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($conn);
}执行脚本确实会创建所需的表,但是它也会给出以下错误
Warning: mysql_query() expects parameter 2 to be resource, boolean given in C:\xampp\htdocs\test\test.php on line 136
Error creating table:现在,根据我的说法,这意味着$create返回"FALSE“--如果没有创建表,这将是一种情况,但是在这种情况下,表已经创建了,那么是什么使它成为布尔值呢?我怎样才能阻止这个错误?
发布于 2013-12-01 11:23:09
条件应该是
if ($create)
{
echo "Table created successfully";
}您已经运行了查询,并且您正在再次调用mysql_query($con,$create)。
发布于 2013-12-01 11:24:26
您正在运行两次$create查询。试试这个:
if ($create)
{
echo "Table created successfully";
}发布于 2013-12-01 11:24:57
为mysql_query()提供第二个参数,这将导致错误。另外,你在$con中拼错了mysql_error()。请使用以下代码:
$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
}https://stackoverflow.com/questions/20311561
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