Question

我创建了一个访问我的数据库的类。简化类如下（我将其命名为dbaccess.php）

class dbaccess {
function read($db) {
    $con = mysqli_connect($db);
    if (mysqli_connect_errno()){
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
        }
    $result = mysqli_query($con,"SELECT * FROM equipment");
    while($row = mysqli_fetch_array($result)){
        $print = $print . $row['ID'] . " " . $row['name'] . " " . $row['new_price'] . " " . $row['residual_value'] . "<br>";
        }
    echo $print;
    mysqli_close($con);
    }
}

要访问该类，我使用此代码

include './dbaccess.php';
//define db address
$add = '"localhost","myuser","mypassword","mydbname"';
$db = new dbaccess;
$db->read($add);

此代码生成

Failed to connect to MySQL: Unknown MySQL server host '"localhost","myuser","mypassword","mydbname"'(2)

我不知道如何修复它，这里的任何人都可以帮助我吗？

Answer 1

您将一个字符串传递给mysqli_connect。您需要将"localhost"，"myuser"，...作为单独的变量传递。

http://php.net/manual/en/function.mysqli-connect.php

Answer 2

我和你的有类似的东西，从dropbox下载我的文件并查看内部 DB Connect

Answer 3

class dbaccess {
    function read($db) {
        $con = mysqli_connect($server, $user, $password, $dbname);
(...)

然后在你的代码中，你应该划分参数。

include './dbaccess.php';
//define db address
$db = new dbaccess;
$db->read("localhost","myuser","mypassword","mydbname");

REF

Answer 4

将主机，用户名，密码和数据库存储在单独的变量中可能更好一点：

$host = "localhost";
$user = "myuser";
$pass = "mypassword";
$data = "mydbname";
$db = new dbaccess(); // <-- It's good practice to use parentheses in the constructor statement.
$db->read($host,$user,$pass,$data);

然后你的dbaccess类的实现可能更像是这样：

$con = mysqli_connect($host,$user,$pass,$data);

这是因为mysqli_connect函数将主机，用户名等视为单独的参数，而不是单个字符串。

无法使用php类访问数据库

4 个答案: