I am using the following formula to yield a 1 if a CS number exists in the cell, and a 0 if it does not. The nested formula works correctly.

The issue is with the outside IF function. What happens is that when it does not find a CS number, it yields `#VALUE!` in the cell. When that happens, I want the cell to say 1.

Using the function below, it still is filling in `#VALUE!` when a CS number is not found.

```
=IF(IF(ISERR(NUMBERVALUE(MID(K2,SEARCH("CS",K2)+M39,1))),MID(K2,SEARCH("CS",K2),9),MID(K2,SEARCH("CS",K2),10)) = "#VALUE!", 1, 0)

```

 Is there a more appropriate way to reference this to yield 1 instead?
['https://Stackoverflow.com/questions/47405426', 'https://Stackoverflow.com', 'https://Stackoverflow.com/users/5389099/']
Ironically, *`#VALUE!` isn't a Value*, so you can't use a comparison function like `If` to deal with it as if it's a string.

​

Change your formula to:

```
=IFERROR(IF(ISERR(NUMBERVALUE(MID(K2,SEARCH("CS",K2)+M39,1))),MID(K2,SEARCH("CS",K2),9),MID(K2,SEARCH("CS",K2),10)),1)

```

...and it will return a `1` instead of `#VALUE!` if the text you're searching for isn't found.

​

* [Here is a handy guide](http://www.exceltactics.com/definitive-guide-excel-error-types-error-handling/) to Excel formula errors and how to handle them:

​
Did you try

```
=IFERROR( your formula here, 1 )

```

Basically, it evaluates your formula as usual, but when *anything* in your formula returns an error, you will get an 1 in return.

Using this idea, you can wrap each part of your formula in an IFERROR to define the default value to return whenever it encounters one. Kinda wasteful though.