请求适用于phpadmin但不是在我的PHP代码
让我请出示的另一个例子...... 我能不能够在PHP中创建这种观点(而我可以在phpMyAdmin)请求适用于phpadmin但不是在我的PHP代码
$sql="CREATE VIEW ratings.rtgemissfitch AS
SELECT derivedtable. ISIN
FROM
(SELECT ISIN, MAX(Date_Notation_Emission) FROM ratings.ratingsemissionfitch as derivedtable GROUP BY ISIN) ";
但我可以在PHP这样做:
$sql="CREATE VIEW ratings.rtgemissfitch AS
(SELECT ISIN, MAX(Date_Notation_Emission) FROM ratings.ratingsemissionfitch as derivedtable GROUP BY ISIN) ";
我真的不understand..First所有的,对不起,我的英语,我是法国人.. 我真的不明白,为什么要求,在phpadmin上运行,不工作在我的php代码.. probaby派生表... 所以,我希望得到最后的评级惠誉: 在phpMyAdmin的SQL请求,其完美的作品:
SELECT `DBFITCH`.`ISIN`, `RATING_FITCH`as FITCH_RTG
FROM
(SELECT `ISIN`, MAX(`RATING_DATE`) as LastUpdate
FROM `ratings`.`ratingsemissionfitch` GROUP BY ISIN) as LAST
INNER JOIN `ratings`.`ratingsemissionfitch` as DBFITCH
ON
DBFITCH.`ISIN`= LAST.`ISIN`
AND DBFITCH.`RATING_DATE`=LAST.LastUpdate
在PHP中,下面的代码至极不运行:
$sql="CREATE VIEW ratings.rtgemissfitch AS
SELECT DBFITCH.ISIN, RATING_FITCH as FITCH_RTG
FROM
(SELECT ISIN, MAX(RATING_DATE) as LastUpdate
FROM ratings.ratingsemissionfitch GROUP BY ISIN) as LAST
INNER JOIN ratings.ratingsemissionfitch as DBFITCH
ON
DBFITCH.ISIN= LAST.ISIN
AND DBFITCH.RATING_DATE=LAST.LastUpdate";
$req = $bdd->exec($sql);
让我请出示的其他例子...
我不能够在PHP中创建这种观点(而我可以在phpMyAdmin)
$sql="CREATE VIEW ratings.rtgemissfitch AS
SELECT derivedtable. ISIN
FROM
(SELECT ISIN, MAX(Date_Notation_Emission) FROM ratings.ratingsemissionfitch as derivedtable GROUP BY ISIN) ";
但我可以在PHP这样做:
$sql="CREATE VIEW ratings.rtgemissfitch AS
(SELECT ISIN, MAX(Date_Notation_Emission) FROM ratings.ratingsemissionfitch as derivedtable GROUP BY ISIN) ";
我真不明白..提前 感谢,
您的数据库用户可能不必须的权限来运行
CREATE VIEW
?
谢谢,但我可以创建视图.. – user2165853 2013-03-13 15:20:44
所以...什么是错误? – djjjuk 2013-03-13 15:21:38
我是新的PHP和MYSQL ..它不创建视图,因为我想.. – user2165853 2013-03-13 15:25:53
您是否在日志中发现任何错误?可能是因为这两个实例中的MySQL用户不同。 – Daedalus 2013-03-13 15:17:56
1)它们不是同一个查询。它们甚至不是相同类型的查询('SELECT' vs'CREATE VIEW')2)它究竟如何失效? – 2013-03-13 15:18:47
在$ bdd类中启用错误报告。没有其他办法。 – 2013-03-13 15:19:06