2014-04-15 61 views
0

我使用PHP在我的數據庫創建MySQL表:MYSQL創建表錯誤「的表必須有1列」

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然而,出現了錯誤,並說:一個表必須至少有1列。

它爲什麼這樣做?

下面是用於創建表的全部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"); 



?> 
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它在這裏工作:http://sqlfiddle.com/#!2/a2f32 –

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它可能與我的代碼有關。我將發佈完整的php文件 – user3505931

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創建表名$ url? –

回答

1

嘗試:

$sql="CREATE TABLE ".$URL."(Image BLOB,Rating INT(255),Id INT KEY AUTO_INCREMENT)"; 
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也許這可能有效.. –

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它仍然會說同樣的錯誤 – user3505931

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不介意!我沒有刷新網頁。萬分感謝! – user3505931