Why Doesn't This Work?

[Dysan]

Hi. The following code, is my attempt of uploading an MP3 file to the server, and then storing the path of the file inside a database. Why doesn't the line of code shown in bold work?

 

if (($_FILES["file"]["type"] == "audio/mpeg3"))
{
  if ($_FILES["file"]["error"] > 0)
  {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br />";
  }
  else
  {
    echo "Upload: " . $_FILES["file"]["name"] . "<br />";
    echo "Type: " . $_FILES["file"]["type"] . "<br />";
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " Kb<br />";
    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br />";

    if (file_exists("upload/" . $_FILES["file"]["name"]))
    {
      echo $_FILES["file"]["name"] . " already exists. ";
    }
    else
    {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file"]["name"];

      $sql="INSERT INTO Music (Path)
      VALUES
[b]      ('"upload/" . $_FILES["file"]["name"]')";[/b]
    }
  }
}
else
{
  echo "Invalid file";
}

[rajivgonsalves]

try this

 

 

    $sql="INSERT INTO Music (Path)
      VALUES
     ('upload/{$_FILES['file']['name']}')";

[Psycho]

The code posted above should work fine, but to answer your question directly the problem was that your quotes were not properly implemented.

 

$sql="INSERT INTO Music (Path)

       VALUES ('"upload/" . $_FILES["file"]["name"]')";

 

The first double quote starts the string. Then your second double quote ends the string. Then you make the opposite error. When appending the variable tot he string you should be outside the quotes, but then you append "text" to the end of the variable without first containing it within quotes.

 

Although the above example will work, here is what I expected you were trying to accomplish:

 

$sql="INSERT INTO Music (Path)
      VALUES ('upload/" . $_FILES['file']['name'] . "')";

[aschk]

My favourite method these days is the following :

 

$sql="INSERT INTO Music (Path) VALUES ('upload/{$_FILES['file']['name']}')";

 

Using {$varname}. N.B. this ONLY works with double quotes (").

 

Just thought i would throw it into the mix...

[Dysan]

How do I create a script, that outputs the file type e.g. "image/gif"?

As the file type I have to upload mp3's "audio/mpeg3".

[aschk]

mime_content_type($filename) will give you what you want.

Please bear in mind this is deprecated now and have been replaced by finfo_file http://uk3.php.net/manual/en/function.finfo-file.php