MYSQL创建表错误“的表必须有1列”

Question

我使用PHP在我的数据库创建MySQL表：

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然而，出现了错误，并说：一个表必须至少有1列。

它为什么这样做？

下面是用于创建表的全部PHP文件：

<?php 

$host="localhost"; // Host name 
$username="root"; // Mysql username 
$password=""; // Mysql password 
$db_name="questapic"; // Database name 
$tbl_name="tabledirector"; 

// Connect to server and select database. 
mysql_connect("$host", "$username", "$password")or die("cannot connect"); 

// get data that sent from form 
$TN = mysql_real_escape_string($_POST['name']); 
$TQ = mysql_real_escape_string($_POST['q']); 
$CR = mysql_real_escape_string($_POST['creator']); 
$datetime=date("d/m/y"); //create date time 
$R1=rand(5000, 15000000); 
$R2=rand(5000, 15000000); 
$R3=rand(5000, 15000000); 

$URL = $TN . $R1 . $TQ . $R2 . $CR . $R3; 

$URL=str_replace(" ","#%","$URL"); 


mysql_select_db("$db_name")or die("cannot select DB"); 
$sql="INSERT INTO $tbl_name(URL,topic)VALUES('$URL','$TQ')"; 
$result=mysql_query($sql); 

$sql="CREATE TABLE $URL(Image BLOB,Rating INT(255),Id INT KEY AUTO_INCREMENT)"; 
$result=mysql_query($sql); 


if (!$result) die (mysql_error()); 
mysql_close(); 

header("location:Your_Special_Code_Is.php?id=$URL"); 



?> 

Answer 1

尝试：

$sql="CREATE TABLE ".$URL."(Image BLOB,Rating INT(255),Id INT KEY AUTO_INCREMENT)";