php将数据库中所有内容生成静态html文档的代码

 更新时间:2010年04月12日 14:44:48   作者:  
比较简单了,而且我的代码优化也很是问题 比较繁琐。下面就直接上代码了
复制代码 代码如下:

<?php
/*
author:www.5dkx.com
done:生成html文档
date:2009-10-27
*/
require_once("conn.php");
if($_GET['all'])
{
/*获取数据库记录,以便于生成html文件有个文件名*/
$sqlquery = "select * from $tbname";
$result = mysql_query($sqlquery,$conn)or die("查询失败!");
$fp = fopen("./template/article.html",r);
$fpcontent = fread($fp,filesize("./template/article.html"));
fclose($fp);
/*写入文件*/
while($row = mysql_fetch_array($result))
{
$fpcontent = str_replace("{thetitle}",$row['title'],$fpcontent);
$fpcontent = str_replace("{chatitle}",$row['title'],$fpcontent);
$fpcontent = str_replace("{bookcontent}",$row['content'],$fpcontent);
$fp = fopen("./html/".$row['id'].".html",w)or die("打开写入文件失败!");
fwrite($fp,$fpcontent)or die("写入文件失败!");
}
echo "<script language=\"javascript\">alert('全部更新');</script>";
}
if($_GET['part'])
{
/*获取最后一条记录的ID,以便于生成html文件有个文件名*/
$sqlquery = "select * from $tbname order by id desc limit 1";
$result = mysql_query($sqlquery,$conn)or die("查询失败!");
$row = mysql_fetch_array($result);
$fp = fopen("./template/article.html",r);
$fpcontent = fread($fp,filesize("./template/article.html"));
fclose($fp);
$fpcontent = str_replace("{thetitle}",$row['title'],$fpcontent);
$fpcontent = str_replace("{chatitle}",$row['title'],$fpcontent);
$fpcontent = str_replace("{bookcontent}",$row['content'],$fpcontent);
$fp = fopen("./html/".$row['id'].".html",w)or die("打开写入文件失败!");
fwrite($fp,$fpcontent)or die("写入文件失败!");
echo "<script language=\"javascript\">alert('部分更新成功!');</script>";
}
?>
<html>
<head>
<title>生成html文档</title>
<script language="javascript">
function btnsubmit(form)
{
theform.submit();
}
</script>
</head>
<body>
<?
echo "<a href=?all=111>全部更新</a><br><a href=?part=111>部分更新</a>";
?>
</body>
</html>

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