包括$ book-> author in echo""

Question

我正在尝试在echo语句中打印$ book-> author.现在我正在使用连接.

echo "book author: " . $book->author . "<br />";

如何将整个内容包含在双引号中？如果我执行以下操作,它会假设我正在尝试回显$book对象,这->author是一个常规字符串.

echo "book author: $book->author<br />";

Answer 1

用花括号包裹它.

echo "book author: {$book->author}<br />";

顺便说一句,我是以下的倡导者.我发现它读得更好,克服了花括号方法的局限性,并且是对字符串连接的微观,微观优化(参见注释).

echo "book author: ", $book->author, "<br />";