在PHP邮件发送HTML表格

问题描述:

我有一个web服务器能够发送来自gmail帐户的电子邮件。我有一个基于SQL查询生成动态表单的页面等。然后,我有一个表单输入您的电子邮件地址,然后重定向到另一个页面,显示“发送消息”,并且该页面上的代码发送电子邮件。但是,如何更改电子邮件的内容,使其成为上一页生成的表格?我不想重新创建整个表并将其设置为一个变量,因为我认为有一个更有效的方法来完成它。任何帮助,将不胜感激。在PHP邮件发送HTML表格

“的search.php”(此页基于由先前的页面执行的查询表)

<html> 
<head> 
<style type="text/css"> 
table { 
background-color: #C0C0C0; 
} 

th{ 
width: 150px; 
text-align:center; 
border-style: solid; 
border-width: 2px; 
border-color: black; 
background-color: #008080; 
font-family: Helvetica; 
} 
td { 
border-style: solid; 
border-width: 2px; 
border-color: black; 
font-family: Helvetica; 
background-color: #FFFF00; 
text-align:center; 
} 
body { 
background-color:#1C2932; 
} 

h1 { 
font-family: Helvetica; 
font-size: 24px; 
color: #989898; 
} 

p { 
font-family: Helvetica; 
font-size: 18px; 
color: #989898; 
} 

</style> 
</head> 
<body> 

<?php 
include 'tablegen.php'; 

if(isset ($_POST['term'])) { 
$x = $_POST['term']; 
    connect($x); 
    tableGen(); 
}//end main if 
//area 52 what is going on... 

echo "<form action='email.php' method = 'post'>"; 
echo "<p><b>Do you want this in an email?</b></p>"; 
echo "<input type='text' name='send'>"; 
echo "<input type='submit' name='submit' value='Send!' />"; 
echo "</form>"; 

?> 
<br></br> 
<form method="LINK" action="landing.php"> 
<input type="submit" value="Go Back!"> 
</form> 

</body> 
</html> 

“email.php”(此页实际发送的电子邮件)

<html> 
<head> 
<style> 
body { 
background-color:#1C2932; 
} 
p { 
font-family: Helvetica; 
font-size: 18px; 
color: #989898; 
} 

</style> 
</head> 
<?php 

$email = $_POST['send']; 

$headers = array(
'From: [email protected]', 
'Content-Type: text/html', 
'Content-Type: text/css' 
); 



mail($email,'HTML Email','I want to send an HTML table!!!',implode("\r\n",$headers)); 
echo "<p>Email Sent!</p>"; 
?> 
</html> 

“tablegen.php”(用于显示表格的功能) - WORKS !!!

<?php 
function connect(){ 

    mysql_connect("localhost","root","water123") or die ('Error Reaching Database'); 
    mysql_select_db("MathGuide"); 


} 
    //Area 51, idk what I'm doing 
function tableGen($x) { 
$term=$x; 
$sql = mysql_query("select * from student_info where ID like '%$term%'"); 
echo "<h1>STUDENT DATA for ID: $search</h1>"; 
echo "<table>"; 
echo "<tr> 
<th>ID</th> 
<th>Project</th> 
<th>Starter Project</th> 
<th>Course</th> 
<th>KDs Completed in your Course</th> 
<th>Projects Completed</th> 
<th>Project 1</th> 
<th>P1KD1</th> 
<th>P1KD2</th> 
<th>P1KD3</th> 
<th>P1KD4</th> 
<th>P1KD5</th> 
<th>Project 2</th> 
<th>P2KD1</th> 
<th>P2KD2</th> 
<th>P2KD3</th> 
<th>P2KD4</th> 
<th>P2KD5</th> 
<th>Project 3</th> 
<th>P3KD1</th> 
<th>P3KD2</th> 
<th>P3KD3</th> 
<th>P3KD4</th> 
<th>P3KD5</th> 
<th>Project 4</th> 
<th>P4KD1</th> 
<th>P4KD2</th> 
<th>P4KD3</th> 
<th>P4KD4</th> 
<th>P4KD5</th> 
</tr>"; 

while ($row = mysql_fetch_array($sql)) 
{ 
echo "<tr><td>"; 
echo $row['ID']; 
echo "</td><td>"; 
echo $row['Project']; 
echo "</td><td>"; 
echo $row['Starter Project']; 
echo "</td><td>"; 
echo $row['Course']; 
echo "</td><td>"; 
echo $row['KDs completed in your course']; 
echo "</td><td>"; 
echo $row['Projects Completed']; 
echo "</td><td>"; 
echo $row['Project 1']; 
echo "</td><td>"; 
echo $row['P 1 KD 1']; 
echo "</td><td>"; 
echo $row['P 1 KD 2']; 
echo "</td><td>"; 
echo $row['P 1 KD 3']; 
echo "</td><td>"; 
echo $row['P 1 KD 4']; 
echo "</td><td>"; 
echo $row['P 1 KD 5']; 
echo "</td><td>"; 
echo $row['Project 2']; 
echo "</td><td>"; 
echo $row['P 2 KD 1']; 
echo "</td><td>"; 
echo $row['P 2 KD 2']; 
echo "</td><td>"; 
echo $row['P 2 KD 3']; 
echo "</td><td>"; 
echo $row['P 2 KD 4']; 
echo "</td><td>"; 
echo $row['P 2 KD 5']; 
echo "</td><td>"; 
echo $row['Project 3']; 
echo "</td><td>"; 
echo $row['P 3 KD 1']; 
echo "</td><td>"; 
echo $row['P 3 KD 2']; 
echo "</td><td>"; 
echo $row['P 3 KD 3']; 
echo "</td><td>"; 
echo $row['P 3 KD 4']; 
echo "</td><td>"; 
echo $row['P 3 KD 5']; 
echo "</td><td>"; 
echo $row['Project 4']; 
echo "</td><td>"; 
echo $row['P 4 KD 1']; 
echo "</td><td>"; 
echo $row['P 4 KD 2']; 
echo "</td><td>"; 
echo $row['P 4 KD 3']; 
echo "</td><td>"; 
echo $row['P 4 KD 4']; 
echo "</td><td>"; 
echo $row['P 4 KD 5']; 
echo "</td></tr>"; 
} 

echo "</table>"; 
}//end main if 
+0

您的mysql_query容易受到SQL注入指出,参见[此链接](http://stackoverflow.com/questions/60174/how-to-prevent- SQL注入在PHP中)的方式来避免它 – dequis 2013-04-30 01:31:40

+0

其实,无论何时你包括用户输入,你应该消毒它以某种方式,请参阅[这个其他链接](http://stackoverflow.com/questions/129677/whats-the-best -method-for-sanitizing-user-input-with-php/130323),因为html中的其他$术语容易受到XSS的影响 – dequis 2013-04-30 01:33:35

+0

非常感谢您指出这一点!本周末我肯定会花一些时间来堵塞安全漏洞。由于数据已公开,因此我并不太担心这一点,因为我只是建立一个门户网站来更快地搜索,而不是下载整个电子表格。无论如何,我可以通过电子邮件发送此表吗? – Carpetfizz 2013-04-30 01:40:42

你应该将其产生的表可以由两个包含一个PHP文件中的代码,把连接到MySQL的代码,这不,在不同的生成表查询和代码的代码功能。

个人而言,我会通过创建一个模板引擎的字符串生成的电子邮件中的表,但保持了“回声”的基础代码,你可以使用ob_start,并与ob_get_clean得到的内容给一个变量。然后,您将发送第二个参数mail()中的此变量的内容,而不是读取“HTML电子邮件”的字符串。

还要注意securityissues我在评论

+0

如果我可以添加注释,您可以稍微改进一下答案:跳过那个更快的部分。这通常是误导性的,只会促使年轻程序员专注于错误的事情。 – 2013-04-30 01:47:14

+0

是的,非常真实。谢谢。 – dequis 2013-04-30 01:50:12

+0

嗨!感谢你的回答,我试着将它们分解成函数,现在我在“search.php”上得到一个空白页面。我是PHP新手,所以请原谅我的任何新手的错误。我更新了“search.php”的代码,并添加了“tablegen.php”的代码。 – Carpetfizz 2013-04-30 01:59:50