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我是 PHP 新手，我想显示我的表的一列名为 music，表名是作曲家，但每次我给出错误：致命错误：调用非对象上的成员函数 pg_fetch_object()在第 35 行的 /Applications/XAMPP/xamppfiles/htdocs/admin/music.php

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";
        }

    ?>
4

2 回答 2

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";
        }
于 2013-10-26T19:09:38.130 回答
0

更改pg_fetch_object为mysql_fetch_object

于 2013-10-26T19:06:36.010 回答