How to print a sqlite table content with genie programming language

问题

Based on previous questions here I managed to create the dataset, print all recipes listed and now I am trying to pick one of the recipes from that list and show its Title, Instructions and Ingredients. The instructions are mapped to the Recipes via the pkID column and the ingredients are mapped to the Recipes through a recipeID column. When I open the database on Sqlite Database Browser I can access this information inside the Tables dropdown list, so I suppose the proper name for them are tables within the database.

I am not being able to "filter" by pkID and by recipeID, so that after picking one recipe, only the appropriate content is shown.

This is the code in Python of what I am trying to do in Genie:

  def PrintSingleRecipe(self,which):
    sql = 'SELECT * FROM Recipes WHERE pkID = %s' % str(which)
    print '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'
    for x in cursor.execute(sql):
      recipeid =x[0]
      print "Title: " + x[1]
      print "Serves: " + x[2]
      print "Source: " + x[3]
    print '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'
    sql = 'SELECT * FROM Ingredients WHERE RecipeID = %s' % recipeid
    print 'Ingredient List:'
    for x in cursor.execute(sql):
        print x[1]
    print ''
    print 'Instructions:'
    sql = 'SELECT * FROM Instructions WHERE RecipeID = %s' % recipeid
    for x in cursor.execute(sql):
      print x[1]
    print '~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'
    resp = raw_input('Press A Key -> ')

I have not been able to improve much of my code, it seems that using the approach I used before of iterating in a step statement cannot be used here. This is how far I got in Genie:

def PrintSingleRecipe(db:Database)
    stmt:Statement = PreparedStatements.select_all( db )
    res:int = UserInterface.raw_input("Select a recipe -> ").to_int()
    cols:int = stmt.column_count ()
    var row = new dict of string, string
    item:int = 1
    print "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"
    while res == ROW
        for i:int = 0 to (cols - 1)
            row[ stmt.column_name( i ) ] = stmt.column_text( i )
        stdout.printf( "%-5s", item.to_string( "%03i" ))
        stdout.printf( "%-30s", row[ "Title" ])
        stdout.printf( "%-20s", row[ "Serves" ])
        stdout.printf( "%-30s\n", row[ "Source" ])
    print "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"
    print "Ingredient list"
    print " "
    stdout.printf("%-5s", item.to_string( "%03i" ))

回答1:

I have found a solution to the problem, maybe it can be optimized. For now it is enough.

Answers from another question helped immensely. The solution I used was to use the exec function and point the callback to the PrintSingleRecipe().

Some adjustments had to be done for it to work as a callback, but I got what I needed.

Here is the code where the function gets called:

while true
    response:string = UserInterface.get_input_from_menu()
    if response == "1" // Show All Recipes
        PrintAllRecipes(db)
    else if response is "2" // Search for a recipe
        pass
    else if response is "3" //Show a Recipe
        res:string = UserInterface.raw_input("Select a recipe -> ")
        sql:string = "SELECT * FROM Recipes WHERE pkID = " + res
        db.exec(sql, PrintSingleRecipe, null)
    else if response is "4"//Delete a recipe
        pass
    else if response is "5" //Add a recipe
        pass
    else if response is "6" //Print a recipe
        pass
    else if response is "0" //Exit
        print "Goodbye"
        break
    else
        print "Unrecognized command. Try again."

Here is how the PrintSingleRecipe looks like:

def PrintSingleRecipe(n_columns:int, values:array of string, column_names:array of string):int
    print "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"
    for i:int = 0 to n_columns
        stdout.printf ("%s = %s\n", column_names[i], values[i])
    print "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~"
    print "Ingredient list"
    print " "
    return 0

来源：https://stackoverflow.com/questions/33446085/how-to-print-a-sqlite-table-content-with-genie-programming-language

标签

sqlite

genie