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±í1
¹¤ºÅ    ÐÕÃû
1      Ëν­
2      ¬¿¡Òå

±í2
¹¤ºÅ    µÃ·Ö    ÆÚÊý
1      50      1
1      60      2
1      64      3

²éѯ½á¹û

ÐÕÃû    µÃ·Ö£¨ÈýÆÚµÄavg£©
Ëν­    58

SQL code:
select ÐÕÃû,µÃ·Ö=avg(µÃ·Ö) from ±í1 A,±í2 B where A.¹¤ºÅ=B.¹¤ºÅ
group by ÐÕÃû


SQL code:
select ÐÕÃû,avg(µÃ·Ö) as µÃ·Ö
from ±í1 inner join ±í2 on ±í1.¹¤ºÅ=±í2.¹¤ºÅ
group by ÐÕÃû



SQL code:
select a.¹¤ºÅ,a.ÐÕÃû,µÃ·Ö=avg(b.µÃ·Ö)
from ±íÒ» a inner join ±í¶þ b on a.¹¤ºÅ=b.¹¤ºÅ
group by a.¹¤ºÅ


SQL code:
1> select * from ±í1;
2> go
¹¤ºÅ |ÐÕÃû
-----------|----------
1|Ëν­
2|¬¿¡Òå

(2 rows


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SQL code:
select distinct *
from tablename
where id in (
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mssql ×Ö·û´®´¦Àí

@str = |108||203|

²éÕÒ·µ»Ø½á¹ûstr = 108 Ö»·µ»ØµÚÒ»¸ö ÆäËüµÄ¶¼¸Éµô
SQL code:
declare @str varchar(100)
set @str = '|108||203|'
select substring(@str,2,charindex('|',stuff(@str,1,1,''))-1)
......

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$myServer = "124.172.125.197"; //Ö÷»ú
$myUser = "jmkjxy"; //Óû§Ãû
$myPass = "jmkjxy023"; //ÃÜÂë
$myDB = "jmkjxy";  //MSSQL¿âÃû ......
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