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

问题描述：

I have the following script to create tables in a mysql database

$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);
                        }

Executing the script does create an the desired table, however it also gives the following error

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

Now according to me this means $create returns "FALSE" which would be the case if the table was not created, however in this case the table has been created, what then is making it a Boolean? and How can I stop this error??

我有以下脚本在mysql数据库中创建表 p>

$ dbhost ='localhost'; $ dbuser ='root'; $ con =“mysql_connect（$ dbhost，$ dbuser，''）”; $ create = mysql_query（“CREATE TABLE` $ table_name `（ID int NOT NULL AUTO_INCREMENT KEY，高TEXT，低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“表创建成功”; } else { echo“创建表错误：”。 mysql_error（$ conn）; } code> pre>

执行脚本会创建所需的表，但是它也会出现以下错误 p> \ n

 警告：mysql_query（）期望参数2是资源，布局在第136行的C：\ xampp \ htdocs \ test \ test.php中给出
创建表：
  code>   pre> 
 
 现在根据我的意思，这意味着$ create返回“FALSE”，如果没有创建表就会出现这种情况，但是在这种情况下表已经创建了，那么它是什么呢？  ？ 以及如何停止此错误？ p> 
  div>

答

the condition should be

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

u have already run the query and u are pasing again the mysql_query($con,$create)

答

Your connection is not a connection, it's a string:

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

Try creating a real connection instead:

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

You are also running mysql_query twice. The first is without a connection parameter. Try running it once instead. And the connection should be the second, not the first, parameter:

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

答

You are running the $create query twice. try this:

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

答

You supply second parameter for mysql_query() which causes an error. Also, you misspelled $con in mysql_error(). Use this code instead:

$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
                        }

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

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