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#1
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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 |
#2
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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 |
#3
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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 |
#4
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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 |
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