Question

这是代码，为什么我会收到此错误？

警告：mysqli_num_rows（）期望参数1为mysqli_result，boolean given

on $ conta_artisti = mysqli_num_rows（$ preferenze1）;

如果查询没有找到项目？

session_start();
$mail = $_SESSION['Mail'];
$query = $mysqli->query("SELECT * FROM valutazione WHERE email='$mail'"); 
$countresults = mysqli_num_rows($query);

if($countresults == 0)
{
$preferenze = $mysqli->query("DROP VIEW IF EXISTS artisti_p"); //recupero id degli artisti preferiti
$preferenze1 = $mysqli->query("CREATE VIEW artisti_p AS SELECT idartista FROM artisti_preferiti WHERE email='$mail'"); 
if($preferenze1)
{

    $conta_artisti = mysqli_num_rows($preferenze1);
    if($conta_artisti > 0)
    {
        $preferenze = $mysqli->query("DROP VIEW IF EXISTS album_p"); 
        $album_artita_p = $mysqli->query("CREATE VIEW album_p AS SELECT idalbum FROM artisti_p JOIN artisti_album ON artisti_p.idartista=artisti_album.idartista"); 
        $preferenze = $mysqli->query("DROP VIEW IF EXISTS info_album_p");
        $info_album_artista_p = $mysqli->query("CREATE VIEW info_album_p AS SELECT * FROM album_p JOIN album ON album_p.idalbum=album.ID'");
        $index_p = 0;
        while($row = mysqli_fetch_array($info_album_artista_p)) 
        {
            $id_p[$index_p]=$row['idalbum'];
            $nome_p[$index_p]=$row['NOME'];
            $casa_p[$index_p]=$row['CASA_DISCOGRAFICA'];
            $anno_p[$index_p]=$row['ANNO_USCITA'];
            $durata_p[$index_p]=$row['DURATA_ALBUM'];
            $index_p++;
        }
    }
}

Answer 1

查看mysqli::query()

的文档

失败时返回FALSE。

您的代码假定不会发生这种情况。它刚刚做到了。处理那个案子：

$query = $mysqli->query("SELECT * FROM valutazione WHERE email='$mail'"); 
if (! $query) {
    die("error");

}
$countresults = mysqli_num_rows($query);

BTW还要看一下：How can I prevent SQL injection in PHP?

php - mysqli_num_rows（）期望参数1为mysqli_result，给定布尔值

1 个答案: