errtile Not Working Correctly

[Scott]

Hi,

I have a lisp routine, in it, the user picks a size from a list, upon pressing OK on the size dialog, the routine calls another routine. This second routine is responsible for ensuring that the user has selected a size. I use the errtile tag in my DCL file to output an error message to the dialog. I have this same setup throughout my routine, with the only difference being the name of the fitting. The size checking routine is called properly, and it does work, the dialog reappears if an invalid size is chosen. The only trouble is that I do not have the message displayed on the dialog box. Here is the size checking routine:

(defun numcheck
(if (or (= number 0) (= number nil))
(progn
(setq runner 0)
(set_tile "error" "Invalid Size")
)
(progn
(setq runner 1)
(done_dialog)
)
)
)

Are there any obvious errors?

TIA

Scott

 

[Rudy Tovar]

Not without reading the entire code.

I'm not seeing the whole picture nor the main dialog.

 

[Scott]

I found the problem, it seems that, because of the way things are done with setting the size, number, the variable that numcheck checks is always nonzero.

I know a way that it can be fixed, the trouble is can I do this? This routine works by, having the user select a size from the list box in the dialog, pressing OK on the dialog box sets the index number of the selected size to the variable number. So if the user chose the first size in the list box number = "0", number is converted from a string to a real, then 1 is added to number. So number now = 1 From that, using a conditional statement, I set another variable called rodia# equal to another variable, in this case ar1#, where the the 1 is number. So say the user choose the 4th size, number would be set to 5 and then rodia# would be set to ar5#.

The ar# variables are (in lisp):

(setq ar3# 4.00)

(action_tile "reddia" "(setq number $value)")
;reddia is a listbox in the dialog, and number is it's index value ;Other bits of the reduce dialog here;

;** Assume that, for this example the user choose the second size down in the list **; So number is set to "2"

(setq number (atof number))
(setq number2 (rtos (+ 1 number) 2 0))
(setq thevar (strcat "ar" number2 "#"))

Now thevar = "ar3#"

(setq rodia# thevar)

Hopefully you can see what I am trying to do, I want to set rodia# to the predefined variable ar3# or = 4.00, but rodia# will = "ar3#" is there a way to make this work, so that I can use a variable to call a variable?

Thanks Scott

 

[Rudy Tovar]

Sorry for not getting back to you sooner Scott.

I didn't have a chance to review the methods you've provided.

And I really need to see the dialog and complete code.

I understand if you don't want to post this, so I you'd like please email it to me directly.

I found the problem, it seems that, because of the way things are done with setting the size, number, the variable that numcheck checks is always nonzero.

I know a way that it can be fixed, the trouble is can I do this? This routine works by, having the user select a size from the list box in the dialog, pressing OK on the dialog box sets the index number of the selected size to the variable number. So if the user chose the first size in the list box number = "0", number is converted from a string to a real, then 1 is added to number. So number now = 1 From that, using a conditional statement, I set another variable called rodia# equal to another variable, in this case ar1#, where the the 1 is number. So say the user choose the 4th size, number would be set to 5 and then rodia# would be set to ar5#.

The ar# variables are (in lisp):

(setq ar3# 4.00)

(action_tile "reddia" "(setq number $value)")
;reddia is a listbox in the dialog, and number is it's index value ;Other bits of the reduce dialog here;

;** Assume that, for this example the user choose the second size down in the list **; So number is set to "2"

(setq number (atof number))
(setq number2 (rtos (+ 1 number) 2 0))
(setq thevar (strcat "ar" number2 "#"))

Now thevar = "ar3#"

(setq rodia# thevar)

Hopefully you can see what I am trying to do, I want to set rodia# to the predefined variable ar3# or = 4.00, but rodia# will = "ar3#" is there a way to make this work, so that I can use a variable to call a variable?

Thanks Scott

Not without reading the entire code.

I'm not seeing the whole picture nor the main dialog.

displayed
on

 

[Rudy Tovar]

I hope I followed you correctly

(action_tile "list1" "(return_list_value)")

(defun return_list_value (/ val)
(setq val (1+ (atoi $value)))
(cond
((= val 1)(Do_this))
((= val 2)(Do_that))
)
(princ)
)
;;In either case you don't have to add a one, just a condition.

Sorry for not getting back to you sooner Scott.

I didn't have a chance to review the methods you've provided.

And I really need to see the dialog and complete code.

I understand if you don't want to post this, so I you'd like please email it to me directly.

I found the problem, it seems that, because of the way things are done with setting the size, number, the variable that numcheck checks is always nonzero.

I know a way that it can be fixed, the trouble is can I do this? This routine works by, having the user select a size from the list box in the dialog, pressing OK on the dialog box sets the index number of the selected size to the variable number. So if the user chose the first size in the list box number = "0", number is converted from a string to a real, then 1 is added to number. So number now = 1 From that, using a conditional statement, I set another variable called rodia# equal to another variable, in this case ar1#, where the the 1 is number. So say the user choose the 4th size, number would be set to 5 and then rodia# would be set to ar5#.

The ar# variables are (in lisp):

(setq ar3# 4.00)

(action_tile "reddia" "(setq number $value)")
;reddia is a listbox in the dialog, and number is it's index value ;Other bits of the reduce dialog here;

;** Assume that, for this example the user choose the second size down in the list **; So number is set to "2"

(setq number (atof number))
(setq number2 (rtos (+ 1 number) 2 0))
(setq thevar (strcat "ar" number2 "#"))

Now thevar = "ar3#"

(setq rodia# thevar)

Hopefully you can see what I am trying to do, I want to set rodia# to the predefined variable ar3# or = 4.00, but rodia# will = "ar3#" is there a way to make this work, so that I can use a variable to call a variable?

Thanks Scott

Not without reading the entire code.

I'm not seeing the whole picture nor the main dialog.

Hi,

I have a lisp routine, in it, the user picks a size from a list, upon pressing OK on the size dialog, the routine calls another routine. This second routine is responsible for ensuring that the user has

selected

a message
to invalid
size

 

[Scott]

Thank you for your input, I just realized that I was doing the error checking before I had set any values!

Thanks,
Scott

I hope I followed you correctly

(action_tile "list1" "(return_list_value)")

(defun return_list_value (/ val)
(setq val (1+ (atoi $value)))
(cond
((= val 1)(Do_this))
((= val 2)(Do_that))
)
(princ)
)
;;In either case you don't have to add a one, just a condition.

Sorry for not getting back to you sooner Scott.

I didn't have a chance to review the methods you've provided.

And I really need to see the dialog and complete code.

I understand if you don't want to post this, so I you'd like please

email

it
to me directly.

then

1

is added to number. So number now = 1 From that, using a conditional statement, I set another variable called rodia# equal to another variable, in this case ar1#, where the the 1 is number. So say the user choose the 4th size, number would be set to 5 and then rodia# would be set to ar5#.

The ar# variables are (in lisp):

(setq ar3# 4.00)

(action_tile "reddia" "(setq number $value)")
;reddia is a listbox in the dialog, and number is it's index value ;Other bits of the reduce dialog here;

;** Assume that, for this example the user choose the second size down in the list **; So number is set to "2"

(setq number (atof number))
(setq number2 (rtos (+ 1 number) 2 0))
(setq thevar (strcat "ar" number2 "#"))

Now thevar = "ar3#"

(setq rodia# thevar)

Hopefully you can see what I am trying to do, I want to set rodia# to the predefined variable ar3# or = 4.00, but rodia# will = "ar3#" is there

a
way

to make this work, so that I can use a variable to call a variable?

Thanks Scott

Not without reading the entire code.

I'm not seeing the whole picture nor the main dialog.

Hi,

I have a lisp routine, in it, the user picks a size from a list, upon pressing OK on the size dialog, the routine calls another routine. This second routine is responsible for ensuring that the user has

selected

a
size. I use the errtile tag in my DCL file to output an error message to the dialog. I have this same setup throughout my routine, with

the
only

difference being the name of the fitting. The size checking

routine
is

called properly, and it does work, the dialog reappears if an invalid size is chosen. The only trouble is that I do not have the message displayed on the dialog box. Here is the size checking routine:

(defun numcheck
(if (or (= number 0) (= number nil))
(progn
(setq runner 0)
(set_tile "error" "Invalid Size")
)
(progn
(setq runner 1)
(done_dialog)
)
)
)

Are there any obvious errors?

TIA

Scott