用PREG_REPLACE替换智能引号的哑引号

我正在尝试将愚蠢的报价交换为它们的智能对等报价。例如，我想将Jane said 'How do we do this?' 'I don't know' replied Sam.变成Jane said ‘How do we do this?’ ‘I don’t know’ replied Sam.。

don't中的撇号很简单。模式为/(w+)'(w+)/的preq_replace将在单词中找到撇号。但是我不能很好地调换演讲词的引语。当前我有：

$singlequotesPattern = "/'(.*)'/";
$singlequotesReplacement = "‘$1’";

$singlequotes = preg_replace($singlequotesPattern, $singlequotesReplacement, $text);

但是上面的语句失败了，并产生Jane said ‘How do we do this?' 'I don't know’ replied Sam.它只匹配最外面的单引号。如何使其替换这两对引号？

---

解决方案

$singlequotesPattern = "/'(.*?)'/";

添加?以使*量词非贪婪。贪婪的量词会找到可能的最长匹配。不贪婪的人会找到最短的人。

贪婪：

Jane said 'How do we do this?' 'I don’t know' replied Sam.
           ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

非贪婪：

Jane said 'How do we do this?' 'I don’t know' replied Sam.
           ^^^^^^^^^^^^^^^^^^   ^^^^^^^^^^^^