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From:Eugene Kosov Date:April 7 2009 1:56pm
Subject:Re: PHP-MYSQL Question
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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.
>  > >>>
>  > >>> _________________________________________________________________
>  > >>> More than messages–check out the rest of the Windows
> Live™.
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>  > >>
>  > >> --
>  > >> С уважением,
>  > >> Евгений Косов
>  > >>
>  > >> --
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>  >
>  >
>  > --
>  > С уважением,
>  > Евгений Косов
> 
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-- 
С уважением,
Евгений Косов
Thread
PHP-MYSQL Questionabdulazeez alugo7 Apr
  • Re: PHP-MYSQL QuestionEugene Kosov7 Apr
    • RE: PHP-MYSQL Questionabdulazeez alugo7 Apr
      • Re: PHP-MYSQL QuestionEugene Kosov7 Apr
        • RE: PHP-MYSQL Questionabdulazeez alugo7 Apr
          • Re: PHP-MYSQL QuestionEugene Kosov7 Apr
            • RE: PHP-MYSQL Questionabdulazeez alugo7 Apr
              • Re: PHP-MYSQL QuestionSpiros Papadopoulos7 Apr
                • RE: PHP-MYSQL Questionabdulazeez alugo7 Apr
                  • RE: PHP-MYSQL Questionabdulazeez alugo7 Apr