I posted an alternative solution in this topic.

I am also posting the formulas here in order to

- acknowledge that this topic was the inspiration for my solution
- highlight aspects of my refactoring that would have cluttered up the other topic
- make it easier to continue this conversation (where the formula writers are)

While this solution was inspired by W_Vann_Hallās post, the formulas are actually quite different.

- This solution does not need a field to count the number of items in the list, which reduces the complexity of the formulas. (The initial
`IF`

statement is not needed.)
- It uses a combination of
`LEFT`

and `RIGHT`

functions, instead of the `MID`

function. When building the functions, it was easier for me to mentally picture using the `RIGHT`

function to identify everything remaining in the string, and then using the `LEFT`

function to pick off the first item.
- It also reduces the overall amount and complexity of code by
- Using the
`CONCATENATE`

function instead of summing multiple `LEN`

functions.
- Subtracting a single number that increases for each item, rather than repeatedly adding 2 for each previous item.

Here is a screen shot of the sample base.

Here is the formula for the field {Item1}.

```
IF( FIND(",", {List}),
LEFT({List},
FIND(",", {List})-1
),
{List}
)
```

Here is the formula for the field {Item2}:

```
LEFT(
RIGHT({List},
LEN({List}) - LEN({Item1}) - 2
),
FIND( ",",
RIGHT({List},
LEN({List}) - LEN({Item1}) - 2
) & ","
)-1
)
```

Here is the formula for the field {Item3}:

```
LEFT(
RIGHT({List},
LEN({List}) - LEN(CONCATENATE({Item1},{Item2})) -4
),
FIND( ",",
RIGHT({List},
LEN({List}) - LEN(CONCATENATE({Item1},{Item2})) -4
) & ","
)-1
)
```

Here is the formula for the field {Item4}:

```
LEFT(
RIGHT({List},
LEN({List}) - LEN(CONCATENATE({Item1},{Item2},{Item3})) -6
),
FIND( ",",
RIGHT({List},
LEN({List}) - LEN(CONCATENATE({Item1},{Item2},{Item3})) -6
) & ","
)-1
)
```

If you need more items, continue the pattern used for {Item3} and {Item4}:

- add the next item to both of the
`CONCATENATE`

functions
- subtract two more from the results of the
`LEN`

functions (e.g., for the 5th item, subtract 8; for the 6th item subtract 10, etc.) The number to subtract is determined by `(2 x the item number) - 2`

. (Note you cannot put this in the actual formula because the item number is not a field.)