CODEWITHSUNDEEP
informatica
Vicky
Vicky

Reputation: 1235

How the expression transformation works?

Suppose, I am getting id, marks as input to expression transformation. I am calculating values like this.

ID--------------------------------------Input/Output Port 
MARKS-----------------------------------Input Port
O_RESULT= V_RESULT----------------------Output Port
V_RESULT=IIF(MARKS > 60,"PASS","FAIL")--Variable Port

When i debug this code, Normally it calculates the values in the sequencial order. In above example, i have assigned V_RESULT to O_RESULT before calculating it. Still it is showing right result. Ideally, It should show NULL value.

Can somebody tell me why is it showing correct result?

Is there any setting in informatica for reference values? Does it store any unknown value reference for it and later replace it?

Would be grateful for help.

Upvotes: 1

Views: 5984

Answers (3)

Ritika
Ritika

Reputation: 104

Because in debug mode, you see the last snapshot of each row. Debugger does not show separate rows for calculation and assignment. like in you case one row where o_result=null and v_result= and other row for o_result= and v_result=. But rather debugger will show last snapshot of each row. i.e. whatever o_result and v_result both have values. but if you run the workflow, o_result will not be having any values.

Upvotes: 0

mira
mira

Reputation: 1

because output port is calculated last. Had o_result been variable port,then it should have shown the result you expected.

Upvotes: 0

momobo
momobo

Reputation: 1775

Output ports are evaluated after variable ports. I think this is the reason.

Upvotes: 3

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