I want to separate out the octets of an IP address using Formulas.

I have tried some things like substitute & find but cannot figure this out.

Example of what I want to achieve, by only starting with Cell `A1`

and Cell `B1`

:

```
10.17.9.192 | 192.168.0.1
10 | 192
17 | 168
9 | 0
192 | 1
```

Here are four formulas you need

```
A2:=LEFT(A1,FIND(".",A1)-1)
A3:=MID(A1,FIND(".",A1)+1,FIND(".",A1,FIND(".",A1)+1)-FIND(".",A1)-1)
A4:=MID(A1,FIND(".",A1,FIND(".",A1)+1)+1,FIND(".",A1,FIND(".",A1,FIND(".",A1)+1)+1)-(FIND(".",A1,FIND(".",A1)+1)+1))
A5:=MID(A1,FIND(".",A1,FIND(".",A1,FIND(".",A1)+1)+1)+1,LEN(A1)-FIND(".",A1,FIND(".",A1,FIND(".",A1)+1)+1))
```

The FIND function has a third argument named *start*. So to find the second decimal point, you FIND a decimal point but you start where the first decimal point is plus one. It would be like this

```
=FIND(".",A1,4)
```

You start in 4 because your first decimal point is in position 3. But you don’t know that, so you have to compute the ‘4’.

```
=FIND(".",A1,FIND(".",A1)+1)
```

Now to get 4, we find the first decimal and add 1. That’s not too bad to find the second. But to find the third, you have to go another level. And the fourth is yet another level. It’s hard to read and hard to maintain.

To make it a little easier, you can use helper columns.

```
A2 =LEFT(A1,C2-1)
C2 =FIND(".",A1)
A3 =MID(A1,C2+1,C3-C2-1)
C3 =FIND(".",A1,C2+1)
A4 =MID(A1,C3+1,C4-C3-1)
C4 =FIND(".",A1,C3+1)
A5 =MID(A1,C4+1,LEN(A1)-C4-1)
```

That way you’re doing your FINDs in C and referring to those numbers in A.

If you don’t like helper columns, and I don’t, you could write a UDF like

```
Public Function FINDi(find_text As String, within_text As String, Optional instance As Long) As Long
Dim lReturn As Long
Dim i As Long
Const lFINDFIRST As Long = 0
If instance = lFINDFIRST Then
lReturn = InStr(1, within_text, find_text)
ElseIf instance < lFINDFIRST Then 'negative numbers finds last
lReturn = InStrRev(within_text, find_text)
Else
lReturn = 0
For i = 1 To instance
lReturn = InStr(lReturn + 1, within_text, find_text)
Next i
End If
FINDi = lReturn
End Function
```

and that gives you formulas like this

```
A2 =LEFT(A1,findi(".",A1)-1)
A3 =MID(A1,findi(".",A1)+1,findi(".",A1,2)-findi(".",A1,1)-1)
A4 =MID(A1,findi(".",A1,2)+1,findi(".",A1,3)-findi(".",A1,2)-1)
A5 =MID(A1,findi(".",A1,3)+1,LEN(A1)-findi(".",A1,3)-1)
```

Not as clean as the helper column, but self contained and definitely better than the built-in FIND.

Another UDF you could write duplicates what VBA’s split function does.

```
Public Function SplitString(ByVal sInput As String, ByVal sDelim As String, ByVal lWhich As Long) As String
SplitString = Split(sInput, sDelim)(lWhich - 1)
End Function
```

That formula looks like

```
A2 =SplitString($A$1,".",ROW()-1)
A3 =SplitString($A$1,".",ROW()-1)
A4 =SplitString($A$1,".",ROW()-1)
A5 =SplitString($A$1,".",ROW()-1)
```

### Answer：

Here is classic one formula solution:

```
=TRIM(MID(SUBSTITUTE(A$1,".",REPT(" ",999)),(ROW()-1)*999-998,999))
```

`"."`

– is the delimiter.

`(ROW()-1)`

– gives nth item in delimited text.

More info at EXCELFOX

### Answer：

As shown in question,

row 1are the IP addresses (IPv4), starting from A1.

```
| A | B
--+-------------+-------------
1 | 10.17.9.192 | 192.168.0.1
2 | 10 | 192
3 | 17 | 168
4 | 9 | 0
5 | 192 | 1
```

## Solution

**1st part**(cell A2):`=VALUE( LEFT(SUBSTITUTE(A1, ".", " "), 3 ))`

**2nd part**(cell A3):`=VALUE( MID(SUBSTITUTE(A1, ".", " "), 8, 5 ))`

**3rd part**(cell A4):`=VALUE( MID(SUBSTITUTE(A1, ".", " "), 15, 7))`

**4th part**(cell A5):`=VALUE(RIGHT(SUBSTITUTE(A1, ".", " "), 3 ))`

Fill the formulas to the right.

Note: You can use `TRIM(...)`

instead of `VALUE(...)`

if you want the result in text format.

## Bonus: Single formula to get formatted IP address

We can get a **formatted IP address** `001.002.003.004`

by the following formula.

```
= TEXT( LEFT(SUBSTITUTE(A1, ".", " "), 3 ), "000") & "."
& TEXT( MID(SUBSTITUTE(A1, ".", " "), 8, 5 ), "000") & "."
& TEXT( MID(SUBSTITUTE(A1, ".", " "), 15, 7), "000") & "."
& TEXT(RIGHT(SUBSTITUTE(A1, ".", " "), 3 ), "000")
```

## Explanation

By `SUBSTITUTE`

ing the dot `.`

with 6 spaces, we get:

```
|123456789|123546789|123456789|
1.1.1.1 -> 1 1 1 1
11.11.11.11 -> 11 11 11 11
111.111.111.111 -> 111 111 111 111
=1= ==2== ===3===
```

- Character 1-3 contains and only contains the first part.
- Character 8-12 contains and only contains the second part.
- Character 15-21 contains and only contains the third part.
- Rightmost 3 characters contains and only contains the fourth part.

Tags: excelexcel