Php 直接从数据库显示图像

显然，我对显示图像有问题，因为它只显示文件的名称，是否仍有它可以立即显示在网页上？ 下面是我的源代码以及保存图像的数据库的2个屏幕以及我希望它显示的位置：

<?php

error_reporting(E_ERROR); 

include("global.php");


session_start();

$receipt = $_GET['receipt'];
$userid = $_SESSION ['userid'];
if (isset($_SESSION['userid']) == false) 
{
    header ("Location: login.php");
}

$mysqli = new mysqli(spf, dbuser, dbpw, db);
$stmt = $mysqli->prepare("SELECT receipt, date, time, pick, dropoff, userid, carno, cost, branch, area, image FROM items where receipt=?");
$stmt->bind_param("s", $receipt);
$stmt->execute();
$stmt->bind_result($receipt, $date, $time, $pick, $dropoff, $userid, $carno, $cost, $branch, $area, $image);

while ($stmt->fetch()) {
    echo "<table border='1' style='width:40%'>";
    echo "<td>";
    echo "<b>Receipt ID: $receipt</b>";
    echo "<br><br>";
    echo "$image";
    echo "<br><br>";
    echo "<b>Date of Travel: $date</b>";
    echo "<br><br>";
    echo "<b>Time of Travel: $time</b>";
    echo "<br><br>";
    echo "<b>Pick Up Location: $pick</b>";
    echo "<br><br>";
    echo "<b>Drop Off Location: $dropoff</b>";
    echo "<br><br>";
    echo "<b>Area of DropOff: $area</b>";
    echo "<br><br>";
    echo "<b>Cost of Trip: $cost</b>";
    echo "<br><br>";
    echo "<b>User ID (NRIC): $userid</b>";
    echo "<br><br>";
    echo "<b>Branch of Officer: $branch</b>";
    echo "</td>";
}

echo "</table>";

$stmt->close();
$mysqli->close();

?>

试试这个：

echo "<img src='$image'>";

echo”“；

了解
使用
”
use”；写一个答案@romanreign！其他人会更容易看到它（因为这是正确的答案）@Romanreave非常感谢它工作得很好：）
 echo "<img src=$image>"