SqlKeyword.COUNT_DISTINCT 是什么意思

SqlKeyword.COUNT_DISTINCT 是一个SQL关键字，用于计算某个列中不重复值的数量。它通常与SELECT语句一起使用，用于统计某个列中的唯一值的数量。

例如，假设有一个名为"students"的表，其中有一个列名为"age"，存储了学生的年龄信息。如果我们想知道这个表中不同年龄的学生数量，就可以使用COUNT_DISTINCT关键字来实现。

示例查询语句如下：
SELECT COUNT_DISTINCT(age) FROM students;

这条查询语句将返回"age"列中不重复值的数量，即不同年龄的学生数量。

相关问题

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;

帮我把“SELECT ppo.supplier_id, ppo.supplier_name, count(distinct ppo.id) FROM pm_purchase_order ppo inner join pm_purchase_order_detail ppod on ppo.id = ppod.purchase_order_id -- and ppod.confirm_status = '1' and ppod.del_flag = '0' WHERE ppo.del_flag = '0' GROUP BY ppo.supplier_id, ppo.supplier_name”和“SELECT ppo.supplier_id, ppo.supplier_name, count(distinct ppo.id) FROM pm_purchase_order ppo inner join pm_purchase_order_detail ppod on ppo.id = ppod.purchase_order_id and ppod.confirm_status = '1' and ppod.del_flag = '0' WHERE ppo.del_flag = '0' GROUP BY ppo.supplier_id, ppo.supplier_name”合并成一个SQL语句

可以使用`OR`运算符将两个`SELECT`语句合并成一个，同时在第二个`SELECT`语句中添加一个条件判断，当`ppod.confirm_status`不等于1时，不计入结果。合并后的 SQL 语句如下：

SELECT   ppo.supplier_id,   ppo.supplier_name,   count(distinct ppo.id) 
FROM   pm_purchase_order ppo 
inner join pm_purchase_order_detail ppod on ppo.id = ppod.purchase_order_id 
WHERE   ppo.del_flag = '0' AND (ppod.confirm_status = '1' OR ppod.confirm_status <> '1' AND ppod.del_flag = '0')
GROUP BY   ppo.supplier_id,   ppo.supplier_name

注意，这个合并的语句中，`ppod.del_flag = '0'`条件是包含在括号里面的，因为它只对第二个`SELECT`语句生效。如果要对两个`SELECT`语句都生效，需要将这个条件移动到`WHERE`语句的外面，并使用`AND`运算符连接两个条件。