怎么生成静态页面

Question

比方说
<div class="tpwzzs">
<ul>
<?php
require_once('gongxiang.php');
$con = lianjie();
   
 $result = mysql_query("SELECT * FROM tupianwenzi left join detail on tupianwenzi.xiaofenleiindex = detail.xiaofenleidetail");

 while($row = mysql_fetch_array($result))
 {
 





echo'<li>
<div class="tp">
<a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"><img src="'.$row['tupianindex'].'" /></a>
</div>
<div class="detail">'.$row['xiaofenleiindex'].'</div>
<div class="detail1"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'">'.$row['detailindex'].'</a></div>
<div class="detailanniu"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"></a></div>
</li>';

 }
?>
 <?php
 mysql_close($con);
 ?>
</ul>
</div>
这个代码 用什么方法生成静态页面 最好 最快

Answer 1

我觉得 直接把 最终的网页源代码复制出来最快了啊 PHP有没有这种写法。。

Answer 2

在文件开始出加入
<?php
function callback($buffer) {
  file_put_contents('你需要的文件名', $buffer);
  return $buffer; //不需显示页面内容就 return '';
}
ob_start("callback");
?>

Answer 3

<?php
function callback($buffer) {
  file_put_contents('111.html', $buffer);
  return $buffer; //不需显示页面内容就 return '';
}
ob_start("callback");
?>
<?php
echo'wee';
?>这样写 不会生成111.html        

Answer 4

噢，php 已被改得面目全非了。
加个参数
ob_start("callback", PHP_OUTPUT_HANDLER_START);
以前是不要的

Answer 5

$buffer 参数写什么   能不能写个能用的给我。。

Answer 6

这样就行，已通过测试
function callback($buffer) {
  file_put_contents('你需要的文件名', $buffer);
  return $buffer; //不需显示页面内容就 return '';
}
ob_start("callback", PHP_OUTPUT_HANDLER_START);

Answer 7

恩 但我改成
<?php
function callback($buffer) {
  file_put_contents('111.html', $buffer);
  return $buffer; //不需显示页面内容就 return '';
}
ob_start("callback");
?><div class="tpwzzs">
<ul>
<?php
require_once('gongxiang.php');
$con = lianjie();
   
 $result = mysql_query("SELECT * FROM tupianwenzi left join detail on tupianwenzi.xiaofenleiindex = detail.xiaofenleidetail");

 while($row = mysql_fetch_array($result))
 {
 





echo'<li>
<div class="tp">
<a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"><img src="'.$row['tupianindex'].'" /></a>
</div>
<div class="detail">'.$row['xiaofenleiindex'].'</div>
<div class="detail1"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'">'.$row['detailindex'].'</a></div>
<div class="detailanniu"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"></a></div>
</li>';

 }
?>
 <?php
 mysql_close($con);
 ?>
</ul>
</div> 又不行了

Answer 8

function callback($buffer) {
  file_put_contents('111.html', $buffer);
  return $buffer; //不需显示页面内容就 return '';
}
ob_start("callback", PHP_OUTPUT_HANDLER_START);要说几遍才能注意到细节呢？

Answer 9

<?php
function callback($buffer) {
  file_put_contents('111.html', $buffer);
  return $buffer; //不需显示页面内容就 return '';
}
ob_start("callback",PHP_OUTPUT_HANDLER_START);
?><div class="tpwzzs">
<ul>
<?php
require_once('gongxiang.php');
$con = lianjie();
   
 $result = mysql_query("SELECT * FROM tupianwenzi left join detail on tupianwenzi.xiaofenleiindex = detail.xiaofenleidetail");

 while($row = mysql_fetch_array($result))
 {
 





echo'<li>
<div class="tp">
<a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"><img src="'.$row['tupianindex'].'" /></a>
</div>
<div class="detail">'.$row['xiaofenleiindex'].'</div>
<div class="detail1"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'">'.$row['detailindex'].'</a></div>
<div class="detailanniu"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"></a></div>
</li>';

 }
?>
 <?php
 mysql_close($con);
 ?>
</ul>
</div> 还是生成不了 一片空白

Answer 10

刚才又测试了一下，对于多段输出需要写作
define('CACHE_FILE', '111.html');
function callback($buffer) {
  file_put_contents(CACHE_FILE, $buffer, FILE_APPEND);
  return $buffer; //不需显示页面内容就 return '';
}
file_put_contents(CACHE_FILE, '');
ob_start("callback", PHP_OUTPUT_HANDLER_START);当然你可用全局变量而不是常量来指定目标文件名

Answer 11

<?php
/* Created on [2013-3-22] Author[Newton] Filename[test20.php]*/function callback($buffer) {
  file_put_contents('111.html', $buffer);
  return $buffer; //不需显示页面内容就 return '';
}
ob_start("callback",PHP_OUTPUT_HANDLER_START);
?><div class="tpwzzs">
<ul>
<?php
//require_once('gongxiang.php');
//$con = lianjie();
//
// $result = mysql_query("SELECT * FROM tupianwenzi left join detail on tupianwenzi.xiaofenleiindex = detail.xiaofenleidetail");
//
// while($row = mysql_fetch_array($result))
// {
$row = array('id'=>'id','tupianindex'=>'tupianindex','xiaofenleiindex'=>'xiaofenleiindex','detailindex'=>'detailindex');
echo'<li>
<div class="tp">
<a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"><img src="'.$row['tupianindex'].'" /></a>
</div>
<div class="detail">'.$row['xiaofenleiindex'].'</div>
<div class="detail1"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'">'.$row['detailindex'].'</a></div>
<div class="detailanniu"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"></a></div>
</li>';// }
?>
<?php //mysql_close($con); ?>
</ul>
</div>
<?php ob_end_flush(); ?>
<!--源码-->
<div class="tpwzzs">
<ul>
<li>
<div class="tp">
<a target="_blank" href="2jiyemian/2jiyemian.php?id=id"><img src="tupianindex" /></a>
</div>
<div class="detail">xiaofenleiindex</div>
<div class="detail1"><a target="_blank" href="2jiyemian/2jiyemian.php?id=id">detailindex</a></div>
<div class="detailanniu"><a target="_blank" href="2jiyemian/2jiyemian.php?id=id"></a></div>
</li></ul>
</div>

Answer 12

你太给力了 搞定了 你好人做到底吧 怎么把这个页面的php地址都改成html 

Answer 13

怎么把页面的PHP改成html啊
比方说
<div class="detail1"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'">'.$row['detailindex'].'</a></div>
 <div class="detailanniu"><a target="_blank" href="2jiyemian/2jiyemian.php?id='.$row['id'].'"></a></div>
 

Answer 14

你在
file_put_contents('111.html', $buffer);
前先执行
$buffer = preg_replace('/(href=.+?)\.php/', '$1.html', $buffer);不过规则可能没那么简单，你想好了再说