SQL select (查询字符串)初级有关问题请问
SQL select (查询字符串)初级问题请教
某字段[class], 字符型, 内容形式为 39,58,61
想求出所有本字段中含有58的所有记录.... 查不出来..不知道哪错啦?
SELECT Id,Title From NewsInfo Where class in ('58') Order By Id DESC
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某字段[class], 字符型, 内容形式为 39,58,61
想求出所有本字段中含有58的所有记录.... 查不出来..不知道哪错啦?
SELECT Id,Title From NewsInfo Where class in ('58') Order By Id DESC
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- SQL code
SELECT Id,Title From NewsInfo Where class like '%58%' Order By Id DESC
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- SQL code
WHERE CHARINDEX(',58,',','+class+',')>0
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- SQL code
SELECT Id,Title From NewsInfo Where class like '%58%' Order By Id DESC
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- SQL code
SELECT Id,Title From NewsInfo Where class like '%58%' Order By Id DESC
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- SQL code
SELECT Id,Title From NewsInfo Where class like '%58%' Order By Id DESC
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- SQL code
SELECT Id,Title From NewsInfo Where charindex('58,',class+',')>0 Order By Id DESC
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- SQL code
SELECT Id,Title From NewsInfo Where ',' + class + ',' like '%,58,%' Order By Id DESC SELECT Id,Title From NewsInfo Where charindex(',58,' , ',' + class + ',') > 0 Order By Id DESC
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- SQL code
declare @t table(a varchar(300)) insert @t select '39,58,61' insert @t select '50,28' insert @t select '66' insert @t select '83,88,89' select * from @t where a like '%58%' a --------------------- (所影响的行数为 2 行)
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- SQL code
SELECT Id,Title From NewsInfo Where class like '%58%' Order By Id DESC
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1、方法一
- SQL code
select id,title from newsinfo wher class like '%58%' order by id desc