我在一个新的PHP网站工作,我需要这个错误的帮助。
警告:mysqli_query()期望参数1为mysqli, null给定在. .
我有类Database.class,和一个抽象类ActiveRecord.php.
class Database {
private static $instance = null;
private function __construct(){}
public static function getInstance(){
if(!self::$instance){
self::$instance = mysqli_connect(DBHOST, DBUSER, DBPASS, DB);
return self::$instance;
}
}
}
和ActiveRecord类在新文件。
abstract class ActiveRecord {
public function save() {
$q = "UPDATE " . static::$table . " SET ";
foreach($this as $k=>$v){
if($k==static::$key) continue;
$q.=$k."='".$v."',";
}
$q = rtrim($q,",");
$keyField = static::$key;
$q.=" WHERE ".static::$key." = " . $this->$keyField;
mysqli_query(Database::getInstance(), $q);
}
}
当我测试save();函数我得到警告mysqli。我创建了一个新文件,并把这个写在里面。
class Category extends ActiveRecord {
public static $table = "categories";
public static $key = "category_id";
}
$cat = Category::get(1);
$cat->name = "Action";
$cat->description = "Action Description";
$cat->save();
问题是在save()函数Database::getInstane()不被识别。但是,如果我把"mysqli_connect("localhost","root",","db")"一切工作正常,脚本更新数据库记录。
有谁知道是什么问题吗?问题是如果实例已经创建,则不返回任何内容。
更新你的getInstance
函数如下
public static function getInstance(){
if(!self::$instance){
self::$instance = mysqli_connect(DBHOST, DBUSER, DBPASS, DB);
}
return self::$instance;
}