Question

我的网站回复为

[{"name":"xxx","phone":"123","email":"a@a.com"},{"name":"yyy","phone":"456","email":"b@a.com"},{"name":"zzz","phone":"678","email":"c@a.com"}...]

我的代码是

$json = '[{"name":"xxx","phone":"123","email":"a@a.com"},{"name":"yyy","phone":"456","email":"b@a.com"},{"name":"zzz","phone":"678","email":"c@a.com"}]';
$json_decoded = json_decode($json);
foreach($json_decoded as $result){
...
}

当我运行此代码时，出现错误Invalid argument supplied for foreach()

这是什么问题？

如何在html表格中显示为表格？

Answer 1

这段代码完全正常 - 只需在本地机器上测试它

<?php
    $json = '[{"name":"xxx","phone":"123","email":"a@a.com"},{"name":"yyy","phone":"456","email":"b@a.com"},{"name":"zzz","phone":"678","email":"c@a.com"}]';
    $json_decoded = json_decode($json);
    foreach($json_decoded as $result){
      print_r($result);
    }
  ?>

要输出您的资料作为表格，请使用：

<?php
        $json = '[{"name":"xxx","phone":"123","email":"a@a.com"},{"name":"yyy","phone":"456","email":"b@a.com"},{"name":"zzz","phone":"678","email":"c@a.com"}]';
        $json_decoded = json_decode($json);
        echo '<table>';
        foreach($json_decoded as $result){
          echo '<tr>';
            echo '<td>'.$result->name.'</td>';
            echo '<td>'.$result->phone.'</td>';
            echo '<td>'.$result->email.'</td>';
          echo '</tr>';
        }
        echo '</table>';
      ?>

确保您的JSON字符串没有任何语法错误...如果是，则json_decode将失败并且foreach（）循环将引发错误。

Answer 2

你在问题中提到的代码是完美的，我得到表中所需的输出。但请确保检查json_encode()是否正确，以便foreach错误隐藏起来。

您也可以使用此方法，因为我在此处建议在表格中打印。

您必须将TRUE应用于json_decode语句，因为std_object数组与此相关，因此会导致打印数组值时产生混淆。

<强> json_decode（）

Syntax: json_decode($jsonstring,TRUE);

json_decode - 解码JSON字符串

以适当的PHP类型返回json中编码的值。值true，false和null分别返回为TRUE，FALSE和NULL。如果无法解码json或编码数据深于递归限制，则返回NULL。

PHP代码：

<?php
$json = '[{"name":"xxx","phone":"123","email":"a@a.com"},{"name":"yyy","phone":"456","email":"b@a.com"},{"name":"zzz","phone":"678","email":"c@a.com"}]';
$json_decoded = json_decode($json,TRUE);
foreach($json_decoded as $result){
    echo $result['name']; // this wil output the values as xxxyyyzzz
}
?>

在json_decode（）语句中输出没有TRUE的print_r（）：

Array ( [0] => stdClass Object ( [name] => xxx [phone] => 123 [email] => a@a.com ) [1] => stdClass Object ( [name] => yyy [phone] => 456 [email] => b@a.com ) [2] => stdClass Object ( [name] => zzz [phone] => 678 [email] => c@a.com ) )

在json_decode（）语句中使用TRUE输出print_r（）

Array ( [0] => Array ( [name] => xxx [phone] => 123 [email] => a@a.com ) [1] => Array ( [name] => yyy [phone] => 456 [email] => b@a.com ) [2] => Array ( [name] => zzz [phone] => 678 [email] => c@a.com ) )

因此上述代码的输出如下：

xxxyyyzzz

没有问题，您可以遍历foreach数据，然后使用它进行打印。

打印表格中的数据，您可以按照下面的说明进行操作。

<?php
$json = '[{"name":"xxx","phone":"123","email":"a@a.com"},{"name":"yyy","phone":"456","email":"b@a.com"},{"name":"zzz","phone":"678","email":"c@a.com"}]';
$json_decoded = json_decode($json,TRUE);
?>
<table border="1">
<thead>
    <tr>
        <th>Name</th>
        <th>Phone</th>
        <th>Email</th>
    </tr>
</thead>    
<tbody>
<?php
    foreach($json_decoded as $result){
        ?>
        <tr>

            <td><?php echo $result['name']; ?></td>
            <td><?php echo $result['phone']; ?></td>
            <td><?php echo $result['email']; ?></td>
        </tr>
<?php   
}
?>
</tbody>
</table>

我希望我的解决方案可以满足您的需求。

快乐编码：）

PHP - 在html表中显示JSON数组

2 个答案: