Regexp and Decimals 1

[vladk]

Hi, how with Regular Rexpressions extract numbers from a string? Including decimals. For example:

"Blallaal $3.4566 UI<><>" should be just 3.4566
"345 hjkhehj###" shuld be 345

Thanks

vladk

 

[tsuji]

How about this?
[tt] .pattern="\d+(\.\d+){0,1}"
[/tt]

 

[vladk]

tsuji,

I tried this code with your pattern, and it did not do the job:

With objRegExp
.Global = True
.MultiLine = True
.IgnoreCase = True
.Pattern = "d+(\.\d+){0,1}"
strReturn = .Replace(pstrText, vbNullString)
End With

vladk

 

[tsuji]

There is a slash leading...
> .Pattern = "d+(\.\d+){0,1}"
[tt] .Pattern = "[red]\[/red]d+(\.\d+){0,1}"[/tt]

 

[vladk]

tsuji,

Hi, thank you for the responce. I actually need to get the numbers, I need to see them.

vladk

 

[tsuji]

>I actually need to get the numbers, I need to see them
>strReturn = .Replace(pstrText, vbNullString)

This is about it. Try it?

[tt]if .test(pstrText) then
strReturn = .execute(pstrText)(0)
else
strReturn = ""
end [/tt]

 

[tsuji]

Mine >end should be read[tt] end if[/tt].

 

[vladk]

tsuji,

Works great!!!

Thank you!!!

vladk

 

[strongm]

>Works great!!!

Works sufficiently, I suspect you mean.

For example, many European countries use . as a date seperator, e.g.

Today is 28.06.2006

which the regexp given above would identify as a decimal 28.06

 

[tsuji]

Also, if the number appears as ".1234" The leading "." lost significance as part of a number and treated semantically as the period of some statment not decimal separator.

I guess there's always thing to improve... But after seeing a working instance, op could build upon it and would soon be better than anybody else on a particular configuration of interest to his work.

 

[vladk]

Ok, works sufficiently great.

 

[BobRodes]

Very funny, vlad.