duanjiu4498 2016-10-09 19:54
浏览 255

fopen PHP错误(期望参数1为资源)

I am trying to create a directory with a PHP file in it (that part works).

Now I am trying to open that PHP file and put in data, but I get error:

Warning: fwrite() expects parameter 1 to be resource, string given in C:\wamp\www\dev 3\1\index2.php on line 13

and

Warning: fclose() expects parameter 1 to be resource, string given in C:\wamp\www\dev 3\1\index2.php on line 14

Here is my code:

<?php
$directory = substr(md5(rand(1000,10000)),2,8);
mkdir($directory);
$dir = './'.$directory.'/index.php';
$upload = fopen($dir,'x');
$data = '
data data data

';
fopen($dir, "w");
fwrite($dir, $data);
fclose($dir);
?>
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1条回答 默认 最新

  • douhai5835 2016-10-09 20:00
    关注

    You need a resource handle and not string, as your error gracefully said :)

    Use $upload instead of $dir:

    <?php
    $directory = substr(md5(rand(1000,10000)),2,8);
    mkdir($directory);
    $dir = './'.$directory.'/index.php';
    $upload = fopen($dir,'x');
    $data = '
    data data data
    
    ';
    /*fopen($dir, "w");
    fwrite($dir, $data);
    fclose($dir);
    */
    fopen($upload, "w");
    fwrite($upload, $data);
    fclose($upload);
    ?>
    
    评论

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