Help on a db query please.

Hi,
I am getting a value submitted from a form.

$val= $_POST['field'];
Then putting it in an array.
$arr= explode("/", $val);

Now I have to open a db and check if this value in that field already exists or not. if not, insert it.
This is what I am not getting.

mysql_connect(localhost,username,password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM $tablename WHERE field='$cat'";
$result=mysql_query($query);
$num=mysql_numrows($result);
mysql_close();

I know that I need to run a foreach loop to automatically check for each item in $arr but I am completely confused by the number of records in $num now.

My question:
What code needs to be put in FOREACH loop now.
It has to check all records selected for the previous query with the current value in $arr.
If a record with $arr value is found, do nothing or else insert a record.

foreach ($arr as $value)
{

}

Thank you for any help here.

 

foreach ($arr as $value)
{
$query="SELECT * FROM $tablename WHERE field='" . $value . "'";
$result=mysql_query($query);
$num=mysql_numrows($result);
if ($num == 0) {
//Insert
}
}

PHP:

 

Hi,
Thanks so much.
It was easy just had to let go the pressure.
Thanks again.
Here's what I came up with. ( a little longer than what you gave but doesn't involve querying the db over and over)
----------------
Code in first post here.

foreach ($arr as $value)
{

$i=0;
$c= "1";
while ($i < $num) {
$f = mysql_result($result,$i,"field");
if ($f == $value)
$c= "2";

++$i;
}

if ($c== "1")
{
INSERT here
}
}
-----------------
Bye