执行此操作后,我以非联合方式选择数据。我得到2个数据,但第三个数据在第二个数据下面,第四个数据在第一个数据下面,但是第三个和第四个数据不是一个接一个地放在一行中。
我的观点
<div class="container col-md-9">
<?php
foreach($unidet as $tempuni)
{
?>
<div class="service-items col-md-6">
<h4 class="title-normal"><img class="img-responsive thumb" src="images/service/service-icon1.png" alt=""><strong><?php echo $tempuni['name'] ?></strong></h4>
<div class="row">
<div class="col-sm-6 service-item-img">
<img class="img-responsive thumbnail" src="admin/<?php echo $tempuni['image'];?>" alt="">
</div>
<div class="col-sm-6 service-item-content">
<p><?php echo $tempuni['description'];?></p>
<button type="button" class="btn btn-warning btn-sm" onclick="more(<?php echo $tempuni['id'];?>)">Read More</button>
</div>
</div>
</div>
<?php
}
?>
</div>
这就是我得到的
答案 0 :(得分:0)
很好的解决方案是使用array_chunk()
<?php foreach (array_chunk($unidet, 2) as $row): ?>
<div class="row">
<?php foreach ($row as $tempuni): ?>
<div class="service-items col-md-6">
<h4 class="title-normal"><img class="img-responsive thumb" src="images/service/service-icon1.png" alt=""><strong><?php echo $tempuni['name'] ?></strong></h4>
<div class="row">
<div class="col-sm-6 service-item-img">
<img class="img-responsive thumbnail" src="admin/<?php echo $tempuni['image'];?>" alt="">
</div>
<div class="col-sm-6 service-item-content">
<p><?php echo $tempuni['description'];?></p>
<button type="button" class="btn btn-warning btn-sm" onclick="more(<?php echo $tempuni['id'];?>)">Read More</button>
</div>
</div>
</div>
<?php endforeach; ?>
</div>
<?php endforeach; ?>