HELP needed with php codes

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;
?>

<?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.)

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

Quote:

---

Originally Posted by boxxertrumps

$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.

Code:

---

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

---

Hope this helps.

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

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.

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;
?>

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.

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

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.

Code:

---

<?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:

Code:

---

<?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.

djr33, it could be a feature of the format that it allows a playlist to be passed instead of actual video data, like RAM does.

sorry about the shoddy array...
this is the product of my tiny attention span...

Many thanks to both thetestingsite and boxxertrumps for your kind input.
I have just noticed that IE7 prompts one to save the media file when the URL is entered on the address bar :
http://domain.com/request.php?file=one

Is there a way to prevent this?