Question

我试图将mysqli转换为PDO：

  `$query = "SELECT * FROM email_list";
   $result = mysqli_query($dbc, $query)
   or die('Error querying database.');

   while ($row = mysqli_fetch_array($result)){
   $to = $row['email'];
   $first_name = $row['first_name'];
   $last_name = $row['last_name'];
   $msg = "Dear $first_name $last_name,\n$text";
   mail($to, $subject, $msg, 'From:' . $from);
   echo 'Email sent to: ' . $to . '<br />';
   } 

   mysqli_close($dbc);

通过阅读其他问题，我尝试了这个：

 $sql = "SELECT * FROM email_list";


 while ($row = fetch(PDO::FETCH_BOTH)){
 $to = $row['email'];
 $first_name = $row['first_name'];
 $last_name = $row['last_name'];
 $msg = "Dear $first_name $last_name,\n$text";
 mail($to, $subject, $msg, 'From:' . $from);
 echo 'Email sent to: ' . $to . '<br />';
 } 

 $conn->exec($sql);

 $conn = null;

但是我收到了这个错误：解析错误：语法错误，意外＆＃39; {＆＃39;在第36行的C：\ xampp \ htdocs \ sendemail.php中。

更新：

<?php
$from = 'elmer@makemeelvis.com';
$subject = $_POST['subject'];
$text = $_POST['elvismail'];

$servername = "localhost";
$username = "davide";
$password = "";

try {
$conn = new PDO("mysql:host=$servername;dbname=elvis_store", $username,   $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connected successfully<br>";
}
catch (PDOException $e) {
echo 'Connection failed: ' . $e->getMessage();
}

//  $dbc = mysqli_connect('data.makemeelvis.com', 'elmer', 'theking',   'elvis_store')
//   or die('Error connecting to MySQL server.');

// $sql = "SELECT * FROM email_list";


// while ($row = fetch(PDO::FETCH_BOTH))
//  {
//  $to = $row['email'];
//    $first_name = $row['first_name'];
//    $last_name = $row['last_name'];
//    $msg = "Dear $first_name $last_name,\n$text";
//    mail($to, $subject, $msg, 'From:' . $from);
//    echo 'Email sent to: ' . $to . '<br />';
//
//   } 
$query = $conn->prepare("SELECT * FROM email_list");
$query->execute();

while($row = $conn->fetch(PDO::FETCH_BOTH)){
$to = $row['email'];
$first_name = $row['first_name'];
$last_name = $row['last_name'];
$msg = "Dear $first_name $last_name,\n$text";
mail($to, $subject, $msg, 'From:' . $from);
echo 'Email sent to: ' . $to . '<br />';
}

$conn->exec($sql);

$conn = null;

?>

成功连接

注意：未定义的变量：第47行的C：\ xampp \ htdocs \ study \ ch03 \ final \ makemeelvis \ sendemail.php中的dbc

致命错误：在null中调用成员函数prepare（）第47行的C：\ xampp \ htdocs \ study \ ch03 \ final \ makemeelvis \ sendemail.php

更新2：

我还没有出现这个错误，但是现在html页面继续加载...没有给出任何错误但也没有消息作为＆＃34;已连接＆＃34;或类似的......

我做错了什么？

感谢

Answer 1

这应该为你做到！

$query = $conn->prepare("SELECT * FROM email_list");
$query->execute();

while($row = $query->fetch(PDO::FETCH_ASSOC)){
    $to = $row['email'];
    $first_name = $row['first_name'];
    $last_name = $row['last_name'];
    $msg = "Dear $first_name $last_name,\n$text";
    mail($to, $subject, $msg, 'From:' . $from);
    echo 'Email sent to: ' . $to . '<br />';
}

编辑：

戴夫，使用上面编辑的代码，你应该没问题。我将解释您为何会收到错误。

我原来的回复是使用变量$ dbc，而你的PDO数据库连接字符串......

$conn = new PDO("mysql:host=$servername;dbname=elvis_store", $username,   $password);

使用变量$ conn作为PDO处理程序。

希望这是有道理的。

PDO相当于：while（$ row = mysqli_fetch_array（$ result））

1 个答案: