如何使用 genie 编程语言打印 sqlite table 内容

Question

根据此处之前的问题，我设法 the dataset, 列出了所有食谱，现在我正尝试从该列表中选择一个食谱并显示其标题、说明和成分。指令通过 pkID 列映射到 Recipes，成分通过 recipeID 列映射到 Recipes。当我在 Sqlite 数据库浏览器上打开数据库时，我可以在表下拉列表中访问此信息，所以我想它们的正确名称是数据库中的表。

我无法通过 pkID 和食谱 ID "filter"，因此在选择一个食谱后，只会显示适当的内容。

这是 Python 中我想在 Genie 中执行的代码：

  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 -> ')

我的很多代码都没有改进，看来我之前使用的在步骤语句中迭代的方法不能在这里使用。这是我在 Genie 中的进展：

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" ))

Answer 1

我找到了解决问题的方法，也许可以优化一下。现在足够了。

另一个的回答帮助很大。我使用的解决方案是使用 exec 函数并将回调指向 PrintSingleRecipe()。

必须进行一些调整才能将其用作回调，但我得到了我需要的东西。

调用函数的代码如下：

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."

这是 PrintSingleRecipe 的样子：

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

如何使用 genie 编程语言打印 sqlite table 内容

How to print a sqlite table content with genie programming language

sqlite

genie