Question

您好我需要一些帮助编写php来连接数据库我的源代码是

TSnackbar.make(findViewById(android.R.id.content),"Hello from TSnackBar.",TSnackbar.LENGTH_LONG).show();

但它显示错误

警告：mysqli_select_db（）需要2个参数，1在第11行的C：\ xampp \ htdocs \ 123 \ tryforum \ main_forum.php中给出无法选择数据库

如何解决

Answer 1

您需要将连接变量传递给mysqli_select_db。 See this link

$host="127.0.0.1"; // Host name 

$username="root"; // Mysql username 

$password=""; // Mysql password 

$db_name="test"; // Database name 

$tbl_name="forum_question"; // Table name 

// Connect to server and select databse.
$con = mysqli_connect("$host", "$username", "$password") or die("cannot connect"); 
mysqli_select_db($con, "$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name ORDER BY id DESC";`

Answer 2

mysqli_select_db()

首先需要两个参数与数据库连接，然后是你的数据库名称

$conn=mysqli_connect("$host", "$username", "$password")or die("cannot connect"); 
mysqli_select_db($conn,"$db_name")or die("cannot select DB");
$sql="SELECT * FROM $tbl_name ORDER BY id DESC";
$result=mysqli_query($conn,$sql);

无法访问数据库

2 个答案: