Posted by Alvaro G. Vicario on 12/17/93 11:54

*** walterbyrd escribió/wrote (2 Aug 2006 10:49:28 -0700):
> I know I can use the file() function to read a file into an array by
> providing a literal file name between single quotes:
>
> $lines = file('literal_file_name');
>
> But, what if I want file() to use a variable name?

From manual:

array file ( string filename [, int use_include_path [, resource context]]
)

You need to provide a string. Period. PHP doesn't know/care how you create
the string.


> $variable_file_name = "literal_file_name";
> $lines = file($variable_file_name);
>
> Of course, that example won't work. But what will?

What error are you getting?


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