如何使用php从mySQL表中获取一行数据？

下面是我所拥有的当前PHP代码的一个示例。我只想从表中获取一行，但它返回以下错误：

Call to a member function fetch_assoc() on a non-object

如有任何见解，将不胜感激

$pathname = "C:\Users\BL\Documents\GitHub\Moozik\music";
$files = scandir($pathname);
$server = "localhost";
$user = "root";
$pass = "";

while ($files[0] == "." || $files[0] == "..") {
    array_shift($files);
}

print_r($files);
$song = $pathname . '\\' . $files[0];

$conn = new mysqli($server, $user, $pass);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT song_path FROM song_data";
$result = mysqli_query($conn, $sql);

while($row = $result->fetch_assoc()) {
    $its = $row["song_path"];
    printf($its);
}

mysqli_close($conn);

你可以用

$row=mysqli_fetch_array($result)

您甚至不必使用while，因为您只想获取一行。如果要获取多行，则可以使用while。

您可以使用此选项

$row=mysqli_fetch_row($result)

它将从结果集中获取一行

---

希望这会有帮助

第1点：

You have mixed mysqli Object-oriented and Procedural methods...Use any one

$conn = new mysqli($server, $user, $pass); 
// Here You missed database to be selected

$result = mysqli_query($conn, $sql); // Used procedural method But Connection is by Object Oriented method

第2点：

You have mixed mysqli Object-oriented and Procedural methods...Use any one

$conn = new mysqli($server, $user, $pass); 
// Here You missed database to be selected

$result = mysqli_query($conn, $sql); // Used procedural method But Connection is by Object Oriented method

第3点：

You have mixed mysqli Object-oriented and Procedural methods...Use any one

$conn = new mysqli($server, $user, $pass); 
// Here You missed database to be selected

$result = mysqli_query($conn, $sql); // Used procedural method But Connection is by Object Oriented method

这是一个完整的面向对象方法

$conn = new mysqli($server, $user, $pass, $database);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT song_path FROM song_data";
$result = $conn->query($sql);

while($row = $result->fetch_assoc()) {
           $its = $row["song_path"];
            echo $its;
    }


$conn->close();

---

我想这是一个开始！我现在收到以下消息

mysqli_fetch_row（）期望参数1为mysqli_result

@vigneshAre您确定列和表的名称正确吗？真的@bayblade567？我相信所有这些都是正确的。我怎样才能重新检查代码？此表中当前只有一行数据。错误

mysqli_fetch_row（）要求参数1为mysqli_result

表示结果无效。这意味着查询出了问题。请转到phpmyadmin，然后键入查询并检查它是否给出了结果……转到phpmyadmin，然后转到SQL，然后键入查询，然后执行……您能试试我的答案吗