请求适用于phpadmin但不是在我的PHP代码

Question

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）

$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，

Answer 1

您的数据库用户可能不必须的权限来运行

CREATE VIEW 

？