Question

我遇到了我创建的数据库类的问题。当我尝试显示表中的数据时，没有数据显示，但它仍然通过数据库 - ＆gt; select（）。

<?php

class database {

private $DBhost  = "localhost";
private $DBuser  = "username";
private $DBpass  = "password";
private $DBname  = "database";

private $PDO    = null;
private $stmt   = null;

function __construct() {
  $this->PDO = new PDO("mysql:host=" . $this->DBhost . ";dbname=" . $this->DBname . ";charset=utf8", $this->DBuser, $this->DBpass);
}

function __destruct() {
  $this->PDO = null;
}

function select($query, $param = array()) {
  try {
    $this->stmt = $this->PDO->prepare($query);
    if (count($param) != 0) {
      $this->stmt->execute($param);
    } else {
      $this->stmt->execute();
    }
  } catch(Exception $ex) {
    $this->error($ex);
    return false;
  }
}



function resultset() {
  return $this->stmt->fetchAll();
}

function error($ex) {
  die('Something broke :(');
}
}
?>

我正在尝试使用它来显示html表中的数据，如下一页所示。

<?php
include 'code/db.php';
$DB = new database();
?>
<html>
<head>
  <title>List</title>
</head>
<body>
<table>
  <tr>
    <th>First Name</th>
    <th>Last Name</th>
    <th>Email</th>
    <th>Phone number</th>
    <th>Mobile number</th>
    <th>Address</th>
    <th>State</th>
    <th>Post Code</th>
    <th>Settings</th>
  </tr>
<?php
$DB->select('SELECT * FROM owners ORDER BY first_name, last_name');

while($line = $DB->resultset()) {
  echo "<tr>";
  echo "<td>" . $line['first_name'] . "</td>";
  echo "<td>" . $line['last_name'] . "</td>";
  echo "<td>" . $line['email'] . "</td>";
  echo "<td>" . $line['phone_number'] . "</td>";
  echo "<td>" . $line['mobile_number'] . "</td>";
  echo "<td>" . $line['address'] . "</td>";
  echo "<td>" . $line['state'] . "</td>";
  echo "<td>" . $line['post_code'] . "</td>";
  echo '<td><a href="item-list?id=' . $line['id'] . '">Details</a><a href="edit-owner?id=' . $line['id'] . '"</a></td>';
  echo "</tr>";
}

?>
</table>
</body>
</html>

如果你能帮我解决这个问题，那就太好了。

Answer 1

PDO已经是一个数据库类，你不需要任何其他类。特别是这样一个有状态的人。你只是在这个无用的课上挖了一个巨大的陷阱，很快就会陷入其中。

所以，保持原始PDO：

<?php

$DBhost  = "localhost";
$DBuser  = "username";
$DBpass  = "password";
$DBname  = "database";

$dsn = "mysql:host=$DBhostdbname=$DBname;charset=utf8";
$PDO = new PDO($dsn, $DBuser, $DBpass);
$PDO->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

您实际需要的是将SQL与HTML分开。因此，首先获取所有数据然后使用任何模板输出它。

<?php
include 'code/db.php';
$stm  = $PDO->query('SELECT * FROM owners ORDER BY first_name, last_name');
$data = $stm->fetchAll();
?>
<html>
<head>
  <title>List</title>
</head>
<body>
<table>
  <tr>
    <th>First Name</th>
    <th>Last Name</th>
    <th>Email</th>
    <th>Phone number</th>
    <th>Mobile number</th>
    <th>Address</th>
    <th>State</th>
    <th>Post Code</th>
    <th>Settings</th>
  </tr>
<?php foreach($data as $line):?>
  <tr>
    <td><?=$line['first_name']?></td>
    <td><?=$line['last_name']?></td>
    <td>
      <a href="item-list?id=<?=$line['id']?>">Details</a>
      <a href="edit-owner?id=<?=$line['id']?>Owner</a>
    </td>
  </tr>
<?php endforeach ?>
</table>
</body>
</html>

Answer 2

您在fetchAll功能中使用resultset。将返回整个选定值数组并将其存储在$line中。然后你必须迭代它。

将功能更改为：

function resultset() {
  return $this->stmt->fetch( PDO::FETCH_ASSOC );
}

Answer 3

更改此行

while($line = $DB->resultset()) {

到

foreach($DB->resultset() as $line ) {

数据库类错误 - php

3 个答案: