Confused On Keys For "nested" One To Many Tables
This isn't a real-world situation (yet), but it would help me a lot to understand the logistics behind table relationships. Assume I have a software company that makes regular revisions to the software titles. I would have a table listing my software names, one for each software version with the date of that revision (one to many), and then a table of details for each revision (one to many).
Figuring out how to do the query isn't necessarily what I'm looking for, it's the DB setup I'd like to understand for now. But, just to clarify the point a little, I'd like to be able to create a report such as:
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Here's what I would imagine the table structure looking like:
1.) A basic table of software titles in the catalog.
CREATE TABLE software (
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
name VARCHAR(255)
);
2.) A table for the version numbers. Since there would be many possible instances of Ƈ.10' but only one instance of 'File Sorter 1.10', I think that using the FOREIGN KEY from the software table and the version number as primary keys make sense.
CREATE TABLE version (
softwareID INT NOT NULL,
versionID DECIMAL(3,2) NOT NULL,
date DATE,
PRIMARY KEY(catalogID,versionID)
);
3.) A table containing details about that version update. Here is where I'm confused. There can be many details for each revision, so I'd need to reference the correct row in the version table, but it has two PRIMARY KEYS. Does that mean that this table should have three PRIMARY KEYS (softwareID, versionID, and version_detailID)?
CREATE TABLE version_details (
?????
versionID INT NOT NULL,
description VARCHAR(255),
summary TEXT
);
My other thought is to add an AUTO INCREMENT row to the versions table and use that as the PRIMARY KEY, but then how do I ensure that there is only one instance of software and version in the table. Yeesh, this stuff is confusing when you first start out.
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For the second time in the past 6 months my hosts have shut down my site because it was apparently causing a shared mySQL 5 server to max out at 100% usage. I'm not much a programmer, know even less about mySQL and obviously don't have access to the server from the back end but I'd appreciate any advice as to whether it is my query that is causing the problem: Set rsMail = objConnC.Execute("SELECT * ,(SELECT COUNT(*) FROM comments WHERE comments.submissionID = submissions.submissionID AND comments.commentInclude = 1) as CommentCount FROM submissions WHERE PigeonHole = 'mailbag' AND Status <> 'hold' ORDER BY submissionID DESC LIMIT 20") Apart from two instances, this query has been working fine on a page that receives between 10,000 and 30,000 visits a day. The site is running on Windows and coded in ASP but is calling a separate mySQL server.
How Can I "see" A Table In A Database On A Remote MySQL Server After Creating It
I used the following SQL to create a new table in a database on a remote MySQL server by copying one already there. I know the table exists SOMEWHERE in cyberspace. I can read its data, write to it, delete from it. But I cannot see it. The only way I know it exists is by running this SQL from Access 97 pass through query: SELECT ALL new_tbl.name FROM new_tbl The database resides on a MySQL server that was created with a single table (named test) in it for testing purposes. I ran the following SQL to create another copy of the table in the same database named: new_tbl CREATE TABLE new_tbl SELECT * FROM test; I cannot see the new table in the Access 97 database window under the Tables Tab. Anybody know how to overcome this? Its a severe drawback to programming efforts not to KNOW what tables are in your database.
Beta Testers Wanted For New "blueshell Data Guy Professional"
We are seeking beta testers for our upcoming "blueshell Data Guy Professional". This new application is an extension to our award-winning universal database editor "blueshell Data Guy" (bDg) valuable for database professionals like DBA's, developers etc. Although bDg serves almost all databases, we're planning to focus the beta on MS SQL Server, Oracle, MySQL, Sybase Anywhere, Advantage DB Server, DB2 and Jet. Nevertheless, if you want to test it against any other DBMS, please let us know. These are the main features of bDg Professional: - All features of bDg Standard like schema browsing using a tree view, table view and edit, table design, import, export, desktop connectivity etc. (See http://www.blueshell.com/Product.asp?Prod=bDg for product details) - "SQL Cook" to assist you in entering valid SQL statements (something between Intellisense and a wizard) - Design (edit) objects like stored procedures, functions, packages, triggers and sequences - Easy design of table relationship (foreign keys).
Multi-table UPDATE Problem (was "mysql Question")
$sql = "UPDATE char_lair , char_main m SET l.zip_id='$zip', l.lair_addr='$address', m.alias='$alias', m.real_name='$name', m.align='$align' WHERE m.id = $cid AND m.lair_id = l.id"; I am using mysql v.3.23, I got error in this code but can't find the solution. The message displayed in browser is: --------------------------------------------------- UPDATE char_lair , char_main USING char_lair l, char_main m SET l.zip_id='z3', l.lair_addr='a3', m.alias='ch3', m.real_name='rn3', m.align='good' WHERE m.id = 2 AND m.lair_id = l.idYou have an error in your SQL syntax near ' char_main USING char_lair l, char_main m SET l.zip_id='z3', l.lair_addr='a3'' at line 1 ========================================= Is there any syntax error?
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