Question

目前，我的代码如下：

SELECT Volunteer.VolunteerID, Volunteer.LastName, Volunteer.FirstName, Observations.VolunteerID, Roost.RoostID, Roost.UTME, Roost.UTMN, Roost.DecimalDegrees, Observations.Date, Location.LocationName, Location.Address, Town.Town
FROM (((Volunteer INNER JOIN Observations ON Volunteer.[VolunteerID]= Observations.[VolunteeriD])
LEFT JOIN Roost ON Roost.[RoostID]= Observations.[RoostID])
LEFT JOIN Location ON Location.[LocationID]= Roost.[LocationID])
LEFT JOIN Town ON Location.[TownID]= Town.[TownID]
ORDER BY Volunteer.LastName, Volunteer.FirstName;

目前，我的表格的一部分如下所示：

    Observations.VolunteerID | Observations.LocationID | Date

    Bob                      | Main St                 | 2015-01-01  
    Bob                      | Main St                 | 2015-02-02
    Sally                    | Fox St                  | 2015-02-02
    Dave                     | Long St                 | 2015-02-02
    Dave                     | Taylor St               | 2015-02-05
    Lindsay                  | New St                  | 2015-02-01
    Lindsay                  | New St                  | 2015-02-08
    Lindsay                  | Ray St                  | 2015-02-10
    Lindsay                  | Main St                 | 2015-02-25
    Lindsay                  | Taylor St               | 2015-02-31

但是，我想添加一个由{em> VolunteerID 和 LocationID 分组的GROUP BY语句（我认为），以便获得以下输出。我想知道志愿者所在的每个位置，如果他们多次访问同一地点，我希望看到最近的日期。我如何将其纳入我的原始代码？

    Observations.VolunteerID | Observations.LocationID | Date
    Bob                      | Main St                 | 2015-02-02
    Sally                    | Fox St                  | 2015-02-02
    Dave                     | Long St                 | 2015-02-02
    Dave                     | Taylor St               | 2015-02-05
    Lindsay                  | New St                  | 2015-02-08
    Lindsay                  | Ray St                  | 2015-02-10
    Lindsay                  | Main St                 | 2015-02-25
    Lindsay                  | Taylor St               | 2015-02-31

更新了代码，但仍然出现语法错误（查询表达式中缺少运算符＆＃39; Volunteer.FirstName GROUP BY Observations.VolunteerID＆＃39;。

SELECT Volunteer.VolunteerID, Volunteer.LastName, Volunteer.FirstName, Observations.VolunteerID, Roost.RoostID, Roost.UTME, Roost.UTMN, Roost.DecimalDegrees, Max([Observations.Date]), Location.LocationName, Location.Address, Town.Town
FROM (((Volunteer INNER JOIN Observations ON Volunteer.[VolunteerID]= Observations.[VolunteeriD])
LEFT JOIN Roost ON Roost.[RoostID]= Observations.[RoostID])
LEFT JOIN Location ON Location.[LocationID]= Roost.[LocationID])
LEFT JOIN Town ON Location.[TownID]= Town.[TownID]
ORDER BY Volunteer.LastName, Volunteer.FirstName
GROUP BY Observations.VolunteerID, Observations.LocationID;

Answer 1

带有max子句的

group by似乎已经足够了：

select Observations.VolunteerID, Observations.LocationID, max([Date])
from <...>
group by Observations.VolunteerID, Observations.LocationID

Answer 2

您需要在日期使用public static string ExternalUrl(this PageData p, bool absoluteUrl, string languageBranch) { var result = ServiceLocator.Current.GetInstance<UrlResolver>().GetUrl( p.ContentLink, languageBranch, new VirtualPathArguments { ContextMode = ContextMode.Default, ForceCanonical = absoluteUrl }); // HACK: Temprorary fix until GetUrl and ForceCanonical works as expected, // i.e returning an absolute URL even if there is a HTTP context that matches the page's // site definition and host. if (absoluteUrl) { Uri relativeUri; if (Uri.TryCreate(result, UriKind.RelativeOrAbsolute, out relativeUri)) { if (!relativeUri.IsAbsoluteUri) { var siteDefinitionResolver = ServiceLocator.Current.GetInstance<SiteDefinitionResolver>(); var siteDefinition = siteDefinitionResolver.GetDefinitionForContent(p.ContentLink, true, true); var hosts = siteDefinition.GetHosts(p.Language, true); var host = hosts.FirstOrDefault(h => h.Type == HostDefinitionType.Primary) ?? hosts.FirstOrDefault(h => h.Type == HostDefinitionType.Undefined); var basetUri = siteDefinition.SiteUrl; if (host != null) { // Try to create a new base URI from the host with the site's URI scheme. Name should be a valid // authority, i.e. have a port number if it differs from the URI scheme's default port number. Uri.TryCreate(siteDefinition.SiteUrl.Scheme + "://" + host.Name, UriKind.Absolute, out basetUri); } var absoluteUri = new Uri(basetUri, relativeUri); return absoluteUri.AbsoluteUri; } } } return result; }和汇总函数GROUP BY。

MAX()

Answer 3

将ORDER BY子句置于GROUP BY之后。

将Max([Observations.Date])更改为Max(Observations.Date)

将SELECT列表中不是聚合表达式的任何字段表达式添加到GROUP BY。

SELECT
    Volunteer.VolunteerID,
    Volunteer.LastName,
    Volunteer.FirstName,
    Observations.VolunteerID,
    Roost.RoostID,
    Roost.UTME,
    Roost.UTMN,
    Roost.DecimalDegrees,
    Max(Observations.Date) AS latest_observation,
    Location.LocationName,
    Location.Address,
    Town.Town
FROM (((Volunteer INNER JOIN Observations ON Volunteer.[VolunteerID]= Observations.[VolunteeriD])
    LEFT JOIN Roost ON Roost.[RoostID]= Observations.[RoostID])
    LEFT JOIN Location ON Location.[LocationID]= Roost.[LocationID])
    LEFT JOIN Town ON Location.[TownID]= Town.[TownID]
GROUP BY
    Volunteer.VolunteerID,
    Volunteer.LastName,
    Volunteer.FirstName,
    Observations.VolunteerID,
    Roost.RoostID,
    Roost.UTME,
    Roost.UTMN,
    Roost.DecimalDegrees,
    Location.LocationName,
    Location.Address,
    Town.Town
ORDER BY Volunteer.LastName, Volunteer.FirstName;

但我不确定该分组是否合适。在这个问题中，您希望对 Observations.VolunteerID 和 Observations.LocationID 进行分组。较大的GROUP BY可能不会给您相同的结果。

如果是这种情况，您可以对SELECT字段使用聚合函数，然后从GROUP BY中删除该字段。例如......

SELECT Max(Volunteer.LastName)

假设 LastName 对于给定的 VolunteerID 始终相同，Max(Volunteer.LastName)将为您提供相同的值。

分组显示唯一字段

3 个答案: