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FindFile - return file path in expression

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What I'm trying to do is relatively simple, but I can't seem to figure it out.


I want to get the absolute path to a file contained in the TestStand search directories so that I can use the absolute path in the code module (to load a configuration file). I need to specify a bunch of name/path pairs in the code module.


I know that the FindFile function does this, but the path is returned as an output parameter, rather than the return value. Is there some clever way to get the output parameter returned in a single expression?



Something equivalent to this:



Edit: I just tried specifying the file name as the path (in case TestStand automatically expanded the paths) but that didn't work.

LabVIEW Champion, CLA, CLED, CTD
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Accepted by topic author Sam_Sharp



Figured it out, this was the expression I used:

(FindFile("EXS_21xxxx.inf",True,Locals.FilePath,2,2)) ? Locals.FilePath : ""

(Obviously you have to create Locals.FilePath as a local variable)


Not sure if there's a cleaner/neater way (feel free to post if there is) but this worked for me.

LabVIEW Champion, CLA, CLED, CTD
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