Question

我有以下脚本在mysql数据库中创建表

$dbhost = 'localhost';
$dbuser = 'root';
$con = "mysql_connect($dbhost, $dbuser, '')";
$create=mysql_query("CREATE TABLE `$table_name` (ID int NOT NULL AUTO_INCREMENT KEY,high TEXT ,low TEXT ,Divergence TEXT ,Change_Percent TEXT ,Low_Price TEXT ,High_Price TEXT ,Low_Proximity TEXT ,Vol_Index TEXT ,Vol TEXT ,Purchase_Value TEXT)");

                    if (mysql_query($con,$create))
                        {
                        echo "Table created successfully";
                        }
                    else
                        {
                        echo "Error creating table: " . mysql_error($conn);
                        }

执行脚本会创建所需的表，但它也会出现以下错误

Warning: mysql_query() expects parameter 2 to be resource, boolean given in C:\xampp\htdocs\test\test.php on line 136
Error creating table:

现在根据我的意思，这意味着$ create返回“FALSE”，如果没有创建表，情况就是这样，但是在这种情况下表已经创建了，那么它是什么呢？以及如何阻止此错误？

Answer 1

您正在运行$create次查询两次。试试这个：

 if ($create)
 {
    echo "Table created successfully";
 }

Answer 2

您为mysql_query()提供了第二个参数，导致错误。另外，您在$con中拼错了mysql_error()。请改用此代码：

$dbhost = 'localhost';
$dbuser = 'root';
$con = mysql_connect($dbhost, $dbuser, ''); // error: string instead of function
$create=mysql_query("CREATE TABLE `$table_name` (ID int NOT NULL AUTO_INCREMENT KEY,high TEXT ,low TEXT ,Divergence TEXT ,Change_Percent TEXT ,Low_Price TEXT ,High_Price TEXT ,Low_Proximity TEXT ,Vol_Index TEXT ,Vol TEXT ,Purchase_Value TEXT)");

                    if ($create) // error: second mysql_query()
                        {
                        echo "Table created successfully";
                        }
                    else
                        {
                        echo "Error creating table: " . mysql_error($con); // error: misspelled con
                        }

Answer 3

条件应该是

                    if ($create)
                    {
                    echo "Table created successfully";
                    }

你已经运行了查询而你再次点击mysql_query($con,$create)

Answer 4

您的连接不是连接，它是一个字符串：

$con = "mysql_connect($dbhost, $dbuser, '')";

尝试创建真正的连接：

$con = mysql_connect($dbhost, $dbuser, '');

您还在运行mysql_query两次。第一个是没有连接参数。尝试运行一次。连接应该是第二个，而不是第一个参数：

$create = "CREATE TABLE ...";
if (mysql_query($create, $con)) {

PHP创建了mysql表，但也给出了“Table not created”错误

4 个答案: