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让我举个例子……我不能在 php 中创建这个视图（而我可以在 phpmyadmin 中）

$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 中做到这一点：

$sql="CREATE VIEW ratings.rtgemissfitch AS
(SELECT ISIN, MAX(Date_Notation_Emission) FROM ratings.ratingsemissionfitch as derivedtable GROUP BY ISIN) ";

我真的不明白..首先，对不起我的英语，我是法国人..我真的不明白为什么在 phpadmin 上运行的请求在我的 php 代码中不起作用..probaby派生表...所以，我希望获得最后的评分 Fitch：phpmyadmin 中的 SQL 请求完美运行：

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，以下代码无法运行：

$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);

让我再举一个例子……

我无法在 php 中创建此视图（而我可以在 phpmyadmin 中）

$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 中做到这一点：

$sql="CREATE VIEW ratings.rtgemissfitch AS
(SELECT ISIN, MAX(Date_Notation_Emission) FROM ratings.ratingsemissionfitch as derivedtable GROUP BY ISIN) ";

我真的不明白..提前谢谢，

4

1 回答 1

0

您的数据库用户可能没有运行权限

CREATE VIEW

?

于 2013-03-13T15:18:31.023 回答