WithAll().leftjoin

时间: 2023-12-02 11:50:19 浏览: 25
在给定的引用内容中，介绍了使用WITH子句和LEFT JOIN来比较两个查询的结果。根据和，这些查询比较了名为t_left和t_right的表中的值。在中，还提供了一个使用(1, NULL)和(2, NULL)进行比较的示例。 根据这些信息，可以得出结论，WithAll().leftjoin是一种将两个查询结果进行左连接的方法。具体来说，它可能是使用WITH子句定义一个或多个临时表，并使用LEFT JOIN将它们连接起来，以便比较它们的值。 然而，在没有更多上下文的情况下，无法提供更具体的答案。可能需要进一步的信息或代码示例来解释WithAll().leftjoin的具体含义和用法。<span class="em">1</span><span class="em">2</span><span class="em">3</span> #### 引用[.reference_title] - *1* *2* *3* [not in vs left join is null](https://blog.csdn.net/lineuman/article/details/127912042)[target="_blank" data-report-click={"spm":"1018.2226.3001.9630","extra":{"utm_source":"vip_chatgpt_common_search_pc_result","utm_medium":"distribute.pc_search_result.none-task-cask-2~all~insert_cask~default-1-null.142^v93^chatsearchT3_2"}}] [.reference_item style="max-width: 100%"] [ .reference_list ]

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with tmpVaa1BaseData as (SELECT DISTINCT t .* FROM ( SELECT A .vaa01 患者编号, A .vaa03 门诊号, A .vaa05 姓名, C .abw02 性别, A .vaa10 || d.aau02 AS 年龄, A .bcq04 床号, a.bdp02 患者类别 FROM vaa1 A JOIN vac1 b ON A .vaa01 = b.vaa01 LEFT JOIN abw1 C ON A .abw01 = C .abw01 LEFT JOIN aau1 d ON A .aau01 = d.aau01 UNION ALL SELECT A .vaa01 患者编号, COALESCE (b.vae94, a.vaa04) AS 住院号, A .vaa05 姓名, C .abw02 性别, A .vaa10 || d.aau02 AS 年龄, A .bcq04 床号, a.bdp02 患者类别 FROM vaa1 A JOIN vae1 b ON A .vaa01 = b.vaa01 LEFT JOIN abw1 C ON A .abw01 = C .abw01 LEFT JOIN aau1 d ON A .aau01 = d.aau01 ) AS t)

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请找出下面POSTGRESQL 语句报错的原因with tmpVaa1BaseData as (SELECT DISTINCT t .* FROM ( SELECT A .vaa01 患者编号, A .vaa03 门诊号, A .vaa05 姓名, C .abw02 性别, A .vaa10 || d.aau02 AS 年龄, A .bcq04 床号, a.bdp02 患者类别 FROM vaa1 A JOIN vac1 b ON A .vaa01 = b.vaa01 LEFT JOIN abw1 C ON A .abw01 = C .abw01 LEFT JOIN aau1 d ON A .aau01 = d.aau01 UNION ALL SELECT A .vaa01 患者编号, COALESCE (b.vae94, a.vaa04) AS 住院号, A .vaa05 姓名, C .abw02 性别, A .vaa10 || d.aau02 AS 年龄, A .bcq04 床号, a.bdp02 患者类别 FROM vaa1 A JOIN vae1 b ON A .vaa01 = b.vaa01 LEFT JOIN abw1 C ON A .abw01 = C .abw01 LEFT JOIN aau1 d ON A .aau01 = d.aau01 ) AS t) SELECT DISTINCT t.* FROM (SELECT b.lte09, b .vaa02, A .vaa05, A .abw02, A .bcq04, A .vaa10, b .lte04, c .abl02, d .abn02, e .bck03, b .lte06, (select t.lte09 as lastTransTime from lte1 t where t.vaa01 = a.vaa01 ORDER BY t.lte09 DESC LIMIT 1) lastTransTime, b.lte07, b.bak05, b.lte48, b.lte40, b.lte47, b.lte49, b.lte50, g.bby05, f.lth05, f.lth08, h.ltl02, h.ltl03, h.ltl04, h.ltl05, h.ltl06, h.ltl07, h.ltl08, h.ltl09, h.ltl10, h.ltl11, h.ltl12, h.ltl13, h.ltl14, h.ltl15, h.ltl16, h.ltl17, h.ltl18, h.bce01a, h.ltl19 FROM tmpVaa1BaseData A JOIN lte1 b ON A .vaa01 = b.vaa01 JOIN ltl1 h ON h.lte01 = b.lte01 JOIN bck1 e ON b.bck01a = e.bck01 JOIN lth1 f ON b.lte01 = f.lte01 JOIN bby1 g ON f.bby01 = g.bby01 LEFT JOIN abl1 c ON b.abl02a = c.abl01 LEFT JOIN abn1 d ON b.abn02a = d.abn01

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with tmpVaa1BaseData as (SELECT DISTINCT t .* FROM ( SELECT A .vaa01 患者编号, A .vaa03 门诊号, A .vaa05 姓名, C .abw02 性别, A .vaa10 || d.aau02 AS 年龄, A .bcq04 床号, a.bdp02 患者类别 FROM vaa1 A JOIN vac1 b ON A .vaa01 = b.vaa01 LEFT JOIN abw1 C ON A .abw01 = C .abw01 LEFT JOIN aau1 d ON A .aau01 = d.aau01 UNION ALL SELECT A .vaa01 患者编号, COALESCE (b.vae94, a.vaa04) AS 住院号, A .vaa05 姓名, C .abw02 性别, A .vaa10 || d.aau02 AS 年龄, A .bcq04 床号, a.bdp02 患者类别 FROM vaa1 A JOIN vae1 b ON A .vaa01 = b.vaa01 LEFT JOIN abw1 C ON A .abw01 = C .abw01 LEFT JOIN aau1 d ON A .aau01 = d.aau01 ) AS t) SELECT DISTINCT t.* FROM (SELECT b.lte09, b .vaa02, A .vaa05, A .abw02, A .bcq04, A .vaa10, b .lte04, c .abl02, d .abn02, e .bck03, b .lte06, (select t.lte09 as lastTransTime from lte1 t where t.vaa01 = a.vaa01 ORDER BY t.lte09 DESC LIMIT 1) lastTransTime, b.lte07, b.bak05, b.lte48, b.lte40, b.lte47, b.lte49, b.lte50, g.bby05, f.lth05, f.lth08, h.ltl02, h.ltl03, h.ltl04, h.ltl05, h.ltl06, h.ltl07, h.ltl08, h.ltl09, h.ltl10, h.ltl11, h.ltl12, h.ltl13, h.ltl14, h.ltl15, h.ltl16, h.ltl17, h.ltl18, h.bce01a, h.ltl19 FROM tmpVaa1BaseData A JOIN lte1 b ON A .vaa01 = b.vaa01 JOIN ltl1 h ON h.lte01 = b.lte01 JOIN bck1 e ON b.bck01a = e.bck01 JOIN lth1 f ON b.lte01 = f.lte01 JOIN bby1 g ON f.bby01 = g.bby01 LEFT JOIN abl1 c ON b.abl02a = c.abl01 LEFT JOIN abn1 d ON b.abn02a = d.abn01 WHERE b.lte01 = <#if form.lte01?? && form.lte01!="">'${form.lte01}'<#else>null</#if>) as t 报错了为啥呢

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1. 使用INNER JOIN代替LEFT JOIN：如果确保两个表中都含有匹配的行，则使用INNER JOIN可以提高查询效率。 2. 使用EXISTS代替IN：EXISTS执行速度比IN更快，因为它只需要找到匹配的第一个结果，而不是找到所有匹配的...

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Sure, here's the modified SQL query with pagination for Oracle database: SELECT * FROM ( SELECT GDI.P_NO AS pNo, GDI.P_NAME AS pName, SO.ORG_NAME AS pCallDepartment, GTI.P_T_NO AS ptno, ...

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with t as ( select a.* from 医保住院结算明细 a left join 医保住院结算明细 b on a.个人编码 =b.个人编码 where a.医保项目名称 like '%心脏彩色多普勒超声%' and (b.医保项目名称 like '%左心功能测定%' or b.医保项目名称 like '%左心功能测定彩色多普勒超声检查%') union all select b.* from 医保住院结算明细 a left join 医保住院结算明细 b on a.个人编码 =b.个人编码 where a.医保项目名称 like '%心脏彩色多普勒超声%' and (b.医保项目名称 like '%左心功能测定%' or b.医保项目名称 like '%左心功能测定彩色多普勒超声检查%') ) select distinct * from t 优化一下

inner join 医保住院结算明细 b on a.个人编码 = b.个人编码 where a.医保项目名称 like '%心脏彩色多普勒超声%' and (b.医保项目名称 like '%左心功能测定%' or b.医保项目名称 like '%左心功能测定彩色多普勒超声...

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优化下面的sql语句，达到最优：DECLARE @dtNow VARCHAR(100) = CONVERT(VARCHAR(100),GETDATE(),23); SELECT TOP 1 k.Name kehuName ,ISNULL(MaskingApply.quanxianSum,0) quanxianSum FROM xykehu k WITH(NOLOCK) LEFT JOIN ( SELECT cs.quanxianSum,cs.AddTime FROM ClearShowApply cs WITH(NOLOCK) WHERE ISNULL(cs.IsDel,0)=0 AND cs.State='审批通过' AND cs.PowerEndTime > @dtNow AND cs.XyKehuId = @kehuId AND cs.UsersId= @userId UNION SELECT cs.quanxianSum,cs.AddTime FROM ClearShowApply cs WITH(NOLOCK) WHERE ISNULL(cs.IsDel,0)=0 AND cs.State='审批通过' AND cs.PowerEndTime > @dtNow AND cs.XyKehuId = @kehuId AND cs.CardId= @cardId ) MaskingApply ON 1 = 1 WHERE k.Id = @kehuId AND ISNULL(k.IsDel,0)=0 ORDER BY MaskingApply.AddTime DESC;

4. 将LEFT JOIN改为INNER JOIN，因为WHERE子句中已经限定了ClearShowApply表中的条件，所以不需要使用LEFT JOIN。 优化后的SQL语句如下： DECLARE @dtNow VARCHAR(100) = CONVERT(VARCHAR(100), GETDATE(), 23...

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已知n个人（以编号0，1，2，3...n-1分别表示）围坐在一张圆桌周围。从编号为0的人开始报数1，数到m的那个人出列；他的下一个人又从1开始报数，数到m+1的那个人又出列(每次报数值加1)；依此规律重复下去，直到圆桌周围的人全部出列。用递归方法解决

这个问题可以使用递归方法解决。下面是一个思路： 1. 定义一个函数，接收三个参数：n、m、i，表示还剩下n个人，每次数到m时出列，当前报数的人是i； 2. 如果n=1，返回i，即最后留下的那个人的编号； 3. 否则，计算出下一个出列的人的编号j，通过递归调用函数解决n-1个人的问题，其结果为k； 4. 如果k < j，即当前i之后出列的人的编号为k，需要将k转换为在i之前出列的编号，返回值为 k+(n-1)； 5. 如果k>=j，即当前i之后出列的人的编号为k，返回值为 k-(j-1)； 下面是对应的Python代码： ```python def josephus(n, m, i):