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his/sql/迁移记录-DB变更记录/202604251957_check_437_duplicate_charge_items.sql
zhangfei 9c3e603b94 Fix Bug #443: 手术计费:点击签发耗材时异常报错
当手术计费弹窗中点击"签发"耗材时,因耗材的locationId(发放库房)为空导致后端异常。
在DoctorStationAdviceAppServiceImpl.handDevice方法中,当locationId为null时,使用登录用户的科室ID作为默认值,
与NurseBillingAppService中的处理方式保持一致。
2026-05-08 09:14:18 +08:00

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SQL
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-- Bug #437 数据排查:查询手术计费中的重复收费项
-- 用法:在生产环境执行前先在测试环境验证
-- 作者:荀彧
-- 1. 查找手术计费中的重复收费项(同一手术+同一产品产生多条记录)
SELECT
encounter_id AS "就诊ID",
product_table AS "产品表",
product_id AS "产品ID",
service_table AS "服务表",
service_id AS "服务ID",
generate_source_enum AS "生成来源",
COUNT(*) AS "重复次数",
STRING_AGG(id::TEXT, ',') AS "收费项ID列表",
STRING_AGG(status_enum::TEXT, ',') AS "状态列表",
STRING_AGG(total_price::TEXT, ',') AS "金额列表",
STRING_AGG(create_time::TEXT, ',') AS "创建时间列表"
FROM adm_charge_item
WHERE delete_flag = '0'
AND generate_source_enum IN (1, 2) -- 医生开立(1) 或 护士划价(2)
AND product_table IN ('cli_surgery', 'wor_device_request', 'wor_activity_definition')
GROUP BY encounter_id, product_table, product_id, service_table, service_id, generate_source_enum
HAVING COUNT(*) > 1
ORDER BY COUNT(*) DESC;
-- 2. 查找同一手术单号下是否有重复的收费项
SELECT
ci.product_id AS "手术ID",
cs.surgery_no AS "手术单号",
cs.surgery_name AS "手术名称",
COUNT(ci.id) AS "收费项数量",
STRING_AGG(ci.id::TEXT || '(' || ci.status_enum || ')' || '=' || ci.total_price, '; ') AS "收费项详情"
FROM adm_charge_item ci
LEFT JOIN cli_surgery cs ON ci.product_id = cs.id
WHERE ci.delete_flag = '0'
AND ci.product_table = 'cli_surgery'
AND ci.generate_source_enum = 1
GROUP BY ci.product_id, cs.surgery_no, cs.surgery_name
HAVING COUNT(ci.id) > 1
ORDER BY COUNT(ci.id) DESC;
-- 3. 统计重复数据量(用于评估影响范围)
SELECT
generate_source_enum AS "生成来源(1=医生开立,2=护士划价)",
product_table AS "产品表",
COUNT(*) AS "重复组数",
SUM(cnt - 1) AS "多余记录数"
FROM (
SELECT
generate_source_enum,
product_table,
COUNT(*) AS cnt
FROM adm_charge_item
WHERE delete_flag = '0'
GROUP BY encounter_id, product_table, product_id, service_table, service_id, generate_source_enum
HAVING COUNT(*) > 1
) sub
GROUP BY generate_source_enum, product_table
ORDER BY SUM(cnt - 1) DESC;