HELP needed with php codes

[jass]

In the following php code to embed wmv file in a webpage, is it possible to list more than one wmv file (so that I do not have to create one php file for each very one wmv file)? Would appreciate if someone could help. Thank you.

<?php
$file = $_REQUEST["file"];

if ($file == "something") {
$open = 'trial.wmv';
}

else {
$open = ' ';
}

Header("Content-type: video/x-ms-wmv");

echo $open;
?>

[boxxertrumps]

<?php
$file = $_REQUEST["file"];
//info about music files is stored
$farray (
one=>a.wmv,
two=>b.wmv,
three=>c.wmv
)
// file not (!=) null ("")
if ($file != "") {
$open = $farray[$file];
}

else {
$open = ' ';
}

Header("Content-type: video/x-ms-wmv");

echo $open;
?>
the url would be request.php?file=(aray key name, one/two/three ect.)

[jass]

Thanks for your input but there is an error here:
Parse error: parse error, unexpected T_DOUBLE_ARROW on line 5

[thetestingsite]

$farray (
one=>a.wmv,
two=>b.wmv,
three=>c.wmv
)

Anything in the array must have quotes around it, or it will cause errors. Also, at the end of the array, you must have a semicolon. All of the corrections to the above quoted code are posted below in red.

$farray (
'one' => 'a.wmv',
'two' => 'b.wmv',
'three' => 'c.wmv'
);

Hope this helps.

[jass]

Thanks for your input but the same error message still appear:
Parse error: parse error, unexpected T_DOUBLE_ARROW on line 5

[thetestingsite]

Did you copy and paste the code that boxxertrumps posted, or did you just add that to a code that you already had? If you copied and pasted, line 5 is the first line of the array. Otherwise, we would need to see the code in order to find a solution.

[jass]

I copied and pasted boxxertrumps' code and replaced the one which you corrected:

<?php
$file = $_REQUEST["file"];
//info about music files is stored
$farray (
'one' => 'a.wmv',
'two' => 'b.wmv',
'three' => 'c.wmv'
);
// file not (!=) null ("")
if ($file != "") {
$open = $farray[$file];
}

else {
$open = ' ';
}

Header("Content-type: video/x-ms-wmv");

echo $open;
?>

[djr33]

Hmm... looks ok to me, not sure, though.

But.... umm... echo $open means echo 'a.wmv', not actually output the file...

There are several other functions in php that can do this.

For example--
echo file_get_contents($open);

However, I think there may be a more preferable function, but not sure.


Also, $file = $_REQUEST["file"]; is overkill. That leaves it open to cookies, post and get data, when you only need get (URL, like ...?var=value). (cookies are stored text files to set preferences for a user, etc., and post data is from a form (get can be too, but it is then sent as part of the url, not hidden, like post).
$file = $_GET["file"]; would do just fine.
REQUEST isn't really hurting, but I don't see any reason it would help.

[jass]

Thank you very much for your input, I greatly appreciate it.

[thetestingsite]

Ok, so I found the reason for your error message that appears with the script. The original is below and the correction is below that in red.

<?php
$file = $_REQUEST["file"];
//info about music files is stored
$farray (
'one' => 'a.wmv',
'two' => 'b.wmv',
'three' => 'c.wmv'
);
// file not (!=) null ("")
if ($file != "") {
$open = $farray[$file];
}

else {
$open = ' ';
}

Header("Content-type: video/x-ms-wmv");

echo $open;
?>

Correction:

<?php
$file = $_REQUEST["file"];
//info about music files is stored
$farray = array(
'one' => 'a.wmv',
'two' => 'b.wmv',
'three' => 'c.wmv'
);
// file not (!=) null ("")
if ($file != "") {
$open = $farray[$file];
}

else {
$open = ' ';
}

Header("Content-type: video/x-ms-wmv");

echo $open;
?>

I completely overlooked that the code had an array, but the array was never started. Hope this helps.