What I wrote is a subpage that receives form data. This subpage receives a data and automatically generates two corresponding files in the directory based on the received data. Now I want to add a judgment that if the file exists, it will not be executed to generate two files. If the code of a file does not exist, execute the code to generate the file (now only output whether the link exists) my code looks like this
<code><html> <head> <title>XXX</title> </head> <body> <?php $filename = @$_POST['数据1']."/index.htm"; if(!file_exists($filename)){mkdir(@$_POST['数据1']);} file_put_contents($filename,'文件内容1 '); ?> <?php $file = @$_POST['数据1']."/233.htm"; file_put_contents($file,'文件内容2 '); ?> <?php $dir = @$_POST['数据1']; $file = @$_POST['数据1']."/index.htm"; if(file_exists($file)) { echo "链接已存在 www.XXX.com/".$dir.""; } else { echo "您的链接是".$dir.""; } ; ?> </body> </html> </code>
Baidu found an exit function that looks like this
<code><?php $site = "http://www.w3school.com.cn/"; fopen($site,"r") or exit("Unable to connect to $site"); ?> </code>
But if you add <?php ?> directly, it will become php nested within php, and you will be stuck in a puzzle
If the answer you gave me is exit or die, I will write it like this
<code><html> <head> <title>XXX</title> </head> <body> <?php $dir = @$_POST['数据1']; $file = @$_POST['数据1']."/index.htm"; if(file_exists($file)) { echo "链接已存在 www.XXX.com/".$dir.""; } else { echo "您的链接是".$dir.""; exit(" <?php $filename = @$_POST['数据1']."/index.htm"; if(!file_exists($filename)){mkdir(@$_POST['数据1']);} file_put_contents($filename,'文件内容1 '); ?> <?php $file = @$_POST['数据1']."/233.htm"; file_put_contents($file,'文件内容2 '); ?> ") }; ?> </body> </html> 这样的话变成php套嵌php 会报错的</code>
What I wrote is a subpage that receives form data. This subpage receives a data and automatically generates two corresponding files in the directory based on the received data. Now I want to add a judgment that if the file exists, it will not be executed to generate two files. If the code of a file does not exist, execute the code to generate the file (now only output whether the link exists) my code looks like this
<code><html> <head> <title>XXX</title> </head> <body> <?php $filename = @$_POST['数据1']."/index.htm"; if(!file_exists($filename)){mkdir(@$_POST['数据1']);} file_put_contents($filename,'文件内容1 '); ?> <?php $file = @$_POST['数据1']."/233.htm"; file_put_contents($file,'文件内容2 '); ?> <?php $dir = @$_POST['数据1']; $file = @$_POST['数据1']."/index.htm"; if(file_exists($file)) { echo "链接已存在 www.XXX.com/".$dir.""; } else { echo "您的链接是".$dir.""; } ; ?> </body> </html> </code>
Baidu found an exit function that looks like this
<code><?php $site = "http://www.w3school.com.cn/"; fopen($site,"r") or exit("Unable to connect to $site"); ?> </code>
But if you add <?php ?> directly, it will become php nested within php, and you will be stuck in a puzzle
If the answer you gave me is exit or die, I will write it like this
<code><html> <head> <title>XXX</title> </head> <body> <?php $dir = @$_POST['数据1']; $file = @$_POST['数据1']."/index.htm"; if(file_exists($file)) { echo "链接已存在 www.XXX.com/".$dir.""; } else { echo "您的链接是".$dir.""; exit(" <?php $filename = @$_POST['数据1']."/index.htm"; if(!file_exists($filename)){mkdir(@$_POST['数据1']);} file_put_contents($filename,'文件内容1 '); ?> <?php $file = @$_POST['数据1']."/233.htm"; file_put_contents($file,'文件内容2 '); ?> ") }; ?> </body> </html> 这样的话变成php套嵌php 会报错的</code>
Wouldn’t it be better to just do this...
<code class="php"><?php $dir = @$_POST['数据1']; $filename = $dir . "/index.htm"; if (file_exists($filename)) { echo "链接已存在 www.XXX.com/" . $dir . "\n"; exit("链接已存在"); } else { mkdir($dir); } file_put_contents($filename, '文件内容1 '); //另外一个文件一样做 //$file = $dir . "/233.htm"; //file_put_contents($file, '文件内容2 '); </code>
update:
<code class="php"><html> <head> <title>XXX</title> </head> <body> <?php $dir = @$_POST['数据1']; //$dir = 'test223'; $filename = $dir . "/index.htm"; if (file_exists($filename)) { echo "链接已存在 www.XXX.com/:2333" . $dir . "\n"; exit("链接已存在"); } else { if (!file_exists($dir)) { mkdir($dir); echo "您的链接是" . $dir . ""; } } file_put_contents($filename, ' 文件内容1 '); $file = $dir . "/233.htm"; file_put_contents($file, ' 文件内容2 '); ?> </body> </html></code>
<code><html> <head> <title>XXX</title> </head> <body> <?php $dir = @$_POST['数据1']; $filename = @$_POST['数据1']."/index.htm"; if(file_exists($filename )) { echo "链接已存在 www.XXX.com/".$dir.""; exit(); } else { echo "您的链接是".$dir.""; mkdir(@$_POST['数据1']);; }; file_put_contents($filename,'文件内容1'); $file = @$_POST['数据1']."/233.htm"; file_put_contents($file,'文件内容2'); ?> </body> </html></code>
Change it to this
If it is already within the <?php
tag, there is no need to write it again. .