以下是我的PHP代码:
$heads = array();
$heads[0] = "<p class='ourmoto'>Your camera solutions...</p>";
$heads[1] = "<a href='index.html'><img id='rentals' src='images/rentalsnew2.jpg' alt='GroupLogo' /></a>";
$heads[2] = "<p class='ourmoto'>...Your camera solutions</p>";
for ($i = 0; $i<3; $i++){
echo"<p>$heads[$i]</p>";
}
CSS:
.ourmoto {
font-style:oblique;
font-size:30px;
display:inline;
margin: 0 auto;
}
据我所知,echo应该在引用中以字符串形式打印出来的内容,即使它的HTML标签也是如此。它设法打印HTML但CSS不适用于它。我做错了什么?
谢谢!
答案 0 :(得分:-3)
遍历每个脑袋,用课堂回声:
foreach ($heads as $head)
echo "<p class='ourmoto'>{$head}</p>";
}