PHP mysqli_query() 期望参数 1 是 mysqli，对象在 [英] PHP mysqli_query() expects parameter 1 to be mysqli, object given in

问题描述

我有这个代码

class Db_conn {
    private $sn = 'localhost';
    private $un = 'root';
    private $up = '';
    public function connect(string $db_n){
        $conn = mysqli_connect($this->sn, $this->un, $this->up, $db_n);
        if (!$conn) {
            die("Připojení se nezdařilo: " . mysqli_error($conn));
        } else {
            return $conn;
        }
    }
}

还有这段代码

public function update($query){
    $dbconn = new Db_conn();
    if (mysqli_query($dbconn, $query)) {
        return True;
    } else {
        return False;
    }
}

在这一行 if (mysqli_query($dbconn, $query)) {它说有一个错误.

And on this line if (mysqli_query($dbconn, $query)) { it says there is an error.

警告:mysqli_query() 期望参数 1 为 mysqli，对象在 D:\xampp\htdocs\purkiada2\content\Db_parser.inc.php 第 21 行

Warning: mysqli_query() expects parameter 1 to be mysqli, object given in D:\xampp\htdocs\purkiada2\content\Db_parser.inc.php on line 21

推荐答案

mysqli_query 需要一个 mysqli 连接，但你给它一个 Db_conn 对象.这就是错误消息所说的.

mysqli_query expects a mysqli connection, but you give it a Db_conn object. This is what the error message says.

您必须先连接，然后将此(mysqli)连接提供给mysqli_query，例如

You must first connect and then give this (mysqli) connection to mysqli_query, e.g.

public function update($query){
    $dbconn = new Db_conn();
    $conn = $dbconn->connect();
    if (mysqli_query($conn, $query)) {
        return True;
    } else {
        return False;
    }
}

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