我想显示数据库中的所有用户帐户照片和名称,并将一个用户照片和名称放在div标签中,同样地将所有帐户放入一个重定向到所点击帐户的链接。但是我在如何显示它时遇到了问题。如果仍然不清楚,请尝试在线购物网站进行描绘,其中每个名称的产品都单独放在自己的包装盒中。请帮忙,我实际上是想为我的教育项目建立一个社交网站。谢谢。继承我的代码:
$sql = "SELECT * FROM user WHERE NOT username = '$username'";
$account = array();
$photo1 = array();
$result = $conn->query($sql);
if($result->num_rows > 0)
{
while($row = $result->fetch_assoc())
{
$account[] = $row['username'];
$a = $row['photo'];
$photo1[] = "<img src = '$a' alt = 'profile photo' width = '100px' height = '100px'>";
}
}
and Heres a section of my html code:
<div>
<?php echo implode(', ',$photo1); ?>
<?php echo implode(', ',$account);?>
</div>
你能为你的任务添加一些线框吗?
通过这种方式,我认为你需要这样的东西:
<div>
<a href="$linkToUserProfile"><img src="$imgSource" alt="" title=""></a>
<a href="$linkToUserProfile">$userName</a>
</div>
#try this and then add html as you need
$sql = "SELECT * FROM user ";
$result = $conn->query($sql);
if($result->num_rows > 0)
{
while($row = $result->fetch_assoc())
{
echo $row['username'];
$a = 'www.yourdomain.com/usriamegefolder/'.$row['photo'];
echo "<img src = '$a' alt = 'profile photo' width = '100px' height = '100px'>";
}
}