php中字符串的正则表达式，以逗号查找单词

Question

I'm trying to write a Regexp for php to scan true files, the keyword is require and the string I want is in brackets "my string" (require would be reserved) example

File1.txt

require "testing/this/out.js"
require "de/test/as.pen"
require "my_love.test"

.....

print "I require coffee in the morning" //problem

• The requires will be at the top of the page.
• Follow the standard require "_String_" format

---

File2.txt

Class Ben extends Name

....

print "My good boss always extends my deadline" //problem

• Extends would only be follow by class name
• Is only one word and only one result

---

// looping true and determining if class or if reg file by folder structure
$subject = "Code above";
$pattern = '/^require+""/i'; // Not sure of the correct pattern
preg_match($pattern, $subject, $matches);
print_r($matches);

I just want testing this out and de/test/as.pen to return in an array for the first example.

Is this possible? will there be a lot of problems with this?

Answer 1

您可以使用此正则表达式：

$pattern = '/^ *require +"([^"]+)"/i';

Answer 2

^require (['"])([.\w /]+)\1

match the result:

preg_match('#^require (['"])([.\w /]+)\1#', $code, $match);

Explanation :

^               #  Start of string
require         #  reserved word with an space after
(['"])          #  Quotations
(               #  Capturing group
    [.\w /]+    #   Any possible characters
)               #  End of capturing group
\1              #  Same quotation

Demo

Answer 3

The idea is to skip content inside quotes:

$pattern = <<<'LOD'
~
(["']) (?> [^"'\\]++ | \\{2} | \\. | (?!\1)["'] )* \1 # content inside quotes
(*SKIP)(*FAIL)  # forces this possibility to fail
|
(?>^|\s)
require \s+ " ([^"]++) "
~xs
LOD;

preg_match_all($pattern, $subject, $matches, PREG_SET_ORDER);
print_r($matches);