3

以下の例によると: 値は A1 にのみ格納され、他のセルは nil を返します。他のマージされたセルから A1'a 値を取得する方法、または単に A1 セルの範囲を確認する方法はありますか?

例

4

2 に答える 2

0

すべてのマージされたフィールドが前と同じ場合、マージされていないフィールドは配列になるはずです

  xlsx = Roo::Excelx.new(__dir__ + "/output.xlsx", { expand_merged_ranges: true })

  parsed = xlsx.sheet(0).parse(headers: true).drop(1)

  parsed_merged = []
    .tap do |parsed_merged|
      parsed.each do |x|
        if parsed_merged.empty?
          parsed_merged << {
            "field_non_merged1" => x["field_non_merged1"],
            "field_merged1"     => [x["field_merged1"]],
            "field_merged2"     => [x["field_merged2"]],
            "field_merged3"     => [x["field_merged3"]],
            "field_merged4"     => [x["field_merged4"]],
            "field_non_merged2" => x["field_non_merged2"],
            "field_non_merged3" => x["field_non_merged3"],
          }
        else
          field_merged1_is_same_as_prev = x["field_non_merged1"] == parsed_merged.last["field_non_merged1"]
          field_merged2_is_same_as_prev = x["field_non_merged2"] == parsed_merged.last["field_non_merged2"]
          field_merged3_is_same_as_prev = x["field_non_merged3"] == parsed_merged.last["field_non_merged3"]

          merged_rows_are_all_same_as_prev = field_non_merged1_is_same_as_prev && field_merged2_is_same_as_prev && field_merged3_is_same_as_prev

          if merged_rows_are_all_same_as_prev
            parsed_merged.last["field_merged1"].push x["field_merged1"]
            parsed_merged.last["field_merged2"].push x["field_merged2"]
            parsed_merged.last["field_merged3"].push x["field_merged3"]
            parsed_merged.last["field_merged4"].push x["field_merged4"]
          else
            parsed_merged << {
              "field_non_merged1" => x["field_non_merged1"],
              "field_merged1"     => [x["field_merged1"]],
              "field_merged2"     => [x["field_merged2"]],
              "field_merged3"     => [x["field_merged3"]],
              "field_merged4"     => [x["field_merged4"]],
              "field_non_merged2" => x["field_non_merged2"],
              "field_non_merged3" => x["field_non_merged3"],
            }
          end
        end
      end
    end
    .map do |x|
      {
        "field_non_merged1" => x["field_non_merged1"],
        "field_merged1"     => x["field_merged1"].compact.uniq,
        "field_merged2"     => x["field_merged2"].compact.uniq,
        "field_merged3"     => x["field_merged3"].compact.uniq,
        "field_merged4"     => x["field_merged4"].compact.uniq,
        "field_non_merged2" => x["field_non_merged2"],
        "field_non_merged3" => x["field_non_merged3"],
      }
    end
于 2020-02-14T13:24:06.297 に答える