I've tried it and i still got the same error. I even dropped table1 and recreated it to
include the ENGINE=InnoDB and it was successful but table2 remains unsuccessful.
> Date: Tue, 7 Apr 2009 17:56:49 +0400
> From: evgeniy@stripped
> To: defatigue@stripped
> CC: mysql@stripped
> Subject: Re: PHP-MYSQL Question
>
> I suppose the problem is that table1.table1_id and table2.table1_id are
> of different types. The first one is INT UNSIGNED and the second is just
> INT.
>
>
> abdulazeez alugo wrote:
> > Table1 is as below:
> >
> > CREATE TABLE table1(table1_id INT UNSIGNED NOT NULL AUTO_INCREMENT,
> > PRIMARY KEY(table1_id),
> > entrytitle VARCHAR(100) NOT NULL,
> > entrytext TEXT NOT NULL,
> > entrydate TIMESTAMP NOT NULL)" );
> >
> > I did not put the engine. Could that be the problem?
> >
> > > Date: Tue, 7 Apr 2009 17:48:16 +0400
> > > From: evgeniy@stripped
> > > To: defatigue@stripped
> > > CC: mysql@stripped
> > > Subject: Re: PHP-MYSQL Question
> > >
> > > # perror 150
> > > MySQL error code 150: Foreign key constraint is incorrectly formed
> > >
> > > What does table1 look like?
> > >
> > > abdulazeez alugo wrote:
> > > >
> > > > Yeah I used the mysql_error and it returned Can't create table
> > '.\website\table2.frm' (errno: 150). So what does that say?
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >
> > > >> Date: Tue, 7 Apr 2009 17:38:59 +0400
> > > >> From: evgeniy@stripped
> > > >> To: defatigue@stripped
> > > >> CC: mysql@stripped
> > > >> Subject: Re: PHP-MYSQL Question
> > > >>
> > > >> Perhaps you don't have permissions to create tables?
> > > >>
> > > >> It would have been much clearer if your script was like this:
> > > >>
> > > >> $result=mysql_query($your_create_table_statement);
> > > >> if($result){ print"Successful";}
> > > >> else {print "Unsuccessful: ".mysql_error()}
> > > >>
> > > >>
> > > >> abdulazeez alugo wrote:
> > > >>> Hi guys,
> > > >>>
> > > >>> Please can anyone tell me what I'm doing wrong with the code
> > below? It keep
> > > >>> returning unsuccessful.
> > > >>>
> > > >>> $result=mysql_query("CREATE TABLE table2(table2_id INT NOT
> NULL
> > PRIMARY KEY
> > > >>> AUTO_INCREMENT,
> > > >>> table1_id INT NOT NULL,
> > > >>> name VARCHAR(100) NOT NULL,
> > > >>> school VARCHAR(100) NOT NULL,
> > > >>> comment TEXT NOT NULL,
> > > >>> entrydate TIMESTAMP NOT NULL,
> > > >>> FOREIGN KEY(table1_id) REFERENCES table1(table1_id))
> > > >>> ENGINE = INNODB" );
> > > >>>
> > > >>> if($result){ print"Successful";}
> > > >>> else {print "Unsuccessful";}
> > > >>>
> > > >>> Thanks in advance. Cheers.
> > > >>>
> > > >>> Alugo Abdulazeez.
> > > >>>
> > > >>>
> _________________________________________________________________
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> > > >>
> > > >> --
> > > >> С уважением,
> > > >> Евгений Косов
> > > >>
> > > >> --
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> > > --
> > > С уважением,
> > > Евгений Косов
> >
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>
> --
> С уважением,
> Евгений Косов
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