php中字符串的正則表達式，以逗號查找單詞

Question

我正在嘗試為php寫一個Regexp來掃描真實文件，關鍵字是require ，而我想要的字符串在方括號"my string" （需要保留）中

FILE1.TXT

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

.....

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

• 需求將在頁面頂部。
• 遵循標准require "_String_"格式

---

FILE2.TXT

Class Ben extends Name

....

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

• 擴展名只能跟隨類名
• 只有一個字，只有一個結果

---

// 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);

我只想testing this out然后de/test/as.pen返回第一個示例的數組。

這可能嗎？ 會有很多問題嗎？

Answer 1

您可以使用此正則表達式：

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

Answer 2

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

匹配結果：

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

說明 ：

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

演示

Answer 3

這個想法是跳過引號內的內容：

$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);