Question

最后第一次跳进一些PHP，我写了这个程序，我被卡住了。我在整个地方搜索了大约2个小时才找到解决方案。

基本上我正在连接到我的本地数据库，并尝试从我的“歌曲”表中获取所有行，并按名称显示它们。而不是得到他们的名字，我收到一条通知，上面写着“注意：第47行的C：\ xampp \ htdocs \ musiclibrary \ index.php中的数组到字符串转换”

我目前的输出如下：

Title   Artist  Genre
Array   Array   Array
Array   Array   Array
Array

然后我的代码就是......

<?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.";
}

    ?>

输出数据库中行名称的任何解决方案？谢谢！

Answer 1

使用以下代码替换您的代码

首先将值提取到数组
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>";

Answer 2

你需要循环 mysqli_fetch_assoc 函数来计算行然后循环行结果以获取值

以下是代码：

if($rows > 0){

echo "<table>";
echo "<tr><th>Title</th><th>Artist</th><th>Genre</th></tr>";

// Loop through each row in the database table
while($row = $result->mysqli_fetch_assoc()){

    echo "<tr>";
    foreach($row as $key => $value){
        echo "<td>" . $row['title'] . "</td>";
        echo "<td>" . $row['artist'] . "</td>";
        echo "<td>" . $row['genre'] . "</td>";
    }        
    echo "</tr>";   
}       

echo "</table>";
}

希望它有所帮助！

Answer 3

mysqli_fetch_assoc（）函数只接受一个参数。

数组mysqli_fetch_assoc（mysqli_result $ result）

正确的方法是：

$query = "SELECT * FROM songs";

if ($result = mysqli_query($db_server, $query)) {

    while ($row = mysqli_fetch_assoc($result)) {
        echo "<tr>";
        echo "<td>" .  $row['title'] . "</td>";
        echo "<td>" .  $row['artist'] . "</td>";
        echo "<td>" .  $row['genre'] . "</td>";
        echo "</tr>";   
    }

    mysqli_free_result($result);
}

PHP注意，返回“数组数组”而不是值

3 个答案: