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Fix Bug #435: 门诊手术安排：编辑弹窗中"费用类别"字段数据未回显

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根因：getSurgeryScheduleDetail 查询未关联 fin_contract 表，导致 feeType 始终为 null
修复：SQL 中添加 adm_encounter → adm_account → fin_contract 三表关联，取 fc.contract_name AS feeType；
      OpScheduleDto 新增 feeType 字段用于接收映射

Co-Authored-By: Claude Opus 4.7 <noreply@anthropic.com>
This commit is contained in:
赵云
2026-05-12 10:03:50 +08:00
committed by 华佗
parent 39fefe9e80
commit a81e7ec980
2 changed files with 9 additions and 0 deletions
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5
openhis-server-new/openhis-application/src/main/java/com/openhis/web/clinicalmanage/dto/OpScheduleDto.java
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@@ -106,4 +106,9 @@ public class OpScheduleDto extends OpSchedule {
* 创建人名称 * 创建人名称
*/ */
private String createByName; private String createByName;
/**
* 费用类别
*/
private String feeType;
} }

4
openhis-server-new/openhis-application/src/main/resources/mapper/clinicalmanage/SurgicalScheduleAppMapper.xml
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@@ -89,11 +89,15 @@
cs.apply_doctor_name AS apply_doctor_name, cs.apply_doctor_name AS apply_doctor_name,
drf.create_time AS apply_time, drf.create_time AS apply_time,
os.surgery_nature AS surgeryType, os.surgery_nature AS surgeryType,
fc.contract_name AS feeType,
COALESCE(pi.identifier_no, ap.bus_no, '') AS identifierNo COALESCE(pi.identifier_no, ap.bus_no, '') AS identifierNo
FROM op_schedule os FROM op_schedule os
LEFT JOIN adm_patient ap ON os.patient_id = ap.id LEFT JOIN adm_patient ap ON os.patient_id = ap.id
INNER JOIN cli_surgery cs ON os.oper_code = cs.surgery_no AND cs.delete_flag = '0' INNER JOIN cli_surgery cs ON os.oper_code = cs.surgery_no AND cs.delete_flag = '0'
LEFT JOIN adm_organization o ON cs.org_id = o.id LEFT JOIN adm_organization o ON cs.org_id = o.id
LEFT JOIN adm_encounter ae ON ae.id = cs.encounter_id AND ae.delete_flag = '0'
LEFT JOIN adm_account aa ON aa.encounter_id = ae.id AND aa.delete_flag = '0'
LEFT JOIN fin_contract fc ON fc.bus_no = aa.contract_no AND fc.delete_flag = '0'
LEFT JOIN doc_request_form drf ON drf.prescription_no=cs.surgery_no LEFT JOIN doc_request_form drf ON drf.prescription_no=cs.surgery_no
LEFT JOIN ( LEFT JOIN (
SELECT patient_id, identifier_no SELECT patient_id, identifier_no