PHP 注意，返回 "array array array"而不是值

Question

终于第一次接触 PHP，我编写了这个程序，但我陷入了困境。我到处搜索了大约2个小时才找到解决方案。

基本上，我正在连接到本地数据库并尝试从“歌曲”表中获取所有行，并按名称显示它们。我没有得到他们的名字，而是收到一条通知，上面写着“注意:第 47 行 C:\xampp\htdocs\musiclibrary\index.php 中的数组到字符串转换”

我当前的输出如下所示:

Title   Artist  Genre
Array   Array   Array
Array   Array   Array
Array   

然后我的代码是...

<?php

        // Require configuration file
        require_once 'config.php';

        // Connect to the database
        $db_server = mysqli_connect($db_hostname, $db_username, $db_password);

        // Check for database connection error
        if(!$db_server)
        {
            die("Unable to connect to MySQL: " . mysql_error());
        }

        // Select a database
        // The mysqli_select_db() function is used to change the default database for the connection.
        mysqli_select_db($db_server, $db_database);

$prompt = array('Story title', 'Time', 'Person');
$prompt = array('Story title', 'Time', 'Person');
// Page title
echo "<h1>My Music Collection</h1>";

// Get music collection
$query = "SELECT * FROM songs";
$result = mysqli_query($db_server, $query);
$rows = mysqli_num_rows($result);

// If rows exist
if($rows > 0)
{
    // Create HTML table
    echo "<table>";
    echo "<tr><th>Title</th><th>Artist</th><th>Genre</th></tr>";

    // Loop through each row in the database table
    for($j = 0; $j < $rows; $j++)
    {
        // Build HTML table row

//PROBLEM LIES HERE ON THESE MYSQL_FETCH_ASSOC PARTS

        echo "<tr>";
        echo "<td>" .  mysqli_fetch_assoc($result,$j,'title') . "</td>";
        echo "<td>" .  mysqli_fetch_assoc($result,$j,'artist') . "</td>";
        echo "<td>" .  mysqli_fetch_assoc($result,$j,'genre') . "</td>";
        echo "</tr>";   
    }       

    echo "</table>";
}

// If there are no songs in the database table
else
{
    echo "There are currently no songs on file.";
}

    ?>

有什么解决方案可以输出数据库中的行名称吗？谢谢!

Answer 1

使用下面的代码替换您的代码

1. 首先将值提取到数组中

2. 显示表格行中的值

   $values = mysqli_fetch_assoc($result);

   echo "<tr>";

   echo "<td>" . $values ['title']. "</td>";

   echo "<td>" . $values ['artist'] . "</td>";

   echo "<td>" . $values ['genre')]. "</td>";