I just needed to look at my sql a bit closer
SELECT * FROM $table_name
needed to be
SELECT $table_name.* FROM $table_name
It is now showing the id of each of the 153 records
--- In php-objects@xxxxxxxxxxxxxxx, "Ken" <ken.poyser@...> wrote:
>
> I have gotten the SQL query to
> get me the result that there were
> 153 records. My problem is how to
> show me the unique ID of each of the
> 153 records that were not matched
> from both tables...
>
> Like I stated, I have the SQL inner join query
> working just trying to display each of the
> 153 records that weren't in both tables
>
> Do I have my While statement messed up?
>
>
>
>
> //build and issue query
> $table_name = "BaseMissingPerson";
> $table_name1 = "Arch_1";
> $sql = "SELECT * FROM $table_name left join $table_name1 on
> $table_name.NIC = $table_name1.NIC where isnull($table_name1.NIC)";
> $result = @mysql_query($sql,$connection) or die(mysql_error());
> $num = mysql_num_rows($result);
>
>
> while ($row=mysql_fetch_array($result,MYSQL_ASSOC)) {
> $ncount++;
> $nic = $row['$table_name1.NIC'];
> $nam = TRIM($row['NAM']);
> $oca = trim($row['OCA']);
> $mke = $row['MKE'];
>
> echo "<br>$nic--$nam--$oca--$mke--$ncount";
>
> }
>
>
> Any assistance would be appreciated
>
>
> Ken
>
