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有人可以帮我修复我的代码吗?通过PHP使用HTML表格发送MySQL表格
我想根据HTML表单检索我的数据库中的一些数据并通过电子邮件(PHP)发送。
HTML表单:(此表工作正常,我在这里看不到任何问题)
<form method="post" action="valid_tasks.php">
<div class="form-group">
<label for="mailTo">To:</label>
<select class="form-control" id="mailTo" name="mailTo">
<?php echo showUsers(); ?>
</select>
</div>
<div class="form-group">
<label for="statusTo">Task Status:</label>
<select class="form-control" id="statusTo" name="statusTo">
<?php echo showStatus(); ?>
</select>
</div>
<input type="submit" name="submitMail" id="submitMail" class="btn btn-info" value="Send" style="margin-bottom: 20px;">
</form>
PHP:(我使用此代码作为另一页上的功能,它似乎工作也没关系,这是一个有点不同,但它的工作原理)
<?php
require_once('db.class.php');
$objDb = new db();
$link = $objDb->conecta_mysql();
if(isset($_POST['submitMail']))
{
$status = $_POST['statusTo'];
$userMail = $_POST['mailTo'];
$id = $_SESSION['id'];
$username = $_SESSION['username'];
$query = "SELECT T.setor, T.taskWhat, T.taskWho, DATE_FORMAT(T.deadLine,'%d/%m/%Y') AS deadLine,";
$query .= "T.taskStatus, U.username, U.email, S.descricao, S.abDescri";
$query .= "FROM tarefas AS T LEFT JOIN status AS S ON T.taskStatus = S.abDescri ";
$query .= "LEFT JOIN users AS U ON U.username = T.taskWho ";
$query .= "WHERE T.taskWho = '$userMail' AND S.abDescri = '$status'";
$result = mysqli_query($link, $query);
while($row = mysqli_fetch_assoc($result)){
$setor = $row['setor'];
$taskWhat = $row['taskWhat'];
$taskWho = $row['taskWho'];
$deadLine = $row['deadLine'];
$taskStatus = $row['taskStatus'];
$userAcao = $row['username'];
$emailAcao = $row['email'];
$statusDescri = $row['descricao'];
$statusAb = $row['statusAb'];
$setor = mysqli_escape_string($link, $setor);
$taskWhat = mysqli_escape_string($link, $taskWhat);
$taskWho = mysqli_escape_string($link, $taskWho);
$deadLine = mysqli_escape_string($link, $deadLine);
$taskStatus = mysqli_escape_string($link, $taskStatus);
$userAcao = mysqli_escape_string($link, $userAcao);
$emailAcao = mysqli_escape_string($link, $emailAcao);
$statusDescri = mysqli_escape_string($link, $statusDescri);
$statusAb = mysqli_escape_string($link, $statusAb);
echo
'<tr>
<td>'.$setor.'</td>
<td>'.$taskWhat.'</td>
<td>'.$deadLine.'</td>
<td>'.$taskWho.'</td>
<td>'.$statusAb.'</td>
</tr>';
}
}
电子邮件:(我已经看到了很多关于如何使用它的例子,但他们都不是有关从HTML表单获取信息并将其放入PHP邮件中)
$to = $email;
$subject = "Tarefas com status ".$status;
$message = "
<html>
<head>
<title>HTML email</title>
<link rel='stylesheet' type='text/css' href='https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css'>
</head>
<body>
<div class='container'>
<center><h1>Hello, ".$username."!</h1></center>
</div>
// I NEED TO PUT THIS INFO HERE
</div>
</body>
</html>
谢谢大家!
您可以发布你遇到的一些错误? – ldrg
看看phpMailer。你将需要包括图书馆。这里是一个例子,从一个HTML联系表格 https://github.com/PHPMailer/PHPMailer/blob/master/examples/contactform.phps –
我不知道如何显示该信息到电子邮件脚本。我认为,我的HTML和PHP工作正常。 –