消除条件中的重复记录

Question

我正在尝试获取具有唯一值的记录.Below是我在使用后获得的输出

VoucherNumber.where(:goodie=> 2917)
[#<VoucherNumber id: 559, vendor_id: 5, voucher_number: "1", trip_id: nil, expiry_date: "2016-03-23 18:30:00", goodie_id: 2917, status: "UN", amount: #<BigDecimal:8f9b7f0,'0.122E3',9(36)>, created_at: "2016-03-04 11:37:22", updated_at: "2016-03-04 11:37:22">, #<VoucherNumber id: 565, vendor_id: 5, voucher_number: "4", trip_id: nil, expiry_date: "2016-03-23 18:30:00", goodie_id: 2917, status: "UN", amount: #<BigDecimal:8faf750,'0.122E3',9(36)>, created_at: "2016-03-04 11:37:22", updated_at: "2016-03-04 11:37:22">, #<VoucherNumber id: 567, vendor_id: 11, voucher_number: "sss", trip_id: nil, expiry_date: "2016-03-22 18:30:00", goodie_id: 2917, status: "UN", amount: #<BigDecimal:8fad7c0,'0.112E3',9(36)>, created_at: "2016-03-04 11:37:22", updated_at: "2016-03-04 11:37:22">, #<VoucherNumber id: 569, vendor_id: 11, voucher_number: "22", trip_id: nil, expiry_date: "2016-03-22 18:30:00", goodie_id: 2917, status: "UN", amount: #<BigDecimal:902d510,'0.112E3',9(36)>, created_at: "2016-03-04 11:37:22", updated_at: "2016-03-04 11:37:22">]..



But expected output is below output
[#<VoucherNumber id: 559, vendor_id: 5, voucher_number: "1", trip_id: nil, expiry_date: "2016-03-23 18:30:00", goodie_id: 2917, status: "UN", amount: #<BigDecimal:8f9b7f0,'0.122E3',9(36)>, created_at: "2016-03-04 11:37:22", updated_at: "2016-03-04 11:37:22"> #<VoucherNumber id: 567, vendor_id: 11, voucher_number: "sss", trip_id: nil, expiry_date: "2016-03-22 18:30:00", goodie_id: 2917, status: "UN", amount: #<BigDecimal:8fad7c0,'0.112E3',9(36)>, created_at: "2016-03-04 11:37:22", updated_at: "2016-03-04 11:37:22">]

有人可以指出我正确的方向吗？

Answer 1

您可以调用uniq方法，确保您不会两次获得相同的记录。

VoucherNumber.where(:goodie=> 2917).uniq

编辑：

VoucherNumber.where(:goodie => 2917).pluck(:vendor_id).uniq

Answer 2

您可以尝试这种方式，但只返回vendor_id

VoucherNumber.where(:goodie=> 2917).uniq.pluck(:vendor_id)

您可以使用goodie和vendor_id uniq

找到所有使用的列。

VoucherNumber.group("goodie,vendor_id")

对于选定的礼物：

VoucherNumber.where(:goodie=> 2917).group(:vendor_id)

Answer 3

使用此代码：

VoucherNumber.where(:goodie=> 2917).group('goodie, vendor_id')

Answer 4

在该结果上尝试group_by vendor_id然后uniq。

VoucherNumber.where(:goodie=> 2917).select("DISTINCT(vendor)")

查询是，

SELECT DISTINCT(vendor) FROM "voucher_numbers" WHERE "voucher_numbers"."goodie" = 2917