From: Scott Haneda (talklistsnewgeo.com)
Date: Wed Apr 29 2009 - 22:52:41 CDT

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On Apr 29, 2009, at 11:29 AM, Antonio PHP wrote:

> This is MySQL data structure. - I underlined where it causes the error
> message. (datetime)
> `id_Company` smallint(6) unsigned NOT NULL AUTO_INCREMENT,
> `Name` varchar(50) CHARACTER SET utf8 COLLATE utf8_unicode_ci NOT NULL
> `Revenue` mediumint(6) NOT NULL,
> `Company_Size` mediumint(6) NOT NULL,
> `Ownership` tinyint(1) NOT NULL,
> `Homepage` varchar(50) CHARACTER SET utf8 COLLATE utf8_unicode_ci
> NOT NULL,
> `Job_Source` varchar(100) CHARACTER SET utf8 COLLATE utf8_unicode_ci
> NOT
> NULL,
> `Updated` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP,
> *`Created` datetime NOT NULL,
> *PRIMARY KEY (`id_Company`),
> KEY `Ownership` (`Ownership`)
> )
> ENGINE=InnoDB DEFAULT CHARSET=utf8
> FOREIGN KEY (`Ownership`) REFERENCES `ownership` (`id_Ownership`) ON
> DELETE
> CASCADE ON UPDATE CASCADE;

Next time can you include unmodified SQL so it is a copy and paste for
me, rather than debugging what changes you made that are causing error.

> Here is php script -----------------------------
> $sql = "INSERT INTO company SET
> Name='$Name',
> Revenue='$Revenue',
> Company_Size='$Company_Size',
> Ownership='$Ownership',
> Homepage='$Homepage',
> Job_Source='$Job_Source'
> *Created=NOW() // if I remove this line it works fine.
> *";
> mysql_query ($sql) or die (mysql_error());

Same here, as I am not sure your edits are just edits, or the lack of
a comma after the job source variable is the issue.

This works on my end:

      $Name = 'Tom';
      $Revenue = '100';
      $Company_Size = '500';
      $Ownership = 'partner';
      $Homepage = 'example.com';
      $Job_Source = 'friend';

      $sql = "INSERT INTO mailing SET
      Name='$Name',
      Revenue='$Revenue',
      Company_Size='$Company_Size',
      Ownership='$Ownership',
      Homepage='$Homepage',
      Job_Source='$Job_Source',
      Created=NOW()";

      echo $sql;

      mysql_query ($sql) or die (mysql_error());

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