a string like this:
page-9000,page-template,page-type,page-category-128,image-195,listing-latest,rss-latest,even-more-info,even-more-tags
I made this regex that I expect to get the whole tags with:
(?<=\,)(rss-latest|listing-latest-no-category|category-128|page-9000)(?=\,)
I want it to match all the ocurrences.
In this case:
page-9000 and rss-latest.
This regex checks whole words between commas just fine but it ignores the first and the last because it's not between commas (obviously).
I've also tried that it checks if it's between commas OR one comma at the beginning OR one comma to the end, however it would give me false positives, as it would match:
category-128
while the string contains:
page-category-128
Any help?
Try using the following pattern:
(?<=,|^)(rss-latest|listing-latest-no-category|category-128|page-9000)(?=,|$)
The only change I have made is to add boundary markers ^ and $ to the lookarounds to also match on the start and end of the input.
Script:
$input = "page-9000,page-template,page-type,page-category-128,image-195,listing-latest,rss-latest,even-more-info,even-more-tags";
preg_match_all("/(?<=,|^)(rss-latest|listing-latest-no-category|category-128|page-9000)(?=,|$)/", $input, $matches);
print_r($matches[1]);
This prints:
Array
(
[0] => page-9000
[1] => rss-latest
)
Here is a non-regex way using explode and array_intersect:
$arr1 = explode(',', 'page-9000,page-template,page-type,page-category-128,image-195,listing-latest,rss-latest,even-more-info,even-more-tags');
$arr2 = explode('|', 'rss-latest|listing-latest-no-category|category-128|page-9000');
print_r(array_intersect($arr1, $arr2));
Output:
Array
(
[0] => page-9000
[6] => rss-latest
)
The (?<=\,) and (?=,) require the presence of , on both sides of the matching pattern. You want to match also at the start/end of string, and this is where you need to either explicitly tell to match either , or start/end of string or use double-negating logic with negated character classes inside negative lookarounds.
You may use
(?<![^,])(?:rss-latest|listing-latest-no-category|category-128|page-9000)(?![^,])
See the regex demo
Here, (?<![^,]) matches the start of string position or a , and (?![^,]) matches the end of string position or ,.
Now, you do not even need a capturing group, you may get rid of its overhead using a non-capturing group, (?:...). preg_match_all won't have to allocate memory for the submatches and the resulting array will be much cleaner.
PHP demo:
$re = '/(?<![^,])(?:rss-latest|listing-latest-no-category|category-128|page-9000)(?![^,])/m';
$str = 'page-9000,page-template,page-type,page-category-128,image-195,listing-latest,rss-latest,even-more-info,even-more-tags';
if (preg_match_all($re, $str, $matches)) {
print_r($matches[0]);
}
// => Array ( [0] => page-9000 [1] => rss-latest )
I have tried the non capturing group option ?:
Here is my data:
hello:"abcdefg"},"other stuff
Here is my regex:
/hello:"(.*?)"}/
Here is what it returns:
Array
(
[0] => Array
(
[0] => hello:"abcdefg"}
)
[1] => Array
(
[0] => abcdefg
)
)
I wonder, how can I make it so that [0] => abdefg and that [1] => doesnt exist?
Is there any way to do this? I feel like it would be much cleaner and improve my performance. I understand that regex is simply doing what I told it to do, that is showing me the whole string that it found, and the group inside the string. But how can I make it only return abcdefg, and nothing more? Is this possible to do?
Thanks.
EDIT: I am using the regex on a website that says it uses perl regex. I am not actually using the perl interpreter
EDIT Again: apparently I misread the website. It is indeed using PHP, and it is calling it with this function: preg_match_all('/hello:"(.*?)"}/', 'hello:"abcdefg"},"other stuff', $arr, PREG_PATTERN_ORDER);
I apologize for this error, I fixed the tags.
EDIT Again 2: This is the website http://www.solmetra.com/scripts/regex/index.php
preg_match_all
If you want a different captured string, you need to change your regex. Here I'm looking for anything not a double quote " between two quote " characters behind a : colon character.
<?php
$string = 'hello:"abcdefg"},"other stuff';
$pattern = '!(?<=:")[^"]+(?=")!';
preg_match_all($pattern,$string,$matches);
echo $matches[0][0];
?>
Output
abcdefg
If you were to print_r($matches) you would see that you have the default array and the matches in their own additional arrays. So to access the string you would need to use $matches[0][0] which provides the two keys to access the data. But you're always going to have to deal with arrays when you're using preg_match_all.
Array
(
[0] => Array
(
[0] => abcdefg
)
)
preg_replace
Alternatively, if you were to use preg_replace instead, you could replace all of the contents of the string except for your capture group, and then you wouldn't need to deal with arrays (but you need to know a little more about regex).
<?php
$string = 'hello:"abcdefg"},"other stuff';
$pattern = '!^[^:]+:"([^"]+)".+$!s';
$new_string = preg_replace($pattern,"$1",$string);
echo $new_string;
?>
Output
abcdefg
preg_match_all is returning exactly what is supposed to.
The first element is the entire string that matched the regex. Every other element are the capture groups.
If you just want the the capture group, then just ignore the 1st element.
preg_match_all('/hello:"(.*?)"}/', 'hello:"abcdefg"},"other stuff', $arr, PREG_PATTERN_ORDER);
$firstMatch = $arr[1];
I have a text like:
This is a {demo} phrase made for {test}
I need to get
demo
test
Note: My text can have more than one block of {}, not always two. Example:
This is a {demo} phrase made for {test} written in {English}
I used this expression /{([^}]*)}/ with preg_match but it returns only the first word, not all words inside the text.
Use preg_match_all instead:
preg_match_all($pattern, $input, $matches);
It's much the same as preg_match, with the following stipulations:
Searches subject for all matches to the regular expression given in
pattern and puts them in matches in the order specified by flags.
After the first match is found, the subsequent searches are continued
on from end of the last match.
Your expression is correct, but you should be using preg_match_all() instead to retrieve all matches. Here's a working example of what that would look like:
$s = 'This is a {demo} phrase made for {test}';
if (preg_match_all('/{([^}]*)}/', $s, $matches)) {
echo join("\n", $matches[1]);
}
To also capture the positions of each match, you can pass PREG_OFFSET_CAPTURE as the fourth parameter to preg_match_all. To use that, you can use the following example:
if (preg_match_all('/{([^}]*)}/', $s, $matches, PREG_OFFSET_CAPTURE)) {
foreach ($matches[1] as $match) {
echo "{$match[0]} occurs at position {$match[1]}\n";
}
}
As the { and } are part of regex matching syntax, you need to escape these characters:
<?php
$text = <<<EOD
this {is} some text {from}
which I {may} want to {extract}
some words {between} brackets.
EOD;
preg_match_all("!\{(\w+)\}!", $text, $matches);
print_r($matches);
?>
produces
Array
(
[0] => Array
(
[0] => {is}
[1] => {from}
[2] => {may}
[3] => {extract}
[4] => {between}
)
... etc ...
)
This example may be helpful to understand the use of curly brackets in regexes:
<?php
$str = 'abc212def3456gh34ij';
preg_match_all("!\d{3,}!", $str, $matches);
print_r($matches);
?>
which returns:
Array
(
[0] => Array
(
[0] => 212
[1] => 3456
)
)
Note that '34' is excluded from the results because the \d{3,} requires a match of at least 3 consecutive digits.
Matching portions between pair of braces using RegEx, is less better than using Stack for this purpose. Using RegEx would be something like «quick and dirty patch», but for parsing and processing input string you have to use a stack. Visit here for the concept and here for applying the same.
I'm still learning a lot about PHP and string alteration is something that is of interest to me. I've used preg_match before for things like validating an email address or just searching for inquiries.
I just came from this post What's wrong in my regular expression? and was curious as to why the preg_match_all function produces 2 strings, 1 w/ some of the characters stripped and then the other w/ the desired output.
From what I understand about the function is that it goes over the string character by character using the RegEx to evaluate what to do with it. Could this RegEx have been structured in such a way as to bypass the first array entry and just produce the desired result?
and so you don't have to go to the other thread
$str = 'text^name1^Jony~text^secondname1^Smith~text^email1^example-
free#wpdevelop.com~';
preg_match_all('/\^([^^]*?)\~/', $str, $newStr);
for($i=0;$i<count($newStr[0]);$i++)
{
echo $newStr[0][$i].'<br>';
}
echo '<br><br><br>';
for($i=0;$i<count($newStr[1]);$i++)
{
echo $newStr[1][$i].'<br>';
}
This will output
^Jony~^Smith~^example-free#wpdevelop.com~JonySmithexample-free#wpdevelop.com
I'm curious if the reason for 2 array entries was due to the original sytax of the string or if it is the normal processing response of the function. Sorry if this shouldn't be here, but I'm really curious as to how this works.
thanks,
Brodie
It's standard behavior for preg_match and preg_match_all - the first string in the "matched values" array is the FULL string that was caught by the regex pattern. The subsequent array values are the 'capture groups', whose existence depends on the placement/position of () pairs in the regex pattern.
In your regex's case, /\^([^^]*?)\~/, the full matching string would be
^ Jony ~
| | |
^ ([^^]*?) ~ -> $newstr[0] = ^Jony~
-> $newstr[1] = Jony (due to the `()` capture group).
Could this RegEx have been structured in such a way as to bypass the first array entry and just produce the desired result?
Absolutely. Use assertions. This regex:
preg_match_all('/(?<=\^)[^^]*?(?=~)/', $str, $newStr);
Results in:
Array
(
[0] => Array
(
[0] => Jony
[1] => Smith
[2] => example-free#wpdevelop.com
)
)
As the manual states, this is the expected result (for the default PREG_PATTERN_ORDER flag). The first entry of $newStr contains all full pattern matches, the next result all matches for the first subpattern (in parentheses) and so on.
The first array in the result of preg_match_all returns the strings that match the whole pattern you passed to the preg_match_all() function, in your case /\^([^^]*?)\~/. Subsequent arrays in the result contain the matches for the parentheses in your pattern. Maybe it is easier to understand with an example:
$string = 'abcdefg';
preg_match_all('/ab(cd)e(fg)/', $string, $matches);
The $matches array will be
array(3) {
[0]=>
array(1) {
[0]=>
string(7) "abcdefg"
}
[1]=>
array(1) {
[0]=>
string(2) "cd"
}
[2]=>
array(1) {
[0]=>
string(2) "fg"
}
}
The first array will contain the match of the entire pattern, in this case 'abcdefg'. The second array will contain the match for the first set of parentheses, in this case 'cd'. The third array will contain the match for the second set of parentheses, in this case 'fg'.
[0] contains entire match, while [1] only a portion (the part you want to extract)...
You can do var_dump($newStr) to see the array structure, you'll figure it out.
$str = 'text^name1^Jony~text^secondname1^Smith~text^email1^example-
free#wpdevelop.com~';
preg_match_all('/\^([^^]*?)\~/', $str, $newStr);
$newStr = $newStr[1];
foreach($newStr as $key => $value)
{
echo $value."\n";
}
This will result in... (weird result, haven't modified expression)
Jony
Smith
example-
free#wpdevelop.com
Whenever you have problems to imagine the function of preg_match_all you should use an evaluator like preg_match_all tester # regextester.net
This shows you the result in realtime and you can configure things like the result order, meta instructions, offset capturing and many more.