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  • 1.  using a parameter and trying to save as an lxa is failing

    Employee
    Posted 08-10-2015 05:59

    Note: This was originally posted by an inactive account. Content was preserved by moving under an admin account.

    Originally posted by: dhrobertson

    hi,

    I want to use the movefile function to rename a file. I can do this no problem but when I try to save the graph as an lxa it fails with:

    error.jpg

    there is a previous posting about this here:
    http://community.lavastorm.com/threa...t=error+saving

    the solution here was changing the parameter type to long.

    In my case the parameters are string values. I tried changing to type str with no effect.

    my code is :

    moveFile( {{^filerename^}} + ".csv" , {{^filerename^}} + "_"+ prepareDate + ".csv" )

    I tried:
    moveFile( str({{^filerename^}}) + ".csv" , str({{^filerename^}}) + "_"+ prepareDate + ".csv" )

    but it still failed to save as an lxa.

    any ideas please?

    thanks

    douglas


  • 2.  RE: using a parameter and trying to save as an lxa is failing

    Employee
    Posted 08-10-2015 06:11

    Note: This was originally posted by an inactive account. Content was preserved by moving under an admin account.

    Originally posted by: stonysmith

    Parameters are always processed as strings. There is however, an oddity in the parser that gets confused.
    Simply put your parameters within quotes:

    moveFile( "{{^filerename^}}" + ".csv" , "{{^filerename^}}" + "_"+ prepareDate + ".csv" )


  • 3.  RE: using a parameter and trying to save as an lxa is failing

    Employee
    Posted 08-10-2015 06:19

    Note: This was originally posted by an inactive account. Content was preserved by moving under an admin account.

    Originally posted by: dhrobertson

    as always, thanks Stony, worked.