合并相同 patient_id 的 JSON 数据为数组

问题

select patient_id,concat('{"itemText":"',item_text,'","itemValue":"',item_value,'"}') from hs_patient_groups where active = 1;

eef41128c47c401abb7f8885a5f9fbdf	{"itemText":"旧","itemValue":"2"}
eef41128c47c401abb7f8885a5f9fbdf	{"itemText":"新","itemValue":"1"}
0564ccc4c07b4df98bf8207f7084b493	{"itemText":"患者1组","itemValue":"patientGroup1"}
e92acabeb871422dacd11199df89bbc3	{"itemText":"患者1","itemValue":"患者1"}
af2b9bb388af484fb99c9d03cc2041ca	{"itemText":"新","itemValue":"1"}
5e4ff36bebd7478892978f875f48f515	{"itemText":"患者1","itemValue":"患者1"}
344035419b3a4c4e8b29d76c627679f9	{"itemText":"患者2","itemValue":"患者2"}
bc266205ca604150b3d800aab059fba8	{"itemText":"患者2","itemValue":"患者2"}
1801a4524b084d2b8197bfcc33849b24	{"itemText":"患者1","itemValue":"患者1"}
1801a4524b084d2b8197bfcc33849b24	{"itemText":"患者2","itemValue":"患者2"}
73cd39179ab147e6968858aecd56f214	{"itemText":"冠心病组","itemValue":"02"}
d54ced0ccbba4bb2b1d3359877fb0620	{"itemText":"瓣膜组","itemValue":"01"}
3c92057be370414b822407b1ca89ad1b	{"itemText":"瓣膜组","itemValue":"01"}
591883a377d6428f985bdb99b762929b	{"itemText":"瓣膜组","itemValue":"01"}
2b1b8ac08c5d4ac39625c20bd206ed00	{"itemText":"患者2","itemValue":"患者2"}
19822b6d38504d1a8af94dbfa32c2c0e	{"itemText":"患者2","itemValue":"患者2"}
ab62ee95526e405e93f117a25db77eee	{"itemText":"瓣膜组","itemValue":"01"}
181e27e744e242719a47d5a86260eeb1	{"itemText":"冠心病组","itemValue":"02"}
e1d862aa6baa474987545776ad8e84db	{"itemText":"冠心病组","itemValue":"02"}

如何将相同patient_id，不同concat的值合并为一行

合并相同 patient_id 的 JSON 数据为数组

要将相同 patient_id 的不同 JSON 数据合并为一行（合并为一个 JSON 数组），你可以使用 PostgreSQL 的 json_agg 或 jsonb_agg 聚合函数。

解决方案（推荐）

SELECT 
    patient_id,
    json_agg(json_build_object('itemText', item_text, 'itemValue', item_value)) AS items
FROM 
    hs_patient_groups 
WHERE 
    active = 1
GROUP BY 
    patient_id;

结果示例

执行上述查询后，结果将类似于：

patient_id                           | items
-------------------------------------+--------------------------------------------------------------------
eef41128c47c401abb7f8885a5f9fbdf    | [{"itemText":"旧","itemValue":"2"}, {"itemText":"新","itemValue":"1"}]
0564ccc4c07b4df98bf8207f7084b493    | [{"itemText":"患者1组","itemValue":"patientGroup1"}]
1801a4524b084d2b8197bfcc33849b24    | [{"itemText":"患者1","itemValue":"患者1"}, {"itemText":"患者2","itemValue":"患者2"}]
...                                  | ...

其他选项

1. 如果你需要保持原始查询中的 JSON 字符串格式：

SELECT 
    patient_id,
    json_agg(concat('{"itemText":"',item_text,'","itemValue":"',item_value,'"}')::json) AS items
FROM 
    hs_patient_groups 
WHERE 
    active = 1
GROUP BY 
    patient_id;

2. 如果你使用的是 PostgreSQL 9.4 或更早版本（没有 json_agg）：

SELECT 
    patient_id,
    '[' || string_agg(concat('{"itemText":"',item_text,'","itemValue":"',item_value,'"}'), ',') || ']' AS items
FROM 
    hs_patient_groups 
WHERE 
    active = 1
GROUP BY 
    patient_id;

注意：最后一个方法生成的是 JSON 字符串而不是真正的 JSON 类型。