php - fatal error : Call to a member function pg_fetch_object()

Question

我是 PHP 新手，我想显示表中名为音乐的一列，表名是 Composer ，但每次我都会报错: fatal error :在第 35 行/Applications/XAMPP/xamppfiles/htdocs/admin/music.php 中的非对象上调用成员函数 pg_fetch_object()

这是我的代码:

<?php

$host ="localhost";  //host
$username="root"; //username
$password ="";//password of the database
$dbName="wikiseda";//database Name
$tbl_name="users";
$tbl_music="music";
//______________________________
$connection=mysql_connect("$host","$username","$password") or die("can't connect sorry");
mysql_select_db($dbName,$connection) or die('can not select db');

$sql ="SELECT * FROM music";
    $result = mysql_query($sql) or die(mysql_error());
    while($row = $result-> pg_fetch_object()){
        $musicname = $row->composer;
        echo "$musicname";
        }

    ?>

Answer 1

1. mysql_fetch_object 返回对象作为结果。
2. mysql_fetch_row 返回数组作为结果

尝试

 while($row = mysql_fetch_object($result)){
        $musicname = $row->composer;
        echo "$musicname";
        }