显示带有用户产品帖子信息的图像?

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 connection

if (!$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();



?>


慕桂英546537
浏览 99回答 1
1回答

智慧大石

您需要对图像使用相同的行,例如:$sql= "SELECT * FROM products";$result = $con->query($sql);if ($result->num_rows > 0) {&nbsp; &nbsp; // output data of each row&nbsp; &nbsp; while($row = $result->fetch_assoc()) {&nbsp; &nbsp; &nbsp; &nbsp; echo "<img src='uploads/".$row["image"]."' width='200'> ";&nbsp; &nbsp; &nbsp; &nbsp; echo "Title: " . $row["title"]. " Price: $" . $row["price"]. " <br> " . $row["description"]. " <br> " . $row["contact"] . " <br><br>";&nbsp; &nbsp; }} else {&nbsp; &nbsp; echo "0 results";}$con->close();
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