无法打开流:无效的参数 [英] failed to open stream: Invalid argument

问题描述

在此代码中:

$path = "C:\NucServ\www\vv\static\arrays\news.php";
  $fp = fopen($path, "w");
  if(fwrite($fp=fopen($path,"w"),$text))
  {
    echo "ok";
  }
  fclose($fp);

我收到此错误消息:

failed to open stream: Invalid argument

我的代码有什么问题?

推荐答案

PHP将您的反斜杠转换为特殊字符.例如，...arrays\news.php变成

Your backslashes is converted into special chars by PHP. For instance, ...arrays\news.php gets turned into

   ...arrays
   ews.php

您应该这样逃避它们:

$path = "C:\\NucServ\\www\\vv\\static\\arrays\\news.php"; 

或使用单打，如下所示:

Or use singles, like this:

$path = 'C:\NucServ\www\vv\static\arrays\news.php'; 

此外，您的if被弄乱了.您不应该再次fopen该文件.只需使用您已经拥有的$fp.

Also, your if is messed up. You shouldn't fopen the file again. Just use your $fp which you already have.

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