2012-03-26 92 views
1

我试图在PHP代码中创建一个表,以便在浏览器中打开以创建表。我需要的表内的变化的阵列,并具有正确的代码中的Oracle来创建它:在MySQL中添加变量数组(VARRAY)

CREATE TYPE ReportEntries_Type AS OBJECT 
(Subject VARCHAR (500)); 
/

CREATE OR REPLACE TYPE ReportEntries_VA AS 
VARRAY (12) OF ReportEntries_Type; 
/

CREATE TABLE ReportDetails 
(ReportID INTEGER NOT NULL UNIQUE, 
StudentID INTEGER NOT NULL UNIQUE, 
ReportEntries ReportEntries_VA, 
DateLastModified DATE NOT NULL, 
CONSTRAINT ReportDetails_PK PRIMARY KEY (ReportID, StudentID), 
CONSTRAINT RDStudentIDSD FOREIGN KEY (StudentID) 
REFERENCES StudentDetails (StudentID)); 

然而,当放置PHP代码中,并加入到一个MySQL数据库,将显示错误消息。我正在使用的代码如下:

<?php 
require "connect_to_mysql.php"; 

$sqlCommand = "CREATE TYPE ReportEntries_Type AS OBJECT 

...cont. code from above... 

REFERENCES StudentDetails (StudentID))"; 

if (mysql_query ($sqlCommand)) { 
echo "The ReportDetails table has been created successfully!"; 
} else { 
echo "There has been an error"; 
} 
?> 

甚至有可能吗?提前感谢:)

回答

1

不,这不是真的有可能,对不起:(

在MySQL中,我们通常使用昏迷分隔值VARCHAR或文本字段来做到这一点。我知道,这很烦人:)

+0

哦bollocks!没关系!感谢您的快速回复@haltabush :) – 2012-03-26 16:16:34