
09.09.2009, 23:44
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Members of Antichat - Level 5
Регистрация: 09.05.2008
Сообщений: 304
С нами:
9477026
Репутация:
2362
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Код:
mysql> SELECT t1.login, t1.value as city, t2.value as regdate, t3.value as email FROM `tst` t1
-> LEFT JOIN `tst` t2 ON t1.login = t2.login AND t2.param='regdate'
-> LEFT JOIN `tst` t3 ON t1.login = t3.login AND t3.param='email'
-> GROUP BY t1.login;
+-------+--------+------------+-----------------+
| login | city | regdate | email |
+-------+--------+------------+-----------------+
| Alex | London | 22-04-2009 | aswer@yahoo.com |
| Nata | Paris | 15-07-2009 | dser@mail.com |
+-------+--------+------------+-----------------+
2 rows in set (0,00 sec)
Код:
mysql> SELECT login,
-> (SELECT `value` FROM `tst` WHERE `login`=t1.login AND `param`='city') as city,
-> (SELECT `value` FROM `tst` WHERE `login`=t1.login AND `param`='regdate') as regdate,
-> (SELECT `value` FROM `tst` WHERE `login`=t1.login AND `param`='email') as email
-> FROM `tst` t1
-> GROUP BY `login`;
+-------+--------+------------+-----------------+
| login | city | regdate | email |
+-------+--------+------------+-----------------+
| Alex | London | 22-04-2009 | aswer@yahoo.com |
| Nata | Paris | 15-07-2009 | dser@mail.com |
+-------+--------+------------+-----------------+
2 rows in set (0,01 sec)
__________________
включи голову
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