这是我的代码,设置了样式,现在就是不知道要怎样把图片读出,图片的名字和路径在数据库中,<img>中src的值要怎么写,数据库查询要怎么查询?求指点
<?php $query=mysql_query("select * from tb_tpsc where type='达人收藏' order by id desc ",$conn);
while($row=mysql_fetch_array($query))
{
$path=$row['tpmc'];
$name=$row['name'];
?>
<ul id="products">
<li><a href="#"><img src="$path" alt="" width="100"
height="75"/></a>
<span><a href="#"><?php echo"$name";?></a></span>
</li>
<li><a href="#"><img src="$path" alt="" width="100"
height="75"/></a><span><a href="#"><?php echo"$name";?></a></span>
</li>
<li><a href="#"><img src="$path" alt="" width="100"
height="75"/></a>
<span><a href="#"><?php echo"$name";?></a></span>
</li>
</ul>
<?php };?>
<?php $query=mysql_query("select * from tb_tpsc where type='达人收藏' order by id desc ",$conn);
while($row=mysql_fetch_array($query))
{
$path=$row['tpmc'];
$name=$row['name'];
?>
<ul id="products">
<li><a href="#"><img src="$path" alt="" width="100"
height="75"/></a>
<span><a href="#"><?php echo"$name";?></a></span>
</li>
<li><a href="#"><img src="$path" alt="" width="100"
height="75"/></a><span><a href="#"><?php echo"$name";?></a></span>
</li>
<li><a href="#"><img src="$path" alt="" width="100"
height="75"/></a>
<span><a href="#"><?php echo"$name";?></a></span>
</li>
</ul>
<?php };?>
<ul id="products"><?php $query=mysql_query("select * from tb_tpsc where type='达人收藏' order by id desc limit 0,3 ",$conn );
while($row=mysql_fetch_array($query))
{
$path=$row['tpmc'];
$name=$row['name'];
echo <<<eof
<li> <a href="#"><img src="{$path}" alt="{$name}" width="100" height="75" /></a>
<span><a href="#">{$name}</a></span> </li>
eof;
}
?>
</ul>http://www.php.net/manual/zh/language.types.string.php
置于php计算换行,这么个思路:
$i = 1;
while ($row = mysql_fetch_array($query)) {
echo $row['src'];
// some other code
$i++;
if ($i % 3 == 0) {
echo '<br />';
}
}