Причина этой ошибки?

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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 'll find a job description example for most common jobs. Starting with a sampl' at line 1

Что я уже пробовал:

$photo=$_REQUEST["imagename"];
$email_id=$_REQUEST["email"];
$jtitle=$_REQUEST["jtitle"];
$loc=$_REQUEST["loc"];
$region=$_REQUEST["region"];
$jtype=$_REQUEST["jtype"];
$des=$_REQUEST["des"];
//$cdate = $_REQUEST["date"];
//echo''.$cdate;
$date = date('Y-m-d');
$c =new mySqli("localhost","root","","rjobs");
$sql="insert into post_job_db value('$date','$photo','$email_id','$jtitle','$loc','$region','$jtype','$des')";
if($c->query($sql)==TRUE)
{
echo "Record inserted<script>window.location.assign('home.php');</script>";
}
else
{
echo mysqli_error($c);
//echo "sorry<script>window.location.assign('post_job.php');</script>";
}
$c->close();

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