Ruby如何格式化嵌套哈希

Question

我有两个查询正在运行和迭代，我的最终哈希如下所示。但是，我想对数据如何存储在我正在创建的散列中或在我完成创建后格式化它进行格式化。但我不确定如何实现所需的格式，其中names 属于相同的id，如下所示

示例数据的所需格式：

[
    {
        id: 1,
        accepted: false,
        trans: 10234
        names: [
            { name: "Joe", amount: "$1,698.00" },
            { name: "Smith", amount: "$674.24" },
        ]
    },
    {
        id: 2,
        accepted: true,
        trans: 10234,
        names: [
            { name: "Joe", amount: "$1,698.00" },
            { name: "Smith", amount: "$674.24" },
        ]
    }
]

我目前的格式

[
    {
               :id => 1,
               :accepted => false,
               :trans => 8,
               :name => "Smith",
               :amount => 36.0
    },
    {
               :id => 1,
               :amount => false,
               :trans => 8,
               :name => "Joe",
               :amount => 6.0
    },
    {
               :id => 3,
               :accepted => false,
               :trans => 8,
               :name => "Tom",
               :amount => 34.0
    },
     {
               :id => 3,
               :accepted => false,
               :trans=> 8,
               :name => "Martha",
               :amount => 4.0
    }
], 
[
    {
               :id => 2,
               :accepted => true,
               :trans => 7,
               :name => "Bob",
               :amount => 35.0
    },
     {
                :id => 2,
                :accepted => true,
                :trans => 7,
                :name => "John",
                :amount => 5.0
    }
]

创建哈希的逻辑

imports = ListImports.limit(20).order(created_at: :DESC)
groups = imports.map{|import| ListImportGroup.where(list_import_id: import.id)}
pub_hash_true = []
pub_hash_false = []
hash = []
imports.map do |import|
  hash << {
     id: import.id,
     trans: import.trans,
     accepted: import.amount
  }
end
  hash.each do |import|
    groups.flatten.each do |group|
      accepted = import[:accepted]
      num_transactions = import[:trans]
      if accepted == false
        pub_hash_false << {id: import[:id], accepted: accepted, trans: num_transactions, name: group.name, amount: group.amount}
      else
        pub_hash_true << {id: import[:id], accepted: accepted, trans: num_transactions, name: group.name, amount: group.amount}
      end
    end
  end

Answer 1

# Note: You didn't specify what is the association between `ListImport` and `ListImportGroup`.
# However, I'm fairly sure you could be fetching this data via a JOIN query like below,
# rather than making up to 20 additional database calls to fetch the associated records.

imports = ListImports.limit(20).order(created_at: :DESC).includes(:list_import_group)

result = imports.map do |import|
  {
    id: import.id,
    trans: import.trans,
    accepted: import.amount,
    names: import.list_import_groups.pluck(:name, :amount)
  }
end

如果您确实需要过滤 accepted 是 true 或 false 的导入，您可以执行以下操作，而不是手动构建单独的数组：

accepted_imports = result.select { |import| import[:accepted] }
# and
rejected_imports = result.reject { |import| import[:accepted] }

# or even:
accepted_imports, rejected_imports = result.partition { |import| import[:accepted] }

Answer 2

您没有指定 desired 和 current 格式之间的确切对应关系。 但我认为

• 对于具有相同id的条目，accepted和trans的值是相同的。
• 当前格式中 Joe 的所需金额与 所需金额 中的相应金额相同。 （在您的示例中，前者是6.0，而后者是"$1,698.00"，这没有意义。）

然后，以下将进行转换。数组ahout 是所需的格式。

 # Let us assume "a1" is the original array in the "current format"
hout = {}
a1.flatten.map{|h|
   h.slice(*(%i(id trans name amount accepted))).values
}.each{ |a|
  hout[a[0]] = {id: a[0], accepted: a[4], trans: a[1], names: []} if !hout.key? a[0]
  hout[a[0]][:names].push(
    {name: a[2], amount: "$"+helper.number_with_precision(a[3], precision: 2, delimiter: ',')}
  )
}

ahout = hout.values

如果您愿意，可以对ahout 进行排序。

请注意，我假设您使用的是 Rails 5+。否则，helper 方法可能不起作用。在这种情况下，您可以使用sprintf 或任何格式化方法。