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html - how to terminate php code

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Jul 06, 2016 pm 01:51 PM

htmlphp

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>

<title>XXX</title>


<?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."";
     
}
; ?>



</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>

<title>XXX</title>


<?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
'); ?>

")
}; ?>


这样的话变成php套嵌php 会报错的</code>

Reply content:

<code>

<title>XXX</title>


<?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."";
     
}
; ?>



</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>

<title>XXX</title>


<?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
'); ?>

")
}; ?>


这样的话变成php套嵌php 会报错的</code>

Wouldn’t it be better to just do this...

<?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 ');

update:

<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>

<title>XXX</title>


<?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'); 
?>

</code>

Change it to this

If it is already within the <?php tag, there is no need to write it again. .

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