2013-03-05 86 views
1

我有以下的PHP文件来创建数据库和表:当它运行时我就返回此PHP错误创建MySQL表

<?php 
$slName = $_POST['slName']; 
$con=mysqli_connect('localhost', 'setlist', 'music'); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

// Create database 
$sql="CREATE DATABASE $slName"; 
if (mysqli_query($con,$sql)) 
    { 
    echo "Database " . $slName . " created successfully"; 
    } 
else 
    { 
    echo "Error creating database: " . mysqli_error(); 
    } 


    // Create table 
$tbl="CREATE TABLE main (name TEXT,orderno INT,Age TEXT)"; 

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

Database created successfully Error creating table:

我通过在命令提示符下执行该成功:

CREATE TABLE main (name TEXT,orderno INT,Age TEXT); 

我在做什么错?

非常感谢!

罗兰

+3

您忘记选择数据库中创建表。 – 2013-03-05 12:30:35

+1

我想你没有选择数据库。它可能是问题 – 2013-03-05 12:31:29

+1

http://www.php.net/manual/en/mysqli.select-db.php – 2013-03-05 12:32:15

回答

1

创建数据库添加mysqli_select_db($slName);

2

后,您必须选择一个数据库中

mysqli_select_db($slName);