Copying random length part of the string.

Copying random length part of the string. - Printable Version

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Copying random length part of the string. - unbeliever - 12-29-2009

Dear All,

I have following outertext for WebElement:

"Change page: < 1 2 3 4 5 6 7 8 9 10 ... > | Displaying page 1 of 426, items 1 to 10 of 4257."

What I need to do is to grab 4257 out of this string and assign it to a variable. The trickiest part is that number 4257 can be different: it can consist of different amount of digits, for example this value can be 25, 2454 or 10983. So I cannot just assume that it is four digits.

I have a solution in mind:
- pick part of the string "items 1 to 10 of "
- calculate amount of subsequent digits after it
- grab sequence of these digits (value) and assign it to the variable.

Can you please advise me how to code it with examples?

Thanks in advance!

RE: Copying random length part of the string. - cashgiga - 12-29-2009

This can be very easily you can handle:

1.Get the text using GetROProeprty method (Ex. webElement.getroproperty("innertext")) and assign it to a variable.
2.Then split the value of variable using Split function (ex. value = Split(text, ,-1)
3.Then exactvalue = CInt(Ubound(value))
You will get final part of the digit.

RE: Copying random length part of the string. - Saket - 12-29-2009

alternately, you can take trim your value upto the first space(" ") in string

Code:
str = "Change page: < 1 2 3 4 5 6 7 8 9 10 ... > | Displaying page 1 of 426, items 1 to 10 of 4257."

num = mid(str,instrrev(str," "),len(str)-1)

msgbox num

correction - you can take trim your value upto the first space(" ") in string from last

RE: Copying random length part of the string. - unbeliever - 12-29-2009

Thanks a lot Smile

I will try both of suggested methods

RE: Copying random length part of the string. - XeNoMoRpH - 01-04-2010

I tested it and was able to get it to return "4257."
Changing it to the following fixed it for me:

Code:
str = "Change page: < 1 2 3 4 5 6 7 8 9 10 ... > | Displaying page 1 of 426, items 1 to 10 of 4257."

num = mid(str,instrrev(str," "),len(str)-instrrev(str," "))

msgbox num

That returned "4257" for me. Since I'm not entirely sure how VBScript functions work (in terms of memory usage and processing), it might make sense to do:

Code:
position = instrrev(str," ")

num = mid(str,position,len(str)-position)

instead.

RE: Copying random length part of the string. - unbeliever - 01-07-2010

Thanks for this update Smile

This situation is resolved.