无法在PHP中创建表,未显示错误消息

无法在PHP中创建表,未显示错误消息,php,mysql,Php,Mysql,我试图在PHP上创建一个表,但它没有创建任何表 $Usernaam = YuYu; $UserAppartement = $Usernaam . "_Appartement"; $Usertable = $Usernaam. "_Huizen"; $immodata = mysql_connect("localhost", "root", "password") or die(mysql_error()); mysql_select_db("immoda

我试图在PHP上创建一个表,但它没有创建任何表

    $Usernaam = YuYu;
    $UserAppartement = $Usernaam . "_Appartement";
    $Usertable = $Usernaam. "_Huizen";
    $immodata = mysql_connect("localhost", "root", "password") or die(mysql_error());

    mysql_select_db("immodata", $immodata);


    $sql = "create table $UserAppartement 
    (
                ID int NOT NULL AUTO_INCREMENT ,
                SoortLokatie_ID int(11) NOT NULL,
                Verdieping varchar(15) NOT NULL,
                GasMN   int(15) ,
                ElektriciteitMN int(15) ,
                WaterMN int(15) ,
                EAN_Gas varchar(15) ,
                EAN_Elektriciteit   varchar(15) ,
                Oppervlakte int(3) ,
                Kamers  int(1),
                Tuin    TINYINT(1) ,
                Terras  TINYINT(1) ,
                Living  TINYINT(1),
                Keuken  TINYINT(1) ,
                Balkon  TINYINT(1) ,
                Badkamer    TINYINT(1) ,
                Toilet  TINYINT(1),
                Beschrijving    varchar(1000) default 'Geen beschrijving',
                Status_ID   int(1),
                Zoek_ID varchar(6),
                Huis_ID int(11),
                PRIMARY KEY (ID),
                FOREIGN KEY (Huis_ID) REFERENCES $Usertable(ID);

    )
            ";


    echo "Table has been created ";
    mysql_query($sql, $immodata);

最后的回显可以工作,但表没有创建。我没有收到任何类型的错误消息。

您可以使用此代码来解决您的问题

  <?php
   $con=mysqli_connect("example.com","peter","abc123","my_db");
  // Check connection
  if (mysqli_connect_errno())
   {
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
   }

   // Create table
     $sql="CREATE TABLE Persons(FirstName CHAR(30),LastName CHAR(30),Age INT)";

     // Execute query
     if (mysqli_query($con,$sql))
    {
   echo "Table persons created successfully";
    }
   else
    {
     echo "Error creating table: " . mysqli_error($con);
    }
?>


使用
mysql\u error()
查看错误。感谢您的帮助,但它不会在您的
外键中显示任何错误。请检查您的表字段是否具有相同的类型,即INT(UN)签名的每个字段必须相同。删除最后一个括号前的分号您的错误是
无法添加外键约束
:-)