PHP和MySQL查詢數據庫

-2

<?php 
    require_once 'login1.php'; 
    $connect = new mysqli($db_hostname, $db_username, $db_password); 
    if ($connect -> connect_error) { 
     die("Unable to connect to MySQL: " . mysql_error()); 
    } 
    else { 
     echo "connected successfully."; 
    } 

    //selecting a database 
    $connects= mysqli_connect("localhost", "debola" , "mypassword"); 
    if (mysqli_connect_errno()){ 
     echo "failed to connect to MySQL: " . mysqli_connect_error(); 
    } 
    else{ 
     echo "Database anuoluwa selected"; 
    } 

    $query = "SELECT * FROM anuoluwa"; 
    $result = mysqli_query($query); 
    if (!$result) die ("Database access failed: " . mysqli_error($connect)); 
    $rows = mysqli_num_rows($result); 
    for ($j = 0 ; $j < $rows ; ++$j) { 
     echo 'Author: ' . mysql_result($result,$j,'author') . '<br>'; 
     echo 'Title: ' . mysql_result($result,$j,'title') . '<br>'; 
     echo 'Category: ' . mysql_result($result,$j,'category') . '<br>'; 
     echo 'Year: '  . mysql_result($result,$j,'year')  . '<br>'; 
     echo 'ISBN: '  . mysqli_result($result,$j,'isbn')  . '<br><br>'; 
    } 
?>

這是返回錯誤的部分：PHP和MySQL查詢數據庫

$query = "SELECT * FROM anuoluwa"; 
$result = mysqli_query($query); 
if (!$result) die ("Database access failed: " . mysqli_error($connect)); 
$rows = mysqli_num_rows($result);"

參數要求，我無法弄清楚。

來源

2017-02-18 Adebola Rehoboth Ajewole

你爲什麼連接兩次？第一次連接調用應該足夠了，作爲優點，它使用適當的面向對象的接口，而不是傳統的過程接口。作爲一個優點，這將避免陷入偶然使用'mysql_result'的陷阱，這裏完全錯誤，因爲它缺少'i'。 – tadman

你的代碼在太多層次上失敗。 –

對不起，但我不明白這個問題如何保證upvote –