我想把愚蠢的引号换成聪明的引号。例如，我想把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中的撇号是容易的。具有/('w+)'('w+)/模式的preq_replace将在单词中找到撇号。但我不能正确地交换演讲引用。目前我有:
$singlequotesPattern = "/'(.*)'/";
$singlequotesReplacement = "‘$1’";
$singlequotes = preg_replace($singlequotesPattern, $singlequotesReplacement, $text);
但是这在上面的句子中失败了，并且产生Jane said ‘How do we do this?' 'I don't know’ replied Sam.。它只匹配最外层的单引号。我如何使它替换两对引号?
$singlequotesPattern = "/'(.*?)'/";
添加?使*量词非贪婪。贪心量词查找可能的最长匹配。非贪心的找到最短的
Jane said 'How do we do this?' 'I don’t know' replied Sam.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
贪婪:
Jane said 'How do we do this?' 'I don’t know' replied Sam.
^^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^