我正在尝试在 PHP 代码中创建一个表，以便在浏览器中打开以创建该表。我需要表中的一个不同的数组，并有正确的代码在 Oracle 中创建它：
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));
但是，当放置在 PHP 代码中并添加到 MySQL 数据库时，会出现错误消息。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";
}
?>
甚至可能吗？提前谢谢了 ：）