查找开始和一个字符串

Question

内结束引号我有一个文本字符串如下：

The nice man said "How's things today" and I replied "All is good thanks". 

我想换成双引号fontawesome图标。如果我做一个字符串替换双qoutes它工作正常。但正如预期的那样，这只是给了我一个字符串中每个双引号的通用左手引号。

我当前的代码：

$str = The nice man said "How's things today" and I replied "All is good thanks". 

$str = str_replace('"', '<i class="fa fa-quote-left" aria-hidden="true"></i>'); 

输出：

The nice man said <i class="fa fa-quote-left" aria-hidden="true"></i>How's things today<i class="fa fa-quote-left" aria-hidden="true"></i> and I replied <i class="fa fa-quote-left" aria-hidden="true"></i>All is good thanks<i class="fa fa-quote-left" aria-hidden="true"></i>. 

所需的输出：

The nice man said <i class="fa fa-quote-left" aria-hidden="true"></i>How's things today<i class="fa fa-quote-right" aria-hidden="true"></i> and I replied <i class="fa fa-quote-left" aria-hidden="true"></i>All is good thanks<i class="fa fa-quote-right" aria-hidden="true"></i>. 

---

示例输出： http://jsfiddle.net/JfGVE/1975/

Answer 1

如果你确信双引号那里一样是我们目标字符串中看到（平衡，里面没有逃脱双引号），然后一个正则表达式将做的工作：

echo preg_replace('~(")([^"]+)(")~', '<i class="fa fa-quote-left" aria-hidden="true"></i>\\2<i class="fa fa-quote-right" aria-hidden="true"></i>', $str); 

Live demo