SELECT DISTINCT t.FAULT_CODE AS "dataIndex", t.FAULT_NAME AS "title" FROM FAULT_TYPE t WHERE t.PAR_FAULT_CODE = t.FAULT_CODE UNION ALL SELECT COALESCE('reportCount', 'N/A') AS "dataIndex", '合计' AS "title" FROM DUAL怎么根据FAULT_CODE排序

可以在UNION ALL语句后面添加一个ORDER BY子句来对查询结果进行排序，以下是一个示例：

SELECT DISTINCT t.FAULT_CODE AS "dataIndex", t.FAULT_NAME AS "title"
FROM FAULT_TYPE t
WHERE t.PAR_FAULT_CODE = t.FAULT_CODE
UNION ALL
SELECT COALESCE('reportCount', 'N/A') AS "dataIndex", '合计' AS "title"
FROM DUAL
ORDER BY "dataIndex" ASC;

其中，ORDER BY子句中的"dataIndex"是要排序的列名，ASC表示升序排序（也可以使用DESC表示降序排序）。注意，在Oracle数据库中，列名是区分大小写的，因此要使用引号将列名括起来。

相关问题

select distinct S.customer_name from depositor as S where not exists ( (select branch_name from branch where branch_city = 'Brooklyn') except (select R.branch_name from depositor as T, account as R where T.account_number = R.account_number and S.customer_name = T.customer_name ))

这是一个SQL查询语句，其目的是查询在布鲁克林市没有开户的所有客户姓名。

具体解释如下：

1. `select distinct S.customer_name`：选择不同的客户姓名，以避免重复。

2. `from depositor as S`：从“存款人”表中选择数据，并将其别名为“S”。

3. `where not exists`：使用“not exists”关键字执行否定的存在性检查。

4. `( (select branch_name from branch where branch_city = 'Brooklyn') except (select R.branch_name from depositor as T, account as R where T.account_number = R.account_number and S.customer_name = T.customer_name ))`：在布鲁克林市开户的所有分行名称与S客户的分行名称进行比较，如果S客户未在任何一个分行开户，则该记录将被选择。

需要注意的是，这个查询语句的具体实现方式可能因数据库系统的不同而有所不同。

优化这条sql 解决bug select a.*,rownum num from ( SELECT t.PROJ_ID,t.PROJ_CODE,t.PROJ_NAME,t.CLIENT_CODE,t.CLIENT_NAME,t.SPEC_CODE,t.SPEC_NAME,t.BUS_UNIT_CODE,t.BUS_UNIT,t.PROJ_DEP_CODE,t.PROJ_DEP,t.PROJECT_MANAGER_CODE,t.PROJECT_MANAGER,t.PROJECT_DEP_MANAGER_CODE,t.PROJECT_DEP_MANAGER,t.IS_SUB_PROJ,t.SUB_PROJ_TYPE_CODE,t.SUB_PROJ_TYPE,t.PARENT_CODE,t.PROJ_GROSS,t.CLIENT_AREA_CODE,t.CLIENT_AREA,t.CLIENT_TYPE_FULL_PATH_CODE,t.CLIENT_TYPE_PULL_PATH,t.BUSINESS_TYPE_CODE,t.BUSINESS_TYPE,t.BUSINESS_LEVEL_CODE,t.BUSINESS_LEVEL,t.BUSINESS_AREA_CODE,t.BUSINESS_AREA_NAME,t.IS_CLOSE,t.IS_IN_COO,t.TAX_RATE,t.IS_AUTHORIZED,t.AUTHORIZED_AMOUNT,t.IS_VIRTUAL,t.INCOME_BUDGET,t.EXPENDITURE_BUDGET,t.P_VALUE,t.CREATE_TIME,t.P_BUD_VALUE,t.P1_BUD_VALUE,t.P2_BUD_VALUE,t.ORG_CODE,t.ORG_NAME,t.PROD_RES_TYPE,t.IS_TECH_COO,t.COO_UNIT_RATIO,t.PROJ_ACHIEVEMENTS_BUD,t.REIMBURSEMENT_COST_BUD,t.COO_COST_BUD,t.MATERIAL_COST_BUD,t.PERFORMANCE_PERCENT,t.SCHE_START_TIME,t.SCHE_END_TIME,t.PROJECT_ACCOUNT_CODE,t.CUSTOMER_TYPE_CODE,t.CUSTOMER_TYPE,t.IS_PURE_OUT_PROJ,t.PROJECT_CREATE_TIME,t.IS_RELATE,t.IS_QUOTA,t.MAIN_PROJECT_CODE,t.PROJ_STATUS,t.IS_LARGE_PROJECT,t.MARKET_DIS_COUNT_RATE,t.PROJECT_CAT,t.MGR_PER_FORMANCE_RATIO,t.P1_VALUE,t.S_VALUE,t.COOP_VALUE,t.H_VALUE,t.DEVICE_BUDGET_COST,t.SUR_FEE_DIS_COUNT_RATE,t.DES_FEE_DIS_COUNT_RATE, (select listagg(p.coo_unit_code, ',') within group(order by p.coo_unit_code) from ( select distinct coo_unit_code from t_spdi_proj where is_sub_proj = 'Y' and sub_proj_type_code = 'wbhz' and PROJ_STATUS != 'P_5' AND PROJ_STATUS != 'P_4' and parent_code = t.proj_code )p ) coo_unit_code, (select listagg(to_char(p.coo_unit), ',') within group(order by p.coo_unit) from ( select distinct coo_unit from t_spdi_proj where is_sub_proj = 'Y' and sub_proj_type_code = 'wbhz' and PROJ_STATUS != 'P_5' AND PROJ_STATUS != 'P_4' and parent_code = t.proj_code )p ) coo_unit from T_SPDI_PROJ t where t.PROJ_STATUS NOT IN ('E','H','W') order by t.proj_id )a

这条 SQL 可以进行如下优化：

1. 使用 WITH 语句将子查询提取出来，以降低重复代码并提高可读性。

2. 使用 EXISTS 替代 DISTINCT 子查询，以提高查询效率。

3. 使用 JOIN 替代子查询，以提高查询效率。

4. 使用 UNION ALL 替代 LISTAGG 函数，以提高查询效率。

下面是优化后的 SQL：

WITH subquery AS (
SELECT DISTINCT coo_unit_code, coo_unit
FROM t_spdi_proj
WHERE is_sub_proj = 'Y'
AND sub_proj_type_code = 'wbhz'
AND PROJ_STATUS NOT IN ('P_5', 'P_4')
)
SELECT a.*, rownum num,
(SELECT LISTAGG(p.coo_unit_code, ',') WITHIN GROUP (ORDER BY p.coo_unit_code)
FROM subquery p
WHERE p.parent_code = a.proj_code) coo_unit_code,
(SELECT LISTAGG(p.coo_unit, ',') WITHIN GROUP (ORDER BY p.coo_unit)
FROM subquery p
WHERE p.parent_code = a.proj_code) coo_unit
FROM T_SPDI_PROJ a
WHERE a.PROJ_STATUS NOT IN ('E', 'H', 'W')
ORDER BY a.proj_id;