我有4个表(ROUTBOM,HARD,MAT,ROUT)。在主表ROUTBOM中,我具有ID,PROD_NO和TYPE列。我必须根据TYPE和(ID或PROD_NO)进行选择。条件如下。可以使用内部联接-在联接之前有条件吗?
数据库是FoxPro
using (OleDbConnection con = new OleDbConnection(@"provider=VFPOLEDB;Data source=c:\test\Data"))
数据: 路由
id qty prod_no Type Name
20322 0.15000 2 ??? get name from HARD
2.00000 0066773 4 ??? get name from ROUT
37500 4.00000 2 ??? get name from HARD
29750 4.00000 1 ??? get name from MAT
硬表
ID NAME
20322 H68NK0005
37500 HAS2-30XX-H HYBRID POWDER
MAT TABLE
ID NAME
29750 NEOPRENE (McMASTER 8694K61)
02125 SPRING STEEL STRIP .008'' X .50'' X 60''
路线表
ID NAME
0066773 L.V. DOOR ARC PROOF J4
000198 DEVEL. L.V. DOOR J4
必填输出
id qty prod_no Type Name
20322 0.15000 2 H68NK0005
2.00000 0066773 4 L.V. DOOR ARC PROOF J4
37500 4.00000 2 HAS2-30XX-H HYBRID POWDER
29750 4.00000 1 NEOPRENE (McMASTER 8694K61)
查询:
条件:
if id is not null and type 2 JOIN ROUTBOM.id with MAT.id to get the name of material
if id is not null and type 1 JOIN ROUTBOM.id with HARD.id to get the name of hadware
if id is empty and type 4 JOIN ROUTBOM.prod_no with ROUT.id to get the name of part
可以在一个查询中完成吗?我不知道必须在哪里应用条件。一旦我进行了第一次JOIN,查询就会从HARD中获取全部数据。
查询行数据
@"SELECT
t0.Id,
t1.Name,
t0.Per_Router,
t0.Prod_No,
t0.Rout_No,
t0.Seq,
t0.Type
FROM {0} t0 , "Routbom", "Hardware", "Material", "Rout");
答案 0 :(得分:0)
从您的有限解释中,我了解到这一点:
string sql = @"SELECT Id,Name,Per_Router,Prod_No,Rout_No,Seq,Type
from
(
Select t0.*, t1.Name as Name
from RoutBom t0
inner join Mat t1 on t0.Id = t1.Id
where t0.type = 2
union
Select t0.*, t1.Name as Name
from RoutBom t0
inner join Hard t1 on t0.Id = t1.Id
where t0.type = 1
union
Select t0.*, t1.Name as Name
from RoutBom t0
inner join Rout t1 on t0.Prod_no = t1.Id
where t0.type = 4 and Empty(Nvl(t0.id,''))
) tmp
";
DataTable tbl = new DataTable();
using (OleDbConnection con = new OleDbConnection(@"provider=VFPOLEDB;Data source=c:\test\Data"))
using (OleDbCommand cmd = new OleDbCommand(sql, con))
{
con.Open();
tbl.Load(cmd.ExecuteReader());
}
// Do something with tbl
使用Linq可能更容易。检查Tom Brother的Linq To VFP和VFP实体框架。
这是另一种方法,并且可以解决工会问题:
void Main()
{
string sql = @"SELECT *,
Cast(icase(
type = '1', (select Name from Mat t1 where t0.Id = t1.Id),
type = '2', (select Name from Hard t1 where t0.Id = t1.Id),
empty(nvl(id,'')) and type='4', (select Name from Rout t1 where t0.Prod_No = t1.Id),
'') as c(100)) as Name
from RoutBom t0";
DataTable tbl = new DataTable();
using (OleDbConnection con = new OleDbConnection(@"provider=VFPOLEDB;Data source=c:\Test\Data"))
using (OleDbCommand cmd = new OleDbCommand(sql, con))
{
con.Open();
tbl.Load(cmd.ExecuteReader());
}
// Do something with tbl
tbl.Dump(); // you said LinqPad above
}
答案 1 :(得分:0)
首先,我建议您学习JOIN的“标准” SQL基础(不是VFP特定)。您可以在Inner and Outer JOINs in SQL
上找到很好的解释。然后将其应用于您的表。 基本上,您将有1个“父”表和3个“子”表,它们通过SQL语法联接到“父”表。
VFP SQL语法与“通用” SQL语法略有不同,但是在大多数情况下,“通用”语法可以正常工作。
也许是这样的:
def permutations(A):
P = []
permutationsI(A, P)
print(P)
def permutationsI(A, perms):
stack = [(A, [])]
while len(stack):
first, last = stack.pop()
if len(first):
for i in range(len(first)):
stack.append((first[:i] + first[i+1:],last + [first[i]]))
else:
perms.append(last)
permutations([1,2,3])
>>[[3, 2, 1], [3, 1, 2], [2, 3, 1], [2, 1, 3], [1, 3, 2], [1, 2, 3]]
请注意,这只是记录下来的,未经测试,但这是“ BASIC” SQL语法,在尝试将其应用于项目之前应了解该语法。
也许您想去:www3schools - SQL SELECT syntax
祝你好运