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Variable search string for Objects.Names
- This topic has 2 replies, 2 voices, and was last updated 12 years, 2 months ago by John Gerhardy.
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AuthorPosts
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October 22, 2012 at 4:13 pm #12788John GerhardyParticipant
Hello,
I would like to have the search string for the Objects.Names property. Is it possible to use the indirect operator ($$) to accomplish this?
Example:
Dim TransscriptNames = \Messdaten\.Objects(“^P1.*\.fld$”).Namesgive me the names of the folders starting with “P1”, which I would like to be variable, so I tried this:
Dim SearchString = “^” : “P1” : “.*\\.fld$” //”^sig.*\.dat$”
Dim TransscriptNames = \Messdaten\.Objects($SearchString$).Namesbut this results in a “Variable not declared” error.
Syntax error or no can do?
Regards,
jgerhardyOctober 22, 2012 at 4:13 pm #8446John GerhardyParticipantHello,
I would like to have the search string for the Objects.Names property. Is it possible to use the indirect operator ($$) to accomplish this?
Example:
Dim TransscriptNames = \Messdaten\.Objects(“^P1.*\.fld$”).Namesgive me the names of the folders starting with “P1”, which I would like to be variable, so I tried this:
Dim SearchString = “^” : “P1” : “.*\\.fld$” //”^sig.*\.dat$”
Dim TransscriptNames = \Messdaten\.Objects($SearchString$).Namesbut this results in a “Variable not declared” error.
Syntax error or no can do?
Regards,
jgerhardyOctober 26, 2012 at 1:54 pm #9253Bernhard KantzParticipantThe Objects.Names property uses a string as select argument. This can be a constant string like “^P1.*\.FLD$” or a variable containing such a value.
If for example someone wants to specify the folder by a varying number, a code fragment likeDim no = 1 Dim select = "^P" : String no : ".*\.FLD$" dim folderNames = \Messdaten\.Objects(select).Names
lists all folders under the root folder Messdaten prefixed with ‘P’ and the number.
If one uses names like “P001”, “P002”, … to distinguish between the different objects, the following expression constructs these strings from a numeric value:Dim no = 1 Dim objName = "P" : StringMid(String (no + 1000), 1)
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