Question

从this question继续现在我需要设置输出样式才发现php不喜欢我放置标签的方式我还试图将其封装在"和.中或<?php $row['field'] ?>但仍然无效。

    <table class="table table-hover">
<caption>List All Customers from Customer Table</caption>
<thead>
<tr>
<th>id</th>
<th>Inital</th>
<th>First Name</th>
<th>Last Name</th>
<th>Mobile</th>
<th>Landline</th>
<th>Email</th>
<th>Address</th>
<th>Post Code</th>
</tr>
 </thead>
<tbody>
<tr>
<?php
foreach ($rows as $row) { 
?>
 <td><?php $row['clientid']; ?></td>
 <td><?php $row['inital']; ?></td> 
 <td><?php $row['firstname']; ?></td>
 <td><?php $row['lastname']; ?></td>
 <td><?php $row['mobile']; ?></td>
 <td><?php $row['landline']; ?></td>
 <td><?php $row['email']; ?></td>
 <td><?php $row['address']; ?></td>
 <td><?php $row['postcode']; ?></td>
 <?php } ?>
</tr>
</tbody>
</table>

它不起作用，显示没有结果。 :(拔出头发！看了另外的问题，但北方足够清楚。

Answer 1

添加echo ..

<td><?php echo $row['clientid']; ?></td>
<td><?php echo $row['inital']; ?></td> 
<td><?php echo $row['firstname']; ?></td>
<td><?php echo $row['lastname']; ?></td>
<td><?php echo $row['mobile']; ?></td>
<td><?php echo $row['landline']; ?></td>
<td><?php echo $row['email']; ?></td>
<td><?php echo $row['address']; ?></td>
<td><?php echo $row['postcode']; ?></td>

Answer 2

尝试从评论中编辑代码：可能你只是没有将值推入$ rows数组。

<?php 
 $rows = array() ;
 $result = $dbc->query('SELECT * FROM customer');
 if ($result){
   while ($row = $query->fetch_assoc()){
    $rows[] = $row ; //Put your data in the array first. 
   }
 }
?>

<table>
<tbody> 
 <tr> 
<?php 
 foreach ($rows as $row) { ?> 
  <td><?php echo $row['clientid']; ?></td> 
  <td><?php echo $row['inital']; ?></td> 
  <td><?php echo $row['firstname']; ?></td> 
<?php } ?> 
 </tr> 
</tbody> 
</table>

foreach（$行为$ row）无法添加html标记

2 个答案: