I have the following table with duplicate family IDs but different number of family members:

```
tbl_family

+------------+--------------+---------+
| familyID   | Members      | Location|
+------------+--------------+---------+
|      100   | 3            |    xyz  |
|      100   | 4            |    xyz  |
|      101   | 1            |    abc  |
|      101   | 2            |    abc  |
|      102   | 5            |    efg  |
|      103   |              |    hij  |
+------------+--------------+---------+

```

I also have a second table where we verified the correct count of family members for the duplicates

```
tbl_verifier

+------------+--------------+---------+
| familyID   | Members      | Location|
+------------+--------------+---------+
|      100   | 3            |    xyz  |
|      101   | 2            |    abc  |
+------------+--------------+---------+

```

I want to create a view in mysql which will display the families without duplicates and maintain the row with the verified count of family members. The results should look as follows:

```
tbl_results
+------------+--------------+---------+
| familyID   | Members      | Location|
+------------+--------------+---------+
|      100   | 3            |    xyz  |
|      101   | 2            |    abc  |
|      102   | 5            |    efg  |
|      103   |              |    hij  |
+------------+--------------+---------+

```

I am breaking the problem into several steps. I want to first select all those with matching Members then those with null Members

```
/* Step 1: Select only those that are matching family members count in 
verifier and family */

select *
from tbl_family f
inner join
tbl_verifier v
ON f.familyID = v.familyID
WHERE f.Members = v.Members;

/* Step 2 : Select only those that have null  number of rooms*/

select *
from tbl_family f
left join
tbl_verifier v
ON f.familyID = v.familyID
WHERE f.Members is null

```

Now am a bit stuck on how to proceed further.