Question

我有这段代码：

    <?php
session_start();
$page_name = 'Send Emails';
    require_once('header.php');
    if(!empty($_POST['send'])){
        $id_mail= "";
        foreach ($_POST['selected'] as $key => $value) {
            echo "$key - $value <br>";
            $id_mail .= "$value, ";
            }
            $id_mail= substr($id_mail, 0, -2);
        }
    if(!empty($_POST['Send2'])){
            $mail= mysqli_query($conn, "SELECT mail FROM students WHERE student_id IN($_POST[id_mail])");
            echo "SELECT mail FROM students WHERE student_id IN($_POST[id_mail]) ";
            while ($row_mail = mysqli_fetch_assoc($mail)) {
                echo "$row_mail[mail] ";
            $to = $row_mail['mail'];
            $subject = $_POST['head'];
            $message = $_POST['body'];
            $headers = 'From: code-week@example.com' . "\r\n" .
            'Please do not reply to this message' . "\r\n" .
            'X-Mailer: PHP/' . phpversion();
            mail($to, $subject, $message, $headers);
            }

        }
?>
<form name='mail2' method='post' action='send_email.php'><br/>
<input type="hidden" name="id_mail" value="<?php echo $id_mail; ?>">
    <br/>

<input type='text' name='head' placeholder='header'>
<br/>
<br/>

<textarea name='body' rows="4" cols="50" placeholder='body'></textarea><br/>
<br/>
<input class='btn btn-danger'type='submit' name="Send2" value='Send'>
</form>

它的目的是从学生表中读取所选行的ID，打印它以便我可以看到它从上一页中选择正确，然后根据它，发送电子邮件给所有人选定的ID。我的问题是，如何在新的Mysql数据库表中编写已发送的电子邮件（只是主题和消息部分），以及所选的学生ID？每当我尝试使用普通的INSERT查询时，它表示所选ID无效，假设它为0，并且写入主题和消息，并且在ID列中，它输入假定值-0。如果我有两个或三个人选择（2-3个ID）它写第一个然后我得到一个错误。如果第一个字段已满（ID为0），则输出错误并且不写入任何内容最终结果应该是一个包含3列的表 - 学生ID，电子邮件主题和电子邮件。目的是通知选定的学生，然后在表格中写下信息，这样我就可以看到女巫学生已经收到通知。

Answer 1

First of all it's better not to use unescaped POST data in your queries, as it might lead to a sql injection.

As for the 'insert' part - if you have all the necessary fields, primary keys and default values, just use an INSERT statement to save the data

Answer 2

1）。为is_notified创建一个列，并将其初始值设置为零。

2）。电子邮件发送给学生后，立即将is_notified列更新为1。

如果要使用PDO进行多次更新，可以使用： -

<?php
try
{
    $s = $conn->prepare("your SQL");
}
catch(PDOException $e)
{
    echo $e->getMessage();
}

foreach(loop through the ids here)
{
    $s->bindParam(':paramname', $paramvalue);
    $s->execute();
}
?>

Answer 3

INSERT INTO tbl_name (a,b,c) VALUES(1,2,3),(4,5,6),(7,8,9);

Answer 4

这对我有用：

if(!empty($_POST['Send2'])){ 
14             $mail= mysqli_query($conn, "SELECT student_id, mail FROM students WHERE student_id IN($_POST[id_mail])"); 
15             echo "SELECT student_id, mail FROM students WHERE student_id IN($_POST[id_mail]) "; 
16             while ($row_mail = mysqli_fetch_assoc($mail)) { 
17                 echo "$row_mail[mail] "; 
18             $to = $row_mail['mail']; 
19             $subject = $_POST['head']; 
20             $message = $_POST['body']; 
21             $headers = 'From: code-week@example.com' . "\r\n" . 
22             'Please do not reply to this message' . "\r\n" . 
23             'X-Mailer: PHP/' . phpversion(); 
24             mail($to, $subject, $message, $headers); 
25             $sent_mails= mysqli_query($conn, "INSERT INTO sent_mails (id_st, mail_head, content) VALUES ('$_POST[id_mail]', '$subject', '$message') "); 
26             if (mysqli_query($conn, $sent_mails)){ 
27              echo "Success"; 
28             } else { 
29              echo "Error ". $sent_mails . "<br/>" . mysqli_error($conn); 
30             } 
31             } 
32         }

如何使用php将多行数据插入Mysql DB

4 个答案: