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Fix Bug #497: 【住院医生工作站-检查申请】检查申请列表缺失"申请单状态"列及全流程闭环状态流转逻辑

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根因：SQL 查询使用 CASE MIN(wsr.status_enum) 计算状态，但聚合函数 MIN() 出现在 WHERE 子句中，
PostgreSQL 语法错误导致状态筛选时查询失败。且计算状态仅映射 5 种值（缺少"待接收"=3、"已出报告"=6）。

修复：改为直接使用 doc_request_form.status 字段（数据库已存在该列），
SELECT 和 WHERE 均使用 drf.status，支持完整 0-7 状态流转。

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
荀彧
2026-05-14 05:18:58 +08:00
parent 1c87c39473
commit 02f2a14178
1 changed files with 2 additions and 20 deletions
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22
openhis-server-new/openhis-application/src/main/resources/mapper/regdoctorstation/RequestFormManageAppMapper.xml
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@@ -13,16 +13,7 @@
drf.requester_id,
drf.create_time,
ap.NAME AS patient_name,
CASE MIN(wsr.status_enum)
WHEN 1 THEN 0
WHEN 2 THEN 1
WHEN 3 THEN 4
WHEN 4 THEN 4
WHEN 5 THEN 5
WHEN 6 THEN 5
WHEN 7 THEN 5
ELSE NULL
END AS status
drf.status
FROM doc_request_form AS drf
LEFT JOIN adm_encounter AS ae ON ae.ID = drf.encounter_id
AND ae.delete_flag = '0'
@@ -40,16 +31,7 @@
AND drf.create_time &lt;= (#{endDate}::date + INTERVAL '1 day' - INTERVAL '1 second')
</if>
<if test="status != null and status != ''">
AND CASE MIN(wsr.status_enum)
WHEN 1 THEN 0
WHEN 2 THEN 1
WHEN 3 THEN 4
WHEN 4 THEN 4
WHEN 5 THEN 5
WHEN 6 THEN 5
WHEN 7 THEN 5
ELSE NULL
END = #{status}::integer
AND drf.status = #{status}::integer
</if>
<if test="keyword != null and keyword != ''">
AND (drf.prescription_no ILIKE '%' || #{keyword} || '%'