CommiC Posted October 5, 2003 Posted October 5, 2003 I've made a script that combines two different scripts that both only perform a find. In the develop environment this works great. But after compiling the DB I find that only the first find script is executed and the second one is omitted. I use FM 6P and the developer of FM 5.5 Does anyone have a clue? Thanks, John
kenneth2k1 Posted October 6, 2003 Posted October 6, 2003 That's kinda like "my car wont start, do you know why that is?" You'll have to tell us what the steps are in your find. One thing I think that 5.5 doesn't have the Perform Find/Replace step so it might be ignored if a 6 file is bound with 5.5. So check for stuff like that. Ken
CommiC Posted October 9, 2003 Author Posted October 9, 2003 The script I'm talking about is this simple script: Go to Layout [ #05 - Rap_upd_gebruikt ] [ Refresh window ] Show All Records Enter Find Mode Set Field [ LOK_InGebruik, "ja" ] Perform Find [ Replace Found Set ] Enter Find Mode Set Field [ Gewijzigd, "ja" ] Perform Find [ Constrain Found Set ] It does a search on two fields and works fine in development environment but only the results of the second search are shown when run as executable.
DonH Posted October 10, 2003 Posted October 10, 2003 Doesn't REPLACE FOUND SET do exactly that? Replace what was just found? If so, then of course you would only see the 2nd results.
CommiC Posted October 10, 2003 Author Posted October 10, 2003 As [ Replace Found Set ] is the parameter of the first find, so is [ Constrain Found Set ] the parameter of the second. How hard can that be. Still gives me no clue why this gives only the results of the second in a compiled db.
kenneth2k1 Posted October 10, 2003 Posted October 10, 2003 How bout just: Enter Find Mode [] Set Field [LOK_InGebruik, "ja"] Set Field [ Gewijzigd, "ja" ] Perform Find That is essentially what you would get down to anyways, right? KEn
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