在引号之间获取字符串

Question

我有一个包含双引号或单引号的字符串.我需要做的是在引用之间回应所有内容:

 $str = "'abc de', xye, jhy, jjou";
 $str2 = "\"abc de\", xye, jhy, jjou";

我不介意使用正则表达式(preg_match),或任何其他内置功能的PHP.

请指教.

问候,

Answer 1

用途preg_match_all:

$str = "'abc de', xye, \"jhy\", blah blah 'bob' \"gfofgok\", jjou";
preg_match_all('/".*?"|\'.*?\'/', $str, $matches);
print_r($matches);

返回:

Array ( 
   [0] => Array ( 
      [0] => 'abc de' 
      [1] => "jhy" 
      [2] => 'bob' 
      [3] => "gfofgok" 
   )
)

正则表达式的解释:

"   -> Match a double quote
.*  -> Match zero or more of any character
?"  -> Match as a non-greedy match until the next double quote
|   -> or
\'  -> Match a single quote
.*  -> Match zero or more of any character
?\' -> Match as non-greedy match until the next single quote.

这$matches[0]是一个包含单引号或双引号内所有字符串的数组.