[peterlaursen]

We fixed such issue in between RC3 and FINAL.

And it works for me now with your example.

Now you please try 5.23 FINAL.

this tokenizer is very complicated!  But still it should not happen of course!  

The reason why this reoccurred is that with a single query only in the Editor 'execute' (single query) will execute the query even if the cursor is positioned after a semicolon.  That was a user request.

[peterlaursen]

Yog cannot 'pick up' the password.  It is your password and you will need to remember it!  What else is the idea of a password?

If you forgot what password you gave when installing, I think the easiest would be to uninstall (from 'Add/Remove Programs') and install again.  And be careful to remember the password this time!

[peterlaursen]

1)

Could you possibly attach a test case for the first situation?  This could be a data specific issue.

Try this:

1) create a table with the 14 rows like “CREATE TABLE mytable SELECT ….”

2) reproduce with that new 'mytable'

3) export that mytable, zip and attach here (or create a ticket if you do not want to expose data in public)

2)

Please try:

1) Open SQLyog and load the table in DATA and/or RESULT tab as you would normally do.   Note that SQLyog uses memory too (depending on your limit setting)!

2) With SQLyog open and data loaded can you still copy with that other tool?

3) Also try close some process using 20 MB of RAM (or more).  Does that make any difference?  How much RAM and Virtual Memory do you have and how much is normally used? 

[peterlaursen]

It can and will be implemented.  And much similar actually (every such query byilder is a Access clone!)

We will probaly finish the internal (1st) build of the query builder tomorrow. Next we will have to decide in detail how to integrate in the GUI.  

We aim for version 5.3 with unicode support and 1st query builder next week.

[peterlaursen]

I do not understand if your problem is solved or not!

Just let me repeat that connection details (host,port, tunnelling details if used) are OK.  

But your user details (user,password) are not!

If the password is blank in the sqlyog.ini file this only tells that SQLyog has stored a blank password for the connection.

It does not tell what password the MySQL server 'expects' for that user.

Did you install MySQL using the .msi installer?  

If so you probably entered a root password in the 'configuration wizard'.  

This is the password you will need to provide.

[peterlaursen]

“For a one-to-many relationship, to create the foreign key in the 'many' table, do you do the create relationship while that table is selected?”

Do do not create ” one-to-many relationship” while creating the Foreing Key.  Whether it is one-to-one or one-to many is decided by the type of the index on the referenced column

To create a Foreign key both the referencing and the referenced column must have an index.  

Where you create the index as a primary key or a unique index there will only one – if the key is not Primary or unique it will be many.

The most common way is to create a PK from a (not-unique)indexed column on the 'child' to the Primary Key column of the 'parent' . An example: the 'orders' table has a column named 'customer'.  Create a non-unique index on that 'customer' and a Foreign key pointing to the (PK) 'id' column of the 'customers' table.  Every customer can only have one entry in the 'customers' tablke but more in the 'orders' table (because he may place more than one order).

[peterlaursen]

It works fine here.

the SQL sent is

Code:

alter table `test`.`a1` add column `xx` float UNSIGNED NULL after `suman`, add column `yy` double UNSIGNED NULL after `xx`

or

Code:

alter table `test`.`a1` change `xx` `xx` float UNSIGNED NULL , change `yy` `yy` double UNSIGNED NULL

 

and the create statement is after that:

Code:

CREATE TABLE `a1` (
`bikas` int(11) default NULL,
`suman` varchar(25) default NULL,
`xx` float unsigned default NULL,
`yy` double unsigned default NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1

and they are checked next I open the dialogue.

Are you sure you tested the correct version (5.23) this time?  what does Program menu .. help .. about … display?

Please copy the ALTER TABLE statement from the HSITORY tab and the 'CREATE STATEMENT' from the OBJECTS tab.

[peterlaursen]

We will consider saving the latest used string for 'find'.

'find next' is ALT+F.  In the 'find' dialogue you se text like “Find next”.  

Underlining a character like this normally means that the Alt+that_letter will p

rform the action.

[peterlaursen]

Now I do not understand if your probles is solved or not!

But the connection settings are correct.  You 'reach' the MySQL server.  But the MySQL server won't give you access with the user details that you provide.

[peterlaursen]

No .. it will connect to remote MySQL servers as well (if direct connection is available).

I think you just do not understand how to use MONyog!  Just position the mouse over one of the 'groups' (the left column) and the counters for that group will display. Do not click!

BTW: moving this to the 'MONyog' category

[peterlaursen]

we are trying to find a NT4 somewhere now!

BTW also the SetMenuInfo() function creates some issues on WINE.  But here it does not prevent the program to start.  But it is due to this that 'favorites' and 'change table type' (from the table context menu) does not work on WINE.

The function is used when we genrate a menu dynamically – for instance with 'change table type' we generate the menu form the returns of a ” SHOW VARIABLES LIKE 'have_%' ” -statement.  If there is a 'have_innoDB' with the value of “YES” then InnoDB will show as an option in the submenu – else InnoDB will not be an option.

Let us research a little a see if we can find some workaround for this. I think we started using this function from version 5.1.

[peterlaursen]

Please tell:

does the program refuse to start or does this error msg. appear when you are doing a very specific thing?

[peterlaursen]

We do not test on NT4 anymore.  Still it should work.

version 5.23 will be available in a few minutes.  Please try this one.

[peterlaursen]

Quote:

When I try to test my connection it says Error no. 1045

Access denied for user 'Root'@'localhost'

… I am currently not trying to connect to anything.

I do not understand! you try to connect and you do not try to connect ????

This error is a authentication error not a connection error (and not a firewall issue either).  It is the MySQL server that sends this error message.  So what about the password?  When you installed MySQL did you use the installer and configuration wizard?  

If so you entered a password and you must use it.

Quote:

My Mysql is connected and green but I still don't have access

How shall I understand that it is green?  What program displays it as green?   The MySQL 'tray monitor' ?  If so it does not tell that there is any connection but just that the server is running

[peterlaursen]

“QLyog reports an error when I try and do the other one”

Please always tell what error!  copy the dialogue if you want and attach here.

But maybe it is so simple that there must be an index on both the referencing as the referenced column.  Try creating a key on PicID here:

Code:

CREATE TABLE `orderitems` (
`ItemID` int(10) unsigned NOT NULL auto_increment,
`OrderID` int(11) unsigned NOT NULL default '0',
`PicID` int(11) unsigned NOT NULL default '0',
`Quantity` int(11) NOT NULL default '0',
`Material` varchar(20) NOT NULL default '',
`Area` int(11) NOT NULL default '0',
PRIMARY KEY (`ItemID`),
KEY `FK_orderitems` (`OrderID`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1

.

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