Recent content by jdgibson

Looking at the original problem you described I would suspect this is a locking issue even though you say you can see no locks. For sysmptons where killing a session reactivates the progam it is hard to think of anything else. It is however likely to be not the data you are viewing in the...

The appserver agent is not properly started until the startup procedure completes with no error. If your repeat loop is in the startup procedure presumably the startup procedure does not complete so your appserver agent hasn't actually started. As it hasn't started this may explain why you can't...

on "\\" anywhere do: return no-apply. end. will result in nothing happening if the user presses \ You can change anywhere to something more specific if you want.

You could try passing in an empty string password which is the 5th parameter to the open method. Instead of checking error status try checking if the workbook handle is a valid handle which it won't be if the workbook hasn't opened

You would also need a case for the empty string, I forgot to test for that.

FUNCTION rev RETURNS CHAR(INPUT ll AS CHAR). IF LENGTH(ll) = 1 THEN RETURN ll. ELSE RETURN rev(SUBSTRING(ll,2)) + SUBSTRING(ll,1,1). END. DISP rev("Yohn").

Although I have hardly used smart objects I seem to recall that version 8 smart objects are significanlty different from version 9 and a conversion utility is provided in version 9 on the pro tools menu to convert v8 smart objects to v9 smart objects. If you have not done this it may explain...

The last-event system handle will give you the x and y cooridinates of the last mouse click. last-event:x last-event:y is the syntax. It can give you a few other things as well

You are confusing the OPSYS function with the OPSYS preprocessor name. If you want to use it with &IF it should be {&OPSYS}. If you want to use OPSYS it should If OPSYS

If you are using an internal procedure the record will be available in all susequent calls in the first buffer you used whether or not you find the record again or define a new buffer. The original record will also be changed if you update it to whatever it was updated to in the last procedure...

The code do while causes the entire update to be one transaction. If you change it to repeat while it will be one transaction for each iteration of the loop and you will no longer have the lock table overflow problem

Unless I am missing something I think you are mistaken as otherwise it is a remarkable conincidence that it failes for a-n and works for p-z and the string you are comparing with begins " O" that happens to be right in the middle of a-n and p-z in the alphabet. The comparison is alphabetic and...

Of course it matters where a varibale is defined, it must be before it is used and definined in such a way thats its use is within scope. Your code example doesn't work because at the line c = "hello" the variable c isn't in the compilers symbol table because the compiler hasn't encountered it...

Variables defininitions are processed at compile time with an entry being created in the compilers symbol table for the the variable when it encounters it in the code and memory being reserved for it in the compiled code. So in theory and practice your if statement makes no differance to the...

Although I do not know Java I think you are making a mistake between some sort of array reference structure with properties mprops(0).Name where 0 is an array reference and the offset from the start position of a block of memory PUT-STRING(mprops,1) = where 1 is position 1 in the...