我想在一行中显示数据库中的三个图像。目前,所有图像都在垂直线上,我不知道如何修复它。任何帮助,将不胜感激。我只学习PHP,检查了其他问题,他们一直没有帮助。
我尝试过使用Bootstrap,但是也没用。我拥有它的方式确实连续显示3个,但是,如果我有查询$show_men="SELECT model, id, price,image FROM products WHERE id='id' "
<div class="col-xs-12 col-sm-4">
<!--selecting products from database and displaying. when user clicks on more info button, information is stored in array and displayed on item.php-->
<?php
include 'mysql.php';
$show_men="SELECT model, id, price,image FROM products WHERE cat='men' ";
$query_men=mysqli_query($conn,$show_men);
while($row=mysqli_fetch_array($query_men,MYSQLI_ASSOC)){
echo '<br><br><table class="table"><tbody>
<tr>
<div><center><img class="img-responsive" src="../images/'.$row['image'].'" width="200" height="200"></center><br>
<center><strong>'.$row['model'].'</strong><br><br><strong>Price:£ '.$row['price'].'</strong><br><br>
<a href="../php/item.php?id1='.$row['id'].'" class="btn btn-default">More Info</a><br><br><strong></center></div><hr>
</tr>
</tbody>
</table>';
}?>
</div>
答案 0 :(得分:0)
您需要迭代列而不是表..
<div class="row">
<?php
include 'mysql.php';
$show_men="SELECT model, id, price,image FROM products WHERE cat='men' ";
$query_men=mysqli_query($conn,$show_men);
while($row=mysqli_fetch_array($query_men,MYSQLI_ASSOC)){
echo '<div class="col-xs-12 col-sm-4"><br><br><table class="table"><tbody>
<tr>
<div><center><img class="img-responsive" src="../images/'.$row['image'].'" width="200" height="200"></center><br>
<center><strong>'.$row['model'].'</strong><br><br><strong>Price:£ '.$row['price'].'</strong><br><br>
<a href="../php/item.php?id1='.$row['id'].'" class="btn btn-default">More Info</a><br><br><strong></center></div><hr>
</tr>
</tbody>
</table></div>';
}
?>
</div>