2014-02-27 35 views
0

我想創建PHP的一個表,但它不會產生任何表不能在PHP創建表,沒有錯誤消息顯示

$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); 

末作品的回聲,但不創建表。我沒有收到任何類型的錯誤消息。

+2

使用'mysql_error()'來查看錯誤。 –

+0

感謝您的幫助,但並未顯示任何錯誤 – user3362367

+0

在您的「外鍵」中留意 - 檢查您的表字段是否具有相同類型,即INT(UN)SIGNED在每個字段中必須相同。 –

回答

0

您可以使用此代碼解決您的問題

<?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); 
    } 
?>