我想做一个小的联系形式,在其中我可以发送电子邮件给多个用户在一个数据库中,我试图只是选择所有的电子邮件,然后使用while循环发送到每一个,但它只是发送到我的数据库中的第一封电子邮件,而不是其余的。我不知道我哪里出错了
if(isset($_POST['submit'])) {
$body= $_POST['body'];
$subject= $_POST['subject'];
$user = $_SESSION["name"];
$sql = "SELECT * FROM $user";
$result = $conn->query($sql);
while($row = mysqli_fetch_array($result)){
$email = $row["contact_email"];
mail($email, $subject, $body,'From: myemail@gmail.com');
}
}
假设您有以下表格:
用户+----+-------+
| id | name |
+----+-------+
| 1 | Boss |
| 2 | Boris |
+----+-------+
create table users(
id int auto_increment primary key,
name varchar(255) not null
);
电子邮件+---------+-----------------+
| user_id | email |
+---------+-----------------+
| 1 | boss1@boss.com |
| 1 | boss2@boss.com |
| 1 | boss3@boss.com |
| 2 | boris1@boss.com |
| 2 | boris2@boss.com |
+---------+-----------------+
create table emails(
user_id int references users(id),
email varchar(255) not null
);
注意user_id对users表的引用。
现在在PHP
if(isset($_POST['submit'])) {
$body = $_POST['body'];
$subject = $_POST['subject'];
$user = mysqli::real_escape_string($_SESSION["name"]);
$result = $conn->query("SELECT id from users where name='".$user."'");
$row = mysqli_fetch_array($result);
$user_id = intval($row["id"]);
$sql = "SELECT email FROM emails where user_id=".$user_id;
$result = $conn->query($sql);
while($row = mysqli_fetch_array($result)){
$email = $row["email"];
mail($email, $subject, $body,'From: myemail@gmail.com');
}
}