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in Technique[技术] by (71.8m points)

php - failed to open stream: Invalid argument

In this code :

$path = "C:NucServwwwvvstaticarrays
ews.php";
  $fp = fopen($path, "w");
  if(fwrite($fp=fopen($path,"w"),$text))
  {
    echo "ok";
  }
  fclose($fp);

I have this error message:

failed to open stream: Invalid argument

What is wrong in my code?

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1 Reply

0 votes
by (71.8m points)

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

   ...arrays
   ews.php

You should escape them like this:

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

Or use singles, like this:

$path = 'C:NucServwwwvvstaticarrays
ews.php'; 

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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