-1
出于某种原因,查询内部的以下代码在我的MySQL命令控制台中工作,但是当我尝试在PHP中将它作为查询运行时,某些事情总是出错,我不确定是什么。这是我迄今为止所做的代码。错误消息:数据库查询失败
//2. Perform database query
$query = "SELECT skills.element_id, content_model_reference.element_id, element_name FROM skills, content_model_reference WHERE (skills.element_id = content_model_reference.element_id)";
$result = mysql_query($query);
//Tests if there was a query error
if(!$result){
die("Database query failed.");
}
是否有某些东西阻止在MySQL(SELECT行)工作的代码,或者我的语法在某种程度上是错误的?
编辑:所以这就是说我没有选择一个数据库。但我认为我有。下面是它上面的代码:
//1. Create a database connection
$dbhost = "host"; //Host: Can be either an IP address, or a domain (like google.com).
$dbuser = "user";//User: The user that is connecting to the database.
$dbpass = "pass";//Password: This is the password that the user is using.
$dbname = "db";//Name: This is the name of the database.
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);//The value, 'handle,' is the connection.
//Test if connection occurred. Die ends the program/php, and in this case, also prints a message
if(mysqli_connect_errno()){
die("Database connection failed: ".
mysqli_connect_error().
" (". mysqli_connect_errno() . ")"
);
}
就像我说的,我得到的是仅与该查询错误消息,服务器是罚款与我的数据库连接。
您需要使用mysql_error()来查看实际的错误消息是什么。 –
不要再使用mysql_ *。使用mysqli_ *或PDO。 – WillardSolutions
这是说我没有选择一个数据库,但我认为我做了。 –