我正在基于该列名称传递动态列名称,以获取值,并且在我的表格下方
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Example1
Table_CandidateInfo
Id Name Age City
1 Mazhar 30 Gulbarga
20 Khan 29 Bidar
我无法通过Declare @ColumnName varchar(100), @Id int
set @ColumnName='Age'
set @Id=20
select * from Table_CandidateInfo where ID=@Id and
查询传递列名,因为列名是通过代码动态传递的。我的输出应该是
and
示例2:如果我的@ ColumnName ='City'和@ Id = 20,则输出应如下所示
29
答案 0 :(得分:3)
我认为您的实际追求如下:
DECLARE @ColumnName sysname, @Id int;
SET @Id = 29;
SET @ColumnName = N'Age';
DECLARE @SQL nvarchar(MAX);
SET @SQL = N'SELECT ' + QUOTENAME(@ColumnName) + N' FROM dbo.Table_CandidateInfo WHERE Id = @Id;';
--PRINT @SQL; --Your debugging friend
EXEC sp_executesql @SQL, N'@Id int', @Id = @Id;
答案 1 :(得分:2)
A,您不能将标识符作为参数传递。您需要使用动态SQL:
f = False
out = []
for i in t:
if i[1] == 'ORGANIZATION':
if not f:
out.append(i[0])
f = True
else:
out[-1] += f' {i[0]}'
f = False
print(out)
# ['Wall Mart', 'Thomas Cook']