通過PHP使用HTML表格發送MySQL表格

我想根據HTML表單檢索我的數據庫中的一些數據並通過電子郵件（PHP）發送。

HTML表單：（此表工作正常，我在這裏看不到任何問題）

<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：（我使用此代碼作爲另一頁上的功能，它似乎工作也沒關係，這是一個有點不同，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>'; 

    } 
}

電子郵件：（我已經看到了很多關於如何使用它的例子，但他們都不是有關從HTML表單獲取信息並將其放入PHP郵件中）

$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>

謝謝大家！

來源

2017-10-11 Guilherme Borlenghi

您可以發佈你遇到的一些錯誤？ – ldrg

看看phpMailer。你將需要包括圖書館。這裏是一個例子，從一個HTML聯繫表格 https://github.com/PHPMailer/PHPMailer/blob/master/examples/contactform.phps –

我不知道如何顯示該信息到電子郵件腳本。我認爲，我的HTML和PHP工作正常。 –

在腳本頂端定義一個變量

$table = '';

裏面的循環中，旁邊的回波，將字符串分配給一個變量：

echo 
'<tr> 
    <td>'.$setor.'</td> 
    <td>'.$taskWhat.'</td> 
    <td>'.$deadLine.'</td> 
    <td>'.$taskWho.'</td> 
    <td>'.$statusAb.'</td> 
</tr>'; 

$table .= "" 
    . "<tr>" 
     . "<td>$setor</td>" 
     . "<td>$taskWhat</td" 
     . "<td>...</td>" 
    . "</tr>" 
;

一旦完成，將它添加到$消息變量：

$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> 
    $table 
</div> 
</body> 
</html>";

來源

2017-10-11 15:59:52 Didhack

非常感謝你們！你救了我的命嘿嘿:) –

通過PHP使用HTML表格發送MySQL表格

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