val a string
 

Hi,

z3 = (Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") + 1)))
creates a string, which is 1.96%

I want to value it so I changed the formula to

z3 = val((Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") + 1))))
but it does not work

Any ideas?

Thanks,
David

val a string
 

Got it.

Thanks

"David" wrote:

Hi,

z3 = (Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") + 1)))
creates a string, which is 1.96%

I want to value it so I changed the formula to

z3 = val((Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") + 1))))
but it does not work

Any ideas?

Thanks,
David

val a string
 

David,

For the benefit of others, it would be a nice gesture to post your solution
as opposed to posting "Got it."

"David" wrote in message
...
Got it.

Thanks

"David" wrote:

Hi,

z3 = (Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") + 1)))
creates a string, which is 1.96%

I want to value it so I changed the formula to

z3 = val((Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") +
1))))
but it does not work

Any ideas?

Thanks,
David

val a string
 

Hi,

The formula I endeup with is:
z3 = Val((Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") + 2))))
/ 100

I could not get Excel to recognize the string a anything other than text
with the % sign it it, so I shortened the Mid to extract one less character.
Then it recognized the number, but not as a %, so it is divided as by 100 to
yield somthing like .07, which was then simply formated correctly.

Thanks,
David

"Skiffle" wrote:

David,

For the benefit of others, it would be a nice gesture to post your solution
as opposed to posting "Got it."

"David" wrote in message
...
Got it.

Thanks

"David" wrote:

Hi,

z3 = (Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") + 1)))
creates a string, which is 1.96%

I want to value it so I changed the formula to

z3 = val((Mid(z, (InStr(1, z, "(") + 1), Len(z) - (InStr(1, z, "(") +
1))))
but it does not work

Any ideas?

Thanks,
David