2012-03-26 63 views
1

我試圖在PHP代碼中創建一個表,以便在瀏覽器中打開以創建表。我需要的表內的變化的陣列,並具有正確的代碼中的Oracle來創建它:在MySQL中添加變量數組(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,當放置PHP代碼中,並加入到一個MySQL數據庫,將顯示錯誤消息。我正在使用的代碼如下:

<?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"; 
} 
?> 

甚至有可能嗎?提前感謝:)

回答

1

不,這不是真的有可能,對不起:(

在MySQL中,我們通常使用昏迷分隔值VARCHAR或文本字段來做到這一點。我知道,這很煩人:)

+0

哦bollocks!沒關係!感謝您的快速回復@haltabush :) – 2012-03-26 16:16:34