我已将db中的图像类型设置为blob，并且图像存储在db中，因为它显示为blob但是当我检索它时，只显示文件名

该代码用于html页面，用于创建表单

<!DOCTYPE HTML>
    <html>
    <head>
    </head>
    <body>
        <h1>Login</h1>
        <form method="post" action="insert.php">
            Name <input name="name" placeholder="Enter your Name" type="text"/><br>
            <br>
            Password <input name="password" type="file"/><br>
            <br>
            <button type="submit" value="Submit">Submit</button>
        </form>
    </body>
    </html>

这是我用来在数据库中插入数据的PHP代码

<?php
    $servername="localhost";
    $username="root";
    $password="";
    $dbname="test";

    $conn=new mysqli($servername, $username, $password, $dbname);

    if($conn->connect_error)
    {
        die($conn->connect_error);
    }
    else 
    {

    }

    $name=$_POST['name'];
    $password=$_POST['password'];

    $sql ="INSERT INTO `login`(`name`, `password`) VALUES ('$name', '$password')";

    if($conn->query($sql) === TRUE)
    {
        echo "New Record Created";
        header("Location: showStudents.php");
    }
    else
    {
        echo "Error " .$sql. "<br>" .$conn->error;
        }
    ?>

这是php的代码，用于显示网页中的文件，但除了显示图像名称外，不显示图像

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "test";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

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

$sql = "SELECT * FROM login";

$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "Name: " .$row["name"]. " Image: " .$row["password"]. "<br>";
    }
} else {
    echo "0 results";
}

?>
php database