我有以下HTML代码:
<div class="content-block1">
<div class="image-block1"></div>
<div class="bear1"></div>
<div class="where-block1"></div>
<div class="intro-block1"></div>
<div class="map-block1"></div>
<div class="extra-block1"></div>
<div class="coordinates-block1"></div>
<a href=""><div class="button-block1">More...</div></a>
</div>
<div class="content-block2">
<div class="image-block2"></div>
<div class="bear2">/div>
<div class="where-block2"></div>
<div class="intro-block2"></div>
<div class="map-block2"></div>
<div class="extra-block2"></div>
<div class="coordinates-block2"></div>
<a href=""><div class="button-block2">More...</div></a>
</div>
..以及以下PHP代码:
$host = "localhost";
$user = "username";
$pass = "password";
$database = "db";
mysql_connect($host,$user,$pass,$database) or die(mysql_error());
mysql_query("SET NAMES 'utf8'");
$query = mysql_query("select image,where,intro,map,extra,coordinates from content limit 2");
mysql_close();
现在,我已经非常努力地做了以下事情:我需要在表格内容中选择前两行,其中包括列图像,where,intro,map,extra和coordinates(不包括在内)。好的,我已经用查询选择了它们,但我有下一个问题 - &gt;我需要将第一行中的内容放在名为content-block1的第一个DIV元素中,因此,columns image,where,intro,map,extra和coordinates需要适合content-block1中的div。
另外,我需要将第二行(图像,where,intro,map,extra和coordinates)中的数据放入名为content-block2的DIV元素中,放入它们所属的DIV中。你能帮我吗?
答案 0 :(得分:1)
我认为这可能就是你要找的......
$host = "localhost";
$user = "username";
$pass = "password";
$database = "db";
mysql_connect($host,$user,$pass,$database) or die(mysql_error());
mysql_query("SET NAMES 'utf8'");
$query = mysql_query("select image,where,intro,map,extra,coordinates from content limit 2");
$i = 0;
while($row = mysql_fetch_array($query , MYSQL_ASSOC))
{
$i++;
echo '
<div class="content-block{$i}">
<div class="image-block{$i}">{$row['yourvaluehere']}</div>
<div class="bear{$i}">{$row['yourvaluehere']}</div>
<div class="where-block{$i}">{$row['yourvaluehere']}</div>
<div class="intro-block{$i}">{$row['yourvaluehere']}</div>
<div class="map-block{$i}">{$row['yourvaluehere']}</div>
<div class="extra-block{$i}"></div>
<div class="coordinates-block{$i}">{$row['yourvaluehere']}</div>
<a href=""><div class="button-block{$i}">More...</div></a>
</div>
';
}
答案 1 :(得分:0)
尝试将您的查询更改为此
$query = mysql_query("select * from content limit 0, 2");
来源= https://dev.mysql.com/doc/refman/5.0/en/select.html
然后执行以下操作以获取查询值
while($row = mysqli_fetch_array($query))
{
$image = $row['image'];
$where = $row['where'];
$map = $row['intro'];
$extra = $row['extra'];
$coordinates = $row['coordinates'];
}
如果您只选择前两名
,这将是您的查询$query = mysql_query("select TOP 2 * from content");