显示带有用户产品帖子信息的图像？

Question

我正在制作一个像craigslist这样的网站，人们可以在其中发布他们想要出售的东西，并带有帖子标题，帖子描述和电子邮件。问题：我可以从数据库中提取除映像之外的所有数据。上传图像时，它们存储在 C：\xampp\htdocs\uploads 中，图像名称与其余信息一起保存在数据库中。下面是我的php。我试图从数据库中提取图像的名称，并用它来像这样抓住它echo "<img src='uploads/".$image."' width='200'> ";我是php的新手，所以任何提示都值得赞赏。这是存储/检索用户上传图像的好方法吗？谢谢<?php$host = "localhost"; /* Host name */$user = "root"; /* User */$password = ""; /* Password */$dbname = "mydb"; /* Database name */$con = mysqli_connect($host, $user, $password,$dbname);// Check connectionif (!$con) {  die("Connection failed: " . mysqli_connect_error());}$sql= "SELECT * FROM products";$result = $con->query($sql);$image = "SELECT image FROM products";if ($result->num_rows > 0) {    // output data of each row    while($row = $result->fetch_assoc()) {        echo "<img src='uploads/".$image."' width='200'> ";        echo "Title: " . $row["title"]. " Price: $" . $row["price"]. " <br> " . $row["description"]. " <br> " . $row["contact"] . " <br><br>";    }} else {    echo "0 results";}$con->close();?>

Answer 1

您需要对图像使用相同的行，例如：

$sql= "SELECT * FROM products";

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

if ($result->num_rows > 0) {

    // output data of each row

    while($row = $result->fetch_assoc()) {

        echo "<img src='uploads/".$row["image"]."' width='200'> ";

        echo "Title: " . $row["title"]. " Price: $" . $row["price"]. " <br> " . $row["description"]. " <br> " . $row["contact"] . " <br><br>";

    }

} else {

    echo "0 results";

}

$con->close();