Expression syntax in TestStand 4.0 tutorial

Hello All
Teststand is new to me, and I am trying its tutorial with "Using TestStand" manual.
At the Using TestSTand page 3-7, Insert a Template Step step 7, we insert a
message popup "Retry if Previous Step Fails". This step has post action "Goto Step",
but it doesn't goto next step and uses an expression "RunState.PreviousStep.TS.Id".
I can't find the expression at the expression browser and don't know what this means.
There is a RunState.PreviousStep at the expression browser but no "TS.Id".
I've attached a pic (teststand_ques.jpg) to show the expression. Could anyone
tell me its meaning.
Also, is there any manual or guide of TestStand expression syntax? I know expression
browser shows its operator and functions, but I'd like to have a detail reference manual.
Thank you
Attachments:
teststand_ques.JPG ‏214 KB

Hi,
The TS is a hidden property of the Step.
You can unhid this property by a setting in the Station Options select the Preference Tab.
You actually dont need to enable this setting, you could of just typed it in (, the .TS.id I mean). But when learning it is handy to see all the properties.
Regards
Ray Farmer
Regards
Ray Farmer

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Hi,
The TS is a hidden property of the Step.
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CASE
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I feel uncomfortable providing support on an unsupported feature. BEA
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appropriateness, nor forward compatibility of any unsupported features.
That being said, what you have looks basically correct although it is
subject to change at any time without notice.
- Neil
Ziad wrote:
I grabbed the policy expression from several roles and tried to immitate it. I came
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Grp(name) or {Usr(name)}
{Grp(name)}
{Grp(name1)|Grp(name2)}
usr(name) or {usr(name)}
combinations of the above
Is that about right?
Neil Smithline <[email protected]> wrote:
Unfortunately, due to timing constraints we could not get this syntax to
a state that we were comfortably making it public. We are trying to
address this in a future release.
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I'm trying to trick SDL into identifying words that are not approved by STE.
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I have all other options in the Utility unchecked.
I am by no means a pro at regular expressions but was able to create a pretty solid command at http://regexlib.com/RETester.aspx.
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I tried your expression in a couple of regex tools and it seems to parse as you wanted it to. I suspect that the SDL implementation doesn't follow the unix/linux standards. I haven't used the tool and the usage documentation is non-existant, except for the limited flash-based demo.
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+"If the ELSE clause is omitted, the system substitutes a default action. For a CASE statement, the default when none of the conditions matches is to raise a CASE_NOT_FOUND exception. For a CASE expression, the default is to return NULL."+
Read For more info on Case Expression and Case Statement
Edited by: Purvesh K on Dec 13, 2012 12:26 PM

Expression syntax

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Message Edited by AllenP on 08-30-2007 01:04 PM

Hi,
The TS is a hidden property of the Step.
You can unhid this property by a setting in the Station Options select the Preference Tab.
You actually dont need to enable this setting, you could of just typed it in (, the .TS.id I mean). But when learning it is handy to see all the properties.
Regards
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Regards
Ray Farmer

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Okay, my HTML is bit rusty. here it is.
So, I ran into this problem that when I put a continued inequality expression (a<x<b) in a Pass/Fail step, it will always pass. This is very easy to duplicate, say you have a local varible x = 3, in a Pass/Fail test step's Data Source Expression you put "0<=Locals.x<=2". Then it will always pass when you run it. If you set b to less than 1, say 0.5, then it will always fail. So I reported this as a bug to NI support. Here is what I got for explanation:
NI support/application engineer:
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me:
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NI support/application engineer:
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me:
"Do you really want me to spell this out to you?! I would feel embarrassed if I were you. But you insisted, so here it goes.
You are correct that these operators are treated the same in any programming language. Here is where you got lost, read this again - "just as a*x*b would find a*x, then apply that result to {result}*b, a<x<b calculates the result of a<x (1 or 0) and applies that result to {result}<b."
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The answer is that "{result}<b is just NOT the same as x<b". Do you see the problem now?
Do you know the difference between a Boolean and an integer? 1 and 0 can be used to present a Boolean, but when you plug it in an integer comparison and it will be used as an integer which throws the whole comparison out of water. I don't think that any other programming language would do that."
NI support/application engineer:
"I am very sorry that you were not satisfied with my explanation of the problem. I understand that this result is not desirable, but the underlying reason is that TestStand does not understand that you are trying to compare a<x and x<b in the same statement. The continued inequality is an implied conjunction of two comparisons, and TestStand (and other programming languages) do not know how to interpret this implication. I realize that this is not intuitive, and though I would agree that a<x<b seems like it would return 1 if x is between a and b and 0 otherwise, this is not a recognized syntax by TestStand and other languages."

Hi CT,
I agree with you completely that this may be undesired behavior depending upon exactly what code is being written. Thanks for bringing to our attention any behavior that seems undesired, we always want to know about these types of things!
I ran the pseudo code "a<x<b" where a, x, and b are numbers in a few different environments:
VB6: This returns true for any x value.
C#: This does not compile, due to a type mismatch.
C++: This returns true for any x value.
C: This returns true for any x value.
LabVIEW: Using comparison functions, this results in a broken wire due to a type mismatch
LabVIEW: Using a MathScript Node, this returns true for any x value.
I would say, generally speaking, it is accepted behavior to
type cast a Boolean result to an integer when the next function
requires integer parameters. In every language, using the logical AND operator, as Jigg suggested, removes any chance of ambiguity between the coder and the compiler.
The good news is that the product suggestion that the support engineer filed on your behalf will still be reviewed by R&D. So, whether this is a bug or a feature request, it will get its due attention.
Message Edited by Evan P. on 10-15-2008 08:25 PM
Evan Prothro
RF Systems Engineer | NI

Report Builder 3.0 Expression i am not able to make it work on Oracle DB Table

Hi,
I am newbie to SSRS Report Builder 3.0 so please be patient with me. I am trying to create a report to query Remedy Ticketing System and it uses Oracle Database installed on a Unix OS...My end objective is to do some calculation on the ticket metrics (i.e.
percentage of tickets closed within SLA, etc) however i am stuck from the very beginning where i am not able to make my expression work...
To begin with. my query statement...
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the data type of the oracle table:
Here is the Query Result of the DataSet:
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Ticket_Status 2 - Means Pending Ticket
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i also used the expression below but it did not work...
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i converted the Ticket_Status to CHAR on the dataset using the TO_CHAR(Ticket_Status) AS Ticket_Status then i tried the expression below but did not work...
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Hi Danny,
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but not 180, right?
I have tested on my local environment and can reproduce the issue, the issue caused by you are using the count() function in the incorrect way, please modify the expression as below and have a test:
=COUNT(IIF(Fields!TICKET_STATUS.Value=4 ,1,Nothing))
or
=SUM(IIF(Fields!TICKET_STATUS=4,1,0))
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Regards,
Vicky Liu
Vicky Liu
TechNet Community Support

Expression syntax in TestStand 4.0 tutorial

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