这是我的疑问：
SELECT ID, [type], naam, adresl1, adresl2, tel, fax, email, contactpersoon
FROM
------------------------ START --------------------------
(SELECT av.personID as [id], 'P' as [type],
av.firstname + ' ' + av.lastname as 'naam',
av.straat as 'adresl1',
c.zipCode + ' ' + c.City as 'adresl2',
av.phone as 'tel',
'' as fax,
av.Email as 'email',
'' as 'website', '' as 'contactpersoon',
coalesce(a1.aliasTitle,'') + '|' + coalesce(a1.aliasTitle2,'') + '|' + coalesce(1.aliasTitle3,'') + '|' +
coalesce(a2.aliasTitle,'') + '|' + coalesce(a2.aliasTitle2,'') + '|' + coalesce(a2.aliasTitle3,'') + '|' +
coalesce(f.FunctionTitle,'') + '|' + coalesce(r.Raad,'') + '|' + coalesce(rci.RedCrossInstitutionName,'') + '|' +
coalesce(av.firstname,'') + ' ' + coalesce(av.lastname,'') + '|' +
coalesce(av.lastname,'') + ' ' + coalesce(av.firstname,'') AS 'wie',
coalesce(a1.aliasTitle,'') + '|' + coalesce(a1.aliasTitle2,'') + '|' + coalesce(a1.aliasTitle3,'') AS 'waar',
coalesce(echelon.Street,'') + '|' + coalesce(echelon.zipcode,'') + '|' +
coalesce(echelon.City,'') + '|' + coalesce(echelon.RedCrossEntityName,'') AS 'waar_E'
FROM RVW_vwAdresboekVrijwilligers av
LEFT JOIN City c ON av.CityID = c.CityID
LEFT JOIN AliasPerson ap ON av.PersonID = ap.PersonID
LEFT JOIN Alias a1 ON ap.AliasID = a1.AliasID
LEFT JOIN FunctionPerson fp ON av.PersonID = fp.PersonID
LEFT JOIN
(SELECT RedCrossEntity.RedCrossEntityID, RedCrossEntity.RedCrossEntityName,
RedCrossEntity.street, City.zipcode, City.city
FROM RedCrossEntity
LEFT JOIN City ON RedCrossEntity.CityID = City.CityID
) AS echelon ON echelon.RedCrossEntityID = fp.RedCrossEntityID
LEFT JOIN [Function] f ON f.FunctionID = fp.FunctionID
LEFT JOIN AliasFunction af ON af.FunctionID = f.FunctionID
LEFT JOIN Alias a2 ON a2.AliasID = af.AliasID
LEFT JOIN FunctionRaad fr ON fr.FunctionID = f.FunctionID
LEFT JOIN Raad r ON r.RaadID = fr.RaadID
LEFT JOIN RedCrossInstitution rci ON rci.RedCrossInstitutionID = fp.RedCrossInstitutionID
WHERE
f.functionid IN (SELECT functionid FROM FunctionResponsibility
WHERE ResponsibilityTypeId = 4)
------------------------- END -----------------------
) data
WHERE
(wie LIKE '%jos%' OR waar LIKE '+++++' )
ORDER BY
[type]
-START-和-END-之间的部分工作正常，并在大约3秒内执行。但是当我添加WHERE (wie LIKE '%jos%' OR waar LIKE '+++++' )时，运行需要30秒。
我也试过这个：
.......
LEFT JOIN RedCrossInstitution rci ON rci.RedCrossInstitutionID = fp.RedCrossInstitutionID
WHERE
1=1
AND (a1.aliasTitle LIKE '%jos%'
OR a1.aliasTitle2 LIKE '%jos%'
OR a1.aliasTitle3 LIKE '%jos%'
OR a1.aliasTitle LIKE '+++++'
OR a1.aliasTitle2 LIKE '+++++'
OR a1.aliasTitle3 LIKE '+++++'
OR a2.aliasTitle LIKE '%jos%'
OR a2.aliasTitle2 LIKE '%jos%'
OR a2.aliasTitle3 LIKE '%jos%'
OR f.FunctionTitle LIKE '%jos%'
OR r.Raad LIKE '%jos%'
OR rci.RedCrossInstitutionName LIKE '%jos%'
OR (av.firstname + ' ' + av.lastname LIKE '%jos%')
OR (av.lastname + ' ' + av.firstname LIKE '%jos%')
)
AND f.functionid IN (SELECT functionid FROM FunctionResponsibility
WHERE ResponsibilityTypeId = 4)
但那甚至更慢。
有没有人看到加快这个速度的方法？
答案 0 :(得分：3)
请删除 IN 并将其替换为如下所述的加入。通过这种方式，我们可以避免由于 IN
而发生的循环SELECT ID, [type], naam, adresl1, adresl2, tel, fax, email, contactpersoon
FROM(
------------------------ START --------------------------
SELECT av.personID as [id], 'P' as [type],
av.firstname + ' ' + av.lastname as 'naam',
av.straat as 'adresl1',
c.zipCode + ' ' + c.City as 'adresl2',
av.phone as 'tel',
'' as fax,
av.Email as 'email',
'' as 'website', '' as 'contactpersoon',
coalesce(a1.aliasTitle,'') + '|' + coalesce(a1.aliasTitle2,'') + '|' + coalesce(a1.aliasTitle3,'') + '|' +
coalesce(a2.aliasTitle,'') + '|' + coalesce(a2.aliasTitle2,'') + '|' + coalesce(a2.aliasTitle3,'') + '|' +
coalesce(f.FunctionTitle,'') + '|' + coalesce(r.Raad,'') + '|' + coalesce(rci.RedCrossInstitutionName,'') + '|' +
coalesce(av.firstname,'') + ' ' + coalesce(av.lastname,'') + '|' +
coalesce(av.lastname,'') + ' ' + coalesce(av.firstname,'') AS 'wie',
coalesce(a1.aliasTitle,'') + '|' + coalesce(a1.aliasTitle2,'') + '|' + coalesce(a1.aliasTitle3,'') AS 'waar',
coalesce(echelon.Street,'') + '|' + coalesce(echelon.zipcode,'') + '|' +
coalesce(echelon.City,'') + '|' + coalesce(echelon.RedCrossEntityName,'') AS 'waar_E'
FROM RVW_vwAdresboekVrijwilligers av
LEFT JOIN City c ON av.CityID = c.CityID
LEFT JOIN AliasPerson ap ON av.PersonID = ap.PersonID
LEFT JOIN Alias a1 ON ap.AliasID = a1.AliasID
LEFT JOIN FunctionPerson fp ON av.PersonID = fp.PersonID
LEFT JOIN
(
SELECT RedCrossEntity.RedCrossEntityID, RedCrossEntity.RedCrossEntityName, RedCrossEntity.street, City.zipcode, City.city
FROM RedCrossEntity
LEFT JOIN City ON RedCrossEntity.CityID = City.CityID
) as echelon ON echelon.RedCrossEntityID = fp.RedCrossEntityID
LEFT JOIN [Function] f ON f.FunctionID = fp.FunctionID
LEFT JOIN AliasFunction af ON af.FunctionID = f.FunctionID
LEFT JOIN Alias a2 ON a2.AliasID = af.AliasID
LEFT JOIN FunctionRaad fr ON fr.FunctionID = f.FunctionID
LEFT JOIN Raad r ON r.RaadID = fr.RaadID
LEFT JOIN RedCrossInstitution rci ON rci.RedCrossInstitutionID = fp.RedCrossInstitutionID
INNER Join FunctionResponsibility FRes on FRes.functionid = f.functionid
WHERE FRes.ResponsibilityTypeId = 4
------------------------- END -----------------------
) data
WHERE (wie LIKE '%jos%' OR waar LIKE '+++++' )
ORDER BY [type]
答案 1 :(得分：1)
问题在于您正在进行通配符搜索，查找包含在您正在搜索的字段内的任何位置的特定字符串。如果这是性能关键，并且无法使您的搜索更具体，请考虑使用全文搜索：
LIKE与全文搜索的比较
与全文搜索相比，LIKE Transact-SQL谓词仅适用于字符模式。此外，您不能使用LIKE谓词来查询格式化的二进制数据。此外，针对大量非结构化文本数据的LIKE查询比针对相同数据的等效全文查询慢得多。对数百万行文本数据的LIKE查询可能需要几分钟才能返回;而对于相同的数据，全文查询只需几秒或更短时间，具体取决于返回的行数。
请注意，全文搜索是SQL Server的一个独立组件，它自行安装和管理，因此如果该开销值得获得可能的性能提升，则必须自行评估。有关详细信息，请参阅Full-Text Search (SQL Server)。
答案 2 :(得分：0)
在查询中使用wie运算符时，SQL Server很难在waar和LIKE列上使用索引。
此处提供更多信息：http://myitforum.com/cs2/blogs/jnelson/archive/2007/11/16/108354.aspx
编辑：如果您尝试实施搜索功能，那么您可以查看Full Text Search
答案 3 :(得分：0)
来自MSDN的这个article讨论了当LIKE子句的WHERE谓词以通配符%开头时的性能。