Question

如果下面的一些代码是多余的，我很抱歉，但我想描绘一下我想要做的一切。

我的意图是

从名为pickupLocations的json数组中选择一个元素。每pickupLocations有许多id_referral。 请参阅下面名为raw_addresses的CTE。和
使用字符串聚合函数（例如string_agg）为每个id_referral创建一个分隔的地址字段。

换句话说，我希望能够从

#id_referral    pickup_addresses
#1  4265 Hillsdale Ave. NE, Grand Rapids, MI 49525
#1  3060 Cheney Ave. NE, Grand Rapids, MI 49525
#2  805 Kendalwood St. NE, Grand Rapids, MI 49505
#2  711 Edgewood St. NE, Grand Rapids, MI 49505

在我的raw_addresses CTE

到

#id_referral    pickup_addresses
#1  4265 Hillsdale Ave. NE, Grand Rapids, MI 49525 | 3060 Cheney Ave. NE, Grand Rapids, MI 49525
#2  805 Kendalwood St. NE, Grand Rapids, MI 49505 | 711 Edgewood St. NE, Grand Rapids, MI 49505

在我的最终选择中。

但是，我现在得到以下结果

#id_referral    pickup_addresses
#1  4265 Hillsdale Ave. NE, Grand Rapids, MI 49525 | 4265 Hillsdale Ave. NE, Grand Rapids, MI 49525
#2  805 Kendalwood St. NE, Grand Rapids, MI 49505 | 805 Kendalwood St. NE, Grand Rapids, MI 49505

使用以下代码时。

 WITH raw_addresses AS (
         SELECT sr.id AS id_referral,
        --parsing a json array which has many pickupLocations for a single id_referral
            json_array_elements(sr."pickupLocations") -> 'pickupLocationAddress' AS pickup_address
           FROM "ServiceReferrals" sr
        )
--want to roll the pickup addresses into a single pipe-delimited field (willing work through an array too as shown with array_agg, but same problem there)
 SELECT raw_addresses.id_referral,
    string_agg(cast(pickup_address as varchar(100)), '|') AS pickup_addresses
   FROM raw_addresses
  GROUP BY raw_addresses.id_referral

出于某种原因，string_agg函数正在重复每个id_referral的第一个值。我也尝试了agg_array并获得相同的行为。

任何关于为什么会被赞赏的想法。

Answer 1

您可以尝试使用中间CTE：

WITH raw_addresses AS (
     SELECT sr.id AS id_referral,
    --parsing a json array which has many pickupLocations for a single id_referral
        json_array_elements(sr."pickupLocations") -> 'pickupLocationAddress' AS pickup_address
       FROM "ServiceReferrals" sr
), transformed_addresses AS (
    SELECT id_referral, 
    cast(pickup_address as varchar(100)) AS pickup_address
    FROM raw_addresses
)
SELECT id_referral,
string_agg(pickup_address, '|') AS pickup_addresses
FROM transformed_addresses
GROUP BY id_referral

第一个结果在postgres中的array_agg和string_agg函数中重复

1 个答案: