PHP添加文件路径到数据库

我已经把这个脚本放在一起，从我上传的文件加载文件路径到数据库。但这似乎不起作用。任何建议都很好。它基本上是一个简单的表单，允许多个文件上传。我需要将此信息发送到数据库的文件路径，以供以后使用。我得到一个输出警告这是一个简单的测试，用于检查变量是否实际张贴

echo $sql;

 INSERT INTO mediamanagement ( `Project_Name`, `Assigned_To`, `Assign_Date`, `Check_Date`, `Due_Date` ) VALUES ( "fvfg df fdh bdfgb", "Ramon", "2013-04-01", "2013-04-18", "2013-04-30", Error:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 13
        Error:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 13

php:

<?php
mysql_connect("MySQLB15.wer.com","york","usa12") or die ('Error:' .mysql_error());
//database connection
    mysql_select_db("mediamanagement");
$project = $_POST['project'];
$assignto = $_POST['assignto'];
$asdate = $_POST['asdate'];
$chdate = $_POST['chdate'];
$ddate = $_POST['ddate'];

$errors = array();
$files = array();
foreach ($_FILES['files'] as $k=>$image) {
    // handle upload errors
    if ($image['error'] != 0 && $image['error'] != 4) {         
        switch ($image['error']) {
            case '1':
            case '2':
                $err = 'The uploaded file exceeds the maximum file size.';
                break;                  
            case '3':
                $err = 'The upload was inturupted, transfer failed.';
                break;
            case '6':
            case '7':
            case '8':
                $err = 'Server error.  Please try again later.';
                break;
        }
        // record error and move on
        $errors[] = array('files'=>$k, 'error'=>$err);
        continue;
    } elseif ($image['error'] == 4) {
        // error 4 means no image was sent
        continue;
    }
    // determine the extension
    $ext = explode('.', $image['name']);
    if (count($ext) != 2) {
        $errors[] = array('files'=>$k, 'error'=>'Could not determine file extension.');
        continue;
    } else {
        switch ($ext[1]) {
            case 'jpg':
            case 'jpeg':
            case 'gif':
            case 'png':
            case 'pdf':
            case 'psd':
            case 'ai':
            case 'pdf':

                break;
            default:
                $errors[] = array('files'=>$k, 'error'=>'Unsupported file extension.');
                continue;
                break;
        }
    }
    // make a random-ish filename
    $filename = time().uniqid(rand(), true) . '.' . $ext[1];
    $path = 'uploads/'.$filename;   // upload directory path is set
    move_uploaded_file($image['tmp_name'], $path);     //  upload the file to the server
    // this is a bad idea right here! Use 775 at least, if possible
    chmod($path,0775);
    $files[] = array('name'=>$filename, 'path'=>$path);
}
// now loop the $files array and put the paths into the database
// you also should do something with the errors listed in $errors
// start building up the SQL query, start with
// some fields that are straightforward

$sql = '
    INSERT INTO mediamanagement (
       `Project_Name`,
        `Assigned_To`,
        `Assign_Date`,
        `Check_Date`,
        `Due_Date`';
// now loop the list of files (5 only), 
// add each needed field
for ($i=1; $i < count($files) && $i < 5; $i++) {
    $sql .= '`files'.$i.'`,';
}
// build out the rest of the query, add values
// for the straightforward fields
$sql .= '
) VALUES (
    "'.$project.'", 
    "'.$assignto.'",
    "'.$asdate.'",
    "'.$chdate.'",
     "'.$ddate.'",
';

// loop the files
$ct = 1;
foreach ($files as $f) {
    $sql .= '"'.$f['name'].'",';
    // only allow 5 files
    if ($ct == 5)
        break;
    $ct++;
}
 ')';


mysql_query($sql) or die ('Error:' .mysql_error());;


?>
<?php
echo("<p><span>Project Name:</span> ".$_POST['project']."</p>");
echo("<p><span>assign to:</span> ".$_POST['assignto']."</p>");
echo("<p><span>Assign Date:</span> ".$_POST['asdate']."</p>");
echo("<p><span>Check Date:</span> ".$_POST['chdate']."</p>");
echo("<p><span>Due Date:</span> ".$_POST['ddate']."</p>");

?>