PHP/mysql error

本文关键字：error mysql PHP | 更新日期: 2023-09-27

我正在尝试制作一个脚本，该脚本将向已经创建的数据库（"alphacrm"）提供一个表。我的错误是，底部附近的回波线将不会显示，并且没有为我的数据库创建表。我已经验证了$dbSuccess是真的，不确定在那之后我哪里出了问题。我确实试着看看它是否被正确地连接在一起，但这似乎不是问题所在。如果你发现我的错误，我将不胜感激！

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 />";
  }
}

发现错误，更改此行：

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

到此行：

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

请告诉我是否修复了它，同时我正在寻找更多。