Из-за чего возникает эта ошибка?

0

Ошибка:

Fatal error: Uncaught Error: Call to undefined function INSERT INTO movie VALUES(null, 'Джой', 'Joy', '2015', '', 'http://c.cineate.cc/media/m/4/5/154054/0.big.jpg', Now(), '1')() in C:\xampp\htdocs\kinomonster\index.php:50 Stack trace: #0 C:\xampp\htdocs\kinomonster\index.php(87): insert(Object(SimpleXMLElement), Object(SimpleXMLElement), Object(SimpleXMLElement), NULL, Object(SimpleXMLElement), 1) #1 {main} thrown in C:\xampp\htdocs\kinomonster\index.php on line 50

Мой код:

<?php
	
	function insert($name, $desc, $year, $rating, $poster, $category_id) {
		$mysqli = new mysqli('localhost', 'root', '', 'kinomonster');

		if(mysqli_connect_errno()){
			print_f('соединение не установлено');
			exit();
		}

		$mysqli->set_charset('utf8');

	$query = "INSERT INTO movie VALUES(null, '$name', '$desc', '$year', '$rating', '$poster', Now(), '$category_id')";


	$result = false;
	if($mysqli-$query($query)){
		$result = true;
	}
	
		return $result;
	}





	$xml = simplexml_load_file("xml_files/movies.xml") or die("Error:Cannot create object");
	

	//echo count($xml);

	$title = null;
	$title_origin = null;
	$post = null;
	$rating = null;
	$year = null;

	foreach ($xml as $movie_key => $movie) {
		$title = $movie->title_russian;
		$title_origin = $movie->title_original;
		$year = $movie->year;

		

		foreach ($movie->poster->big->attributes() as $poster_key => $poster) {
			$post = $poster;
		}

		if($movie->imdb){
			$rating = $movie->imdb->attributes()['rating'];
		} 

		insert($title, $title_origin, $year, $rating, $post, 1);
	}

	echo "<pre>";
	print_r($xml);
	echo "</pre>";


?>
php
mysql

1 ответов

1

Подключение к базе данных точно есть?

Тут есть разбор примера insert с xml от преподавателя:

https://fructcode.com/ru/qa/ne-dobavlaetsa-zapis-iz-xml-v-bd-cto-delau-nepravilno-jlsfdn/

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