代码如下:
<?php
// For use in creating individual page
$tpl_file = "submission.php";
$tpl_path = "templates/";
$submissions_path = "submissions/";
// For use in querying submitter name
$username = $_GET['username'];
session_start();
$_SESSION['username'] = $username;
//Database Information
$dbhost = "";
$dbname = "";
$dbuser = "";
$dbpass = "";
//Connect to database
mysql_connect ( $dbhost, $dbuser, $dbpass)or die("Could not connect: ".mysql_error());
mysql_select_db($dbname) or die(mysql_error());
$name = $_POST['name'];
$filename = $_POST['filename'];
$submitter = $username;
list($width, $height) = getimagesize("$filename");
$type = exif_imagetype($_POST['filename']);
$checkuser = mysql_query("SELECT filename FROM images WHERE filename='$filename'");
$filename_exist = mysql_num_rows($checkuser);
if($filename_exist > 0){
echo "I'm sorry but this image has already been submitted. Please feel free to try another.";
unset($filename);
include 'upload.php';
exit();
}
if (exif_imagetype($_POST['filename']) == IMAGETYPE_GIF) {
echo "Sorry, but we can't accept GIFs. Please feel free to try uploading another.";
unset($filename);
include 'upload.php';
exit();
}
$query = "INSERT INTO images (name, filename, submitter, width, height, type)
VALUES('$name', '$filename', '$submitter', '$width', '$height', $type)";
mysql_query($query) or die(mysql_error());
mysql_close();
echo "Thanks for your submission!<br/> Upload another <a href='/~lyons/upload.php'>here</a>!";
$tpl = file_get_contents($tpl_path.$tpl_file);
$php_file_name = $name.".php";
$fh = fopen($submissions_path.$php_file_name, "w");
fwrite($fh, $tpl);
fclose($fp);
?>
当用户提交图片时，它应该自动基于模板创建页面。这是模板的代码:
<html>
<title><?php echo $name; ?></title>
<head>
</head>
<body>
<h1><?php echo $name ?></h1>
Posted by: <?php echo $username ?>
<br/>
<img src="<?php echo $filename ?>"/>
</body>
</html>
正如您可能已经猜到的那样，我希望它放入在提交图片的第一个脚本中派生的名称、用户名和文件名值。然而，它们似乎并没有延续下去。页面已创建，但在应该回显变量值的地方却是空白的。 如何包含我想要在创建的页面中使用的变量的值？
预先感谢任何可以帮助我的人。
最佳答案
我建议使用 %name%、%username% 等字符串来标记变量的占位符。
然后，在写入文件之前，尝试如下操作:
$tpl = preg_replace("(%([a-z_][a-z0-9_]*)%)ie",'$$1',$tpl);
这将找到例如 %filename% 并将其替换为变量 $filename 的内容。
关于php - 使用 fwrite() 写入文件时如何将一页中的变量包含在另一页中？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/8032004/