假设我有一个包含两个输入字段的表单，标题和评论。用户填写这两个字段并提交数据后，将自动创建一个页面，其中包含用户键入的标题和评论，位于根目录的文件夹中，例如 www.root.com/page/ 将包含自动生成的页面。
更新：
我希望将表单数据发送到 SQL 表行，其中自动生成页面的 url 与标题和评论位于同一 SQL 行中。
我如何通过 PHP 做到这一点？
【问题讨论】：
<?php
//Template for basic page
$template = <<<EOD
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title><!--TITLE--></title>
</head>
<body>
<!--COMMENT-->
</body>
</html>
EOD;
//handle the posted form
if(isset($_POST['title'])&&isset($_POST['comment'])){
//replace the areas of the template with the posted values
$page = str_replace('<!--TITLE-->',htmlentities($_POST['title']),$template);
$page = str_replace('<!--COMMENT-->',htmlentities($_POST['comment']),$page);
//create a name for the new page
$pagename = md5($_POST['title']).'.html';
//db connect & select
$db=mysql_connect('localhost','user','pass');
mysql_select_db('yourdb');
//check if page already exists
$result = mysql_query('SELECT pagename from yourtable WHERE url="'.mysql_real_escape_string($pagename).'"');
if(mysql_num_rows($result)>=1){
$notice = '<p>Page already created <b>./pages/'.$pagename.'</b></p>';
}else{
//inset new page into db
mysql_query('INSERT into yourtable (`id`,`title`,`comment`,`url`)VALUES("",
"'.mysql_real_escape_string(htmlentities($_POST['title'])).'",
"'.mysql_real_escape_string(htmlentities($_POST['comment'])).'",
"'.$pagename.'")');
//put the created content to file
file_put_contents('./pages/'.$pagename,$page);
//make a notice to show the user
$notice = '<p>New Page created <b>./pages/'.$pagename.'</b></p>';
}
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Language" content="en-gb">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Make page example</title>
</head>
<body>
<?php
//if the notice is set then display it
if(isset($notice)){echo $notice;} ?>
<form method="POST" action="">
<p>Title:<input type="text" name="title" size="31"></p>
<p>Comment:</p>
<p><textarea rows="5" name="comment" cols="21"></textarea></p>
<p><input type="submit" value="Submit"></p>
</form>
</body>
</html>
【讨论】：