我有一个表,用于显示JSON中的数据。我能够在表中显示数据,但不能很好地呈现表。
表的结构变形。它不会按我希望的显示顺序显示数据。
<table class="table table-bordered" id="dataTable" width="100%" cellspacing="0">
<thead>
<tr>
<th>Name</th>
<th>Number</th>
<th>Amount</th>
</tr>
</thead>
<?php
$url = 'xxxxxxxxxxxxxxxxxxxxxx'; // path to your JSON file
//$url = 'data.json'; // path to your JSON file
$data = file_get_contents($url); // put the contents of the file into a variable
$characters = json_decode($data);
$i = 0;
?>
<tbody>
<tr>
<?php
foreach ($characters as $character)
{
?>
<td>
<?php echo $character->name; ?>
</td>
<td>
<?php echo $character->phoneNumber; ?>
</td>
<td>
<?php echo $character->amount; ?>
</td>
<?php }
?>
</tr>
</tbody>
</table>
我希望只有三列。数据以正确的顺序显示
答案 0 :(得分:1)
尝试
<table class="table table-bordered" id="dataTable" width="100%" cellspacing="0">
<thead>
<tr>
<th>Name</th>
<th>Number</th>
<th>Amount</th>
</tr>
</thead>
<?php
$url = 'xxxxxxxxxxxxxxxxxxxxxx'; // path to your JSON file
//$url = 'data.json'; // path to your JSON file
$data = file_get_contents($url); // put the contents of the file into a variable
$characters = json_decode($data);
$i = 0;
?>
<tbody>
<?php
foreach ($characters as $character)
{
?>
<tr>
<td>
<?php echo $character->name; ?>
</td>
<td>
<?php echo $character->phoneNumber; ?>
</td>
<td>
<?php echo $character->amount; ?>
</td>
</tr>
<?php }
?>
</tbody>
</table>