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表。我需要表中的变量数组,并在Oracle中创建正确的代码:
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?非常感谢提前:))
答案 0 :(得分:1)
不,这不太可能,抱歉:(
在MySQL中,我们通常使用具有逗号分隔值的VARCHAR或TEXT字段来执行此操作。我知道,这很烦人:)