PHP/mysql 错误

Question

我正在尝试编写一个脚本，该脚本将一个表添加到已经创建的数据库（'alphacrm'）。我的错误是最底部附近的回显线不会显示，并且没有为我创建表数据库。我已经验证了 $dbSuccess 是真的，不知道在那之后我哪里出错了。我确实尝试查看它是否正确连接，但这似乎不是问题。如果您发现我的错误，将不胜感激！

if ($dbSuccess) {

  $createCoyTable_SQL = "CREATE TABLE alphacrm.tCompany ( ";
  $createCoyTable_SQL .= "ID INT( 11 ) NOT NULL AUTO_INCREMENT PRIMARY ";
  $createCoyTable_SQL .= "preName VARCHAR( 50 ) , ";
  $createCoyTable_SQL .= "Name VARCHAR( 250 ) NOT NULL, ";
  $createCoyTable_SQL .= "RegType VARCHAR( 50 ) NULL, ";
  $createCoyTable_SQL .= "SreetA VARCHAR( 150 ) NULL, ";
  $createCoyTable_SQL .= "SreetB VARCHAR( 150 ) NULL, ";
  $createCoyTable_SQL .= "SreetC VARCHAR( 150 ) NULL, ";
  $createCoyTable_SQL .= "Town VARCHAR( 150 ) NULL, ";
  $createCoyTable_SQL .= "County VARCHAR( 150 ) NULL, ";
  $createCoyTable_SQL .= "Postcode VARCHAR( 150 ) NULL, ";

  $createCoyTable_SQL .= "COUNTRY VARCHAR( 250 ) NOT NULL ";
  $createCoyTable_SQL .= ")";

  if (mysql_query($createCoyTable_SQL)) {
    echo "Creation of TABLE tCompany -- Succesful <br />";
  }

  $createPersonTable_SQL = "CREATE TABLE alphacrm.tPerson ( ";
  $createPersonTable_SQL .= "ID INT( 11 ) NOT NULL AUTO_INCREMENT PRIMARY ";
  $createPersonTable_SQL .= "Salutation VARCHAR( 20 ) , ";
  $createPersonTable_SQL .= "FirstName VARCHAR( 50 ) , ";
  $createPersonTable_SQL .= "LastName VARCHAR( 50 ) NOT NULL, ";
  $createPersonTable_SQL .= "CompanyID VARCHAR( 11 ) NOT NULL ";
  $createPersonTable_SQL .= ")";

  if (mysql_query($createPersonTable_SQL)) {
    echo "Creation of tPerson table was succesful <br />";
  }
}

Answer 1

发现错误，改下这一行：

   $createCoyTable_SQL .= "ID INT( 11 ) NOT NULL AUTO_INCREMENT PRIMARY ";

到这一行：

   $createCoyTable_SQL .= "ID INT( 11 ) NOT NULL AUTO_INCREMENT PRIMARY KEY,";

如果解决了问题，请告诉我，同时我正在寻找更多。