php将数据库中所有内容生成静态html文档的代码

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我是靠谱客的博主成就斑马，最近开发中收集的这篇文章主要介绍php将数据库中所有内容生成静态html文档的代码，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

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> 