Use file() function to read a variable name, not a literal name.

[walterbyrd]

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_f ile_name');

But, what if I want file() to use a variable name?

$variable_file_ name = "literal_file_n ame";
$lines = file($variable_ file_name);

Of course, that example won't work. But what will?

[Alvaro G. Vicario]

Re: Use file() function to read a variable name, not a literal name.

*** 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_f ile_name');
>
But, what if I want file() to use a variable name?

From manual:

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

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

$variable_file_ name = "literal_file_n ame";
$lines = file($variable_ file_name);
>
Of course, that example won't work. But what will?

What error are you getting?


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--

[walterbyrd]

Re: Use file() function to read a variable name, not a literal name.


Alvaro G. Vicario wrote:

What error are you getting?
>

This is what I'm using. It is copied almost verbatim from the php.net
examples for using the file() function. One difference: I'm using a
variable ($userfile), instead of a literal. The variable does contain a
file name, I've checked. If I replace the variable, with a literal,
this will work.

// Get a file into an array. In this example we'll go through HTTP to
get
// the HTML source of a URL.
$lines = file($userfile) ;

// Loop through our array, show HTML source as HTML source; and line
numbers too.
foreach ($lines as $line_num =$line) {
echo "Line #<b>{$line_num} </b: " . htmlspecialchar s($line) . "<br
/>\n";
};

Here is the error that I'm getting:

Warning: Invalid argument supplied for foreach() . . .

[walterbyrd]

Re: Use file() function to read a variable name, not a literal name.


Alvaro G. Vicario wrote:

What error are you getting?
>

Opps. I'm sorry to have wasted your time. It was just a syntax error.
I've fixed it.

*sigh* I've been away from programming for too long.