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#1 2010-05-04 07:52:48

Timo
Member

bug: script variable not substituted

I have the following script:

define var_pers = 3267 CHAR

SELECT * FROM import_arbeidsrelatie WHERE iar_pers_nr = $(var_pers)

after pressing F5 I get a popup with the given value in the script. After pressing ENTER
the variable is not substituted.

When I retyp the value in the popup and pressing ENTER the value does get substituted.

What I expect is that the value should be substituted without retyping the value because
the value is already given in the popup. So why should I retyp it.

Regards. Timo

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#2 2010-05-04 07:53:51

Timo
Member

Re: bug: script variable not substituted

PS the version is 3.45 PE

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#3 2010-05-04 21:49:25

sqldbxhelp
Administrator

Re: bug: script variable not substituted

Yes, it seems there's some inconsistency in how script variables work.
This will be addressed in a next release.

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