连接数据库的正确方法是什么?我没有收到错误但也许我的连接没有正确使用我使用mysqli 这是我的代码
<?php
require_once("configur.php");
$mysqli = new mysqli(localhost, root, password, user);
$query_image = 'INSERT INTO fun_table (images1, images2, images3, images4)
values( "' . $_FILES['file1']['name'] . '",
"' . $_FILES['file2']['name'] . '",
"' . $_FILES['file3']['name'] . '",
"' . $_FILES['file4']['name'] . '"
)';
if ($mysqli->query($query_image) === TRUE) {
echo "<script language='javascript'>\n";
echo "alert('Upload successful!')";
echo "</script>\n";
} else {
echo "Error updating record: " . $conn->error;
}
$mysqli->close();
?>
答案 0 :(得分:1)
$ mysqli = new mysqli(localhost,root,password,user);
您需要定义它们或将它们放在引号中。
像:
$servername = 'localhost';
$username = 'user123';
$password = 'helloworld';
$dbname ="testdatabase";
$conn = mysqli_connect($servername, $username, $password, $dbname);
或者
$conn = mysqli_connect("localhost", "user123", "helloworld", "testdatabase");
答案 1 :(得分:0)
mysqli
已过时,请勿使用。相反,您可以使用PDO进行连接,例如:
<?php
try
{
$db = new PDO("mysql:host=".localhost."; dbname=".db_name, root, "");
$db->setAttribute(PDO::ATTR_ERRMODE, PDO:: ERRMODE_EXCEPTION);
}
catch (PDOException $exception)
{
die('Error: '.$exception);
}
?>