php - 从 mysql 转换为 mysqli

Question

我们很难用这段代码将 mysql_query 转换为 mysqli，错误是

Warning: mysqli_query() expects parameter 1 to be mysqli, null given in C:\xampp\htdocs\Projects\MeterReading\include\functions.php on line 45

这是出现错误的 function.php 代码

//Display Data
function DisplayBLLData($db) {
    //global $conn;
    $sql = "SELECT * from `bl_inv_discon`";
    $result = $db->query($sql);
    var_dump($result); die();
    return $result;
}

这是另一个用于查询的代码

<?php
include_once 'config.php';

class DB extends mysqli {
    function DB() {
        global $dbhost, $dbuser, $dbpwd, $dbname;
        $conn = mysqli_connect($dbhost, $dbuser, $dbpwd, $dbname) or die("Failed to connect to server!" .  mysqli_error());
        return $conn;
    }
    /*function selectDB($dbname) {
        global $dbhost, $dbuser, $dbpwd;
        $conn   = mysqli_connect($dbhost, $dbuser, $dbpwd) or die("Failed to connect to server!" . mysqli_error());
        $result = mysqli_select_db($dbname, $conn) or die("Failed to select database!" . mysqli_error());
        return $result;
    }*/
    function query($sql) {
        global $dbhost, $dbuser, $dbpwd, $dbname;
    //  global $conn;
        $conn = mysqli_connect($dbhost, $dbuser, $dbpwd, $dbname) or die("Failed to connect to server!" .  mysqli_error());

        $result = mysqli_query($conn, $sql) or die(mysqli_error($conn));
        //var_dump($conn); die();
        return $result;
    }
    function num_rows($data) {
        return mysqli_num_rows($data);
    }
    function fetch_array($result) {
        return mysqli_fetch_array($result);
    }
    function fetch_assoc($data) {
        return mysqli_fetch_assoc($data);
    }
    function escape_string($data) {
        return mysqli_real_escape_string($data);
    }
}
$db = new DB;
?>

Answer 1

您在这里所做的只是创建一个 DB 类的对象，而不是调用 DB 函数来进行连接。我假设您正在考虑将 DB() 作为构造函数，但在 php 构造函数中以不同的方式创建(使用 __construct 关键字)。您需要做的很简单。

$obj=new DB();
$db=$obj->DB();

然后将 $db 作为参数传递给您的函数，我希望它能正常工作。