MySQLi -将表值作为字符串而不是对象

Question

我正在尝试使用php从我的 use MySQL表中检索name和userEmail值：

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我使用以下代码试图获取当前登录的用户的名称和userEmail

<?php
session_start();
#connect to MySQL database
require_once("settings.php");
$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

#get username of current session
$username = $_SESSION['username'];

#get userEmail of logged-in user from database
$sql = "SELECT userEmail from Users WHERE username LIKE '{$username}' LIMIT 1";
$result = $mysqli->query($sql);
$replyTo = mysqli_fetch_field($result);

#get name of logged-in user from database
$sql2 = "SELECT name from Users WHERE username LIKE '{$username}' LIMIT 1";
$result2 = $mysqli->query($sql2);
$name = mysqli_fetch_field($result2);
?>

然后，我尝试将$replyTo和$name值传递到只接受字符串的函数(确切地说，是来自PHPmailer的setFrom()函数)：

$mail->setFrom($replyTo, $name);

...and得到以下错误：

/var/www/phpmailer/class.phpmailer.php警告：trim()期望参数1是string，在489 (这个对象是$replyTo)中给出的对象

可捕获的致命错误：/var/www/phpmailer/class.phpmailer.php string 490 (该对象为$name)中stdClass类的对象无法转换为字符串

使用Chrome Logger进行调试，我发现这些值是$userEmail和$name的值。

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Answer 1

我认为您可以通过单次查询获得电子邮件和名称--出于安全原因使用准备好的语句(SQLIA)

#get userEmail of logged-in user from database
$sql = "SELECT userEmail,name from Users WHERE username = ? LIMIT 1";
// this is prepared statement and prevent form sql injection attack
$statement = $mysqli->prepare($sql);
$statement->bind_param('s',$username);
$statement->execute();
$result = $statement->get_result();
// fetch first record in associative array
$userDetail = $result->fetch_assoc();
if($userDetail)
{
   $replyTo = $userDetail['userEmail'];
   $name = $uerDetail['name'];
   $mail->setFrom($replyTo, $name);
}
else
{
   echo 'user not found';
}