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Everyone has his/her own driving license, and I need to know what kind =
of
=93person=94 (client or user) is.
mysql> select userID, clientID from client, user where
(clientCodeDrivingLicense=3D 321321321 || userCodeDrivingLicense =3D =
321321321);
+--------+-------+
| userID | clientID |
+--------+-------+
| 1 | 2 |
| 2 | 2 |
| 3 | 2 |
| 4 | 2 |
| 5 | 2 |
+--------+-------+
5 rows in set (0.00 sec)
But, what I want is something like that:
+--------+-------+
| userID | clientID |
+--------+-------+
| Null | 2 |
+--------+-------+
I tried something like this:
select COUNT(DISTINCT u.userID), userID, clientID from client, user =
where
(clientCodeDrivingLicense =3D 321321321 || userCodeDrivingLicense =3D
321321321);
+--------------------------+--------+-------+
| COUNT(DISTINCT u.userID) | userID | clientID |
+--------------------------+--------+-------+
| 5 | 1 | 2 |
+--------------------------+--------+-------+
1 row in set (0.00 sec)
But it wont be efficient enough in the future.
I suppose my solution is an Join, but they have no intersection, so, I =
cant
imagine how do it
Thank you!!
Regards
Roc=EDo G=F3mez Escribano
<mailto:r.sanchez [at] ingenia-soluciones.com> =
r.gomez [at] ingenia-soluciones.com
Descripci=F3n: cid:image002.jpg [at] 01CB8CB6.ADEBA830
Pol=EDgono Campollano C/F, n=BA21T
02007 Albacete (Espa=F1a)
Tlf:967-504-513 Fax: 967-504-513
www.ingenia-soluciones.com
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</o:shapelayout></xml><![endif]--></head><body lang=3DES link=3Dblue =
vlink=3Dpurple><div class=3DWordSection1><p class=3DMsoNormal><span =
lang=3DEN-US>Everyone has his/her own driving license, and I need to =
know what kind of “person” (client or user) =
is.<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>mysql> select userID, clientID from client, user where =
(clientCodeDrivingLicense=3D 321321321 || userCodeDrivingLicense =3D =
321321321);<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>+--------+-------+<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>| userID | clientID =
|<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>+--------+-------+<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>| 1 =
| 2 |<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>| 2 =
| 2 |<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>| 3 =
| 2 |<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>| 4 =
| 2 |<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>| 5 =
| 2 |<o:p></o:p></span></p><p =
class=3DMsoNormal><span =
lang=3DEN-US>+--------+-------+<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>5 rows in set (0.00 =
sec)<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>But, what I want is something like =
that:<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US> =
+--------+-------+<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US> | userID | clientID =
|<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US> =
+--------+-------+<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US> | =
Null | 2 |<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US> =
+--------+-------+<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>I tried something like this:<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>select COUNT(DISTINCT u.userID), =
userID, clientID from client, user where (clientCodeDrivingLicense =
=3D 321321321 || userCodeDrivingLicense =3D =
321321321);<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>+--------------------------+--------+-------+<o:p></o:p></sp=
an></p><p class=3DMsoNormal><span lang=3DEN-US>| COUNT(DISTINCT =
u.userID) | userID | clientID |<o:p></o:p></span></p><p =
class=3DMsoNormal><span =
lang=3DEN-US>+--------------------------+--------+-------+<o:p></o:p></sp=
an></p><p class=3DMsoNormal><span =
lang=3DEN-US>| =
; =
5 | 1 | 2 =
|<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>+--------------------------+--------+-------+<o:p></o:p></sp=
an></p><p class=3DMsoNormal><span lang=3DEN-US>1 row in set (0.00 =
sec)<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>But it wont be efficient enough in the =
future.<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-US>I suppose my solution is an Join, but they have no =
intersection, so, I cant imagine how do it<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>Thank you!!<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US>Regards<o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US><o:p> </o:p></span></p><p =
class=3DMsoNormal><b><span lang=3DEN-US =
style=3D'font-size:10.0pt;color:#0070C0'>Roc=EDo G=F3mez =
Escribano<o:p></o:p></span></b></p><p class=3DMsoNormal><span =
lang=3DES-PE style=3D'font-size:10.0pt;color:#9D9D9D'><a =
href=3D"mailto:r.sanchez [at] ingenia-soluciones.com"><span lang=3DEN-US =
style=3D'color:blue'>r.gomez [at] ingenia-soluciones.com</span></a></span><spa=
n lang=3DEN-US =
style=3D'font-size:10.0pt;color:#9D9D9D'><o:p></o:p></span></p><p =
class=3DMsoNormal><span lang=3DEN-US =
style=3D'color:blue'><o:p> </o:p></span></p><p =
class=3DMsoNormal><span style=3D'font-size:9.0pt;color:#9D9D9D'><img =
border=3D0 width=3D181 height=3D74 id=3D"Imagen_x0020_1" =
src=3D"cid:image001.jpg [at] 01CC08F1.3AC45E30" alt=3D"Descripci=F3n: =
cid:image002.jpg [at] 01CB8CB6.ADEBA830"></span><span =
style=3D'font-size:12.0pt;color:blue'><o:p></o:p></span></p><p =
class=3DMsoNormal><span =
style=3D'font-size:10.0pt;color:#A6A6A6'>Pol=EDgono Campollano C/F, =
n=BA21T</span><span =
style=3D'font-size:10.0pt;color:blue'><o:p></o:p></span></p><p =
class=3DMsoNormal><span style=3D'font-size:10.0pt;color:#9D9D9D'>02007 =
Albacete (Espa=F1a)<o:p></o:p></span></p><p class=3DMsoNormal><span =
lang=3DEN-GB =
style=3D'font-size:10.0pt;color:#9D9D9D'>Tlf:967-504-513 Fax: =
967-504-513<o:p></o:p></span></p><p class=3DMsoNormal><span =
style=3D'font-size:10.0pt;color:#A6A6A6'><a =
href=3D"www.ingenia-soluciones.com" =
title=3D"blocked::www.ingenia-soluciones.com"><span lang=3DEN-GB =
style=3D'color:#9D9D9D'>www.ingenia-soluciones.com</span></a></span><span=
lang=3DEN-GB =
style=3D'font-size:10.0pt;color:#A6A6A6'><o:p></o:p></span></p><p =
class=3DMsoNormal><span =
lang=3DEN-US><o:p> </o:p></span></p></div></body></html>
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