Question

我在名为 json.php 的php中有一个Json数据，现在我需要回显我的数据表中的每个数据，其中包含 index.php 。

我的问题是：

从另一个php文件传递json数据的最佳做法是什么？
之后，如何回显每个数据？

这是 json.php ，php文件生成我的查询到编码的JSON：

<?php
include "connect2.php";
$select = $con->prepare("SELECT id, dateApply, school, city, name, xType, xLevel, phone, status, lastUpdated FROM t_applicants WHERE dateApply between '2018-01-01' AND '2018-12-31'");
$select->execute();

$data = array();

    while($row=$select->fetch(PDO::FETCH_ASSOC)){
        $data['data'][] = $row;
    }
    echo json_encode($data);
?>

index.php ＆gt; 这是正文数据库代码：

<tbody>
   <?php
      //How to pass json.php to this code?
      //how to fetch the data from json.php?
      {?>
   <tr>
      <td><?php echo $data['id']; ?></td>
      <td><?php
         echo $apply = (new DateTime($data['dateApply']))->format('Y-m-d');
         ?></td>
      <td><?php echo $data['name']; ?></td>
      <td><?php echo $data['school']; ?></td>
      <td><?php echo $data['city']; ?></td>
      <td><?php echo $data['status']; ?></td>
      <td>
         <?php
            $timestamp = strtotime($data['lastUpdated']);
            $date = date('Y-m-d H:i:s', $timestamp);
            echo $date;
            ?>
      </td>
      <td>
         <button type="button" data-a="<?php echo $data['id'];?>" data-b="<?php echo $yearX; ?>" href='#detailUpdate' class="modalDetailUpdate btn btn-primary btn-xs" data-toggle="modal" data-backdrop='static' data-keyboard='false' title='Editar usuario'><span class="glyphicon glyphicon-edit"></span></button>
      </td>
      <?php
         }?>
   </tr>
</tbody>

感谢您的帮助。我真的很欣赏它......

Answer 1

<?php
$json = json_decode($data, true);
foreach($json['data'] as $row) { ?>
  <td><?php echo $row['id']; ?></td>
   ......
<?php } ?>

如何在数据表中回显JSON数据？

1 个答案: