Question

我遇到一个问题，显示我编写的脚本。我无法获得表中显示的代码。我究竟做错了什么？无法在某处找到答案......

错误代码：警告：mysqli_fetch_array（）期望参数1为mysqli_result，布尔值在

中给出

$con=mysqli_connect("localhost","database_connected","password");
// Check connection 
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($con,"SELECT * FROM links");

echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";

while($row = mysqli_fetch_array($result)) {
  echo "<tr>";
  echo "<td>" . $row['links'] . "</td>";
  echo "<td>" . $row['url'] . "</td>";
  echo "</tr>";
}

echo "</table>";

mysqli_close($con);

Answer 1

我建议远离mysqli并改用PDO;除了拥有更明智的模型外，安全使用起来也相当容易一些。以下是一个快速示例，说明如何将上面的代码重写为PDO代码：

<?php
$db = new PDO('mysql:host=localhost;dbname=SOME_DB_NAME_HERE;charset=utf8', 'database_connected', 'password');
$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$db->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
try {
    $rows = $db->query('SELECT * FROM links');
} catch(PDOException $ex) {
    echo "An Error occured: ", $ex->getMessage(); //user friendly message
    exit();
}

foreach($rows as $row) {
    echo '<tr>';
    foreach($row as $column=>$value) { echo '<td>', $value, '</td>'; }
    echo '</tr>';
}

有一个相当不错的教程可以帮助您学习将mysql（i）构造映射到the HashPHP wiki的等效PDO。

警告：mysqli_fetch_array（）错误显示SQL代码

1 个答案: