PHP MySQLi 如何在 html 表中显示用户个人资料图像？

Question

我正在尝试在 HTML 表格中显示用户的个人资料图像，其中列出了用户的其余信息。但我无法让它显示，而是在表格中得到Broken Image。如果我右键单击“损坏的图像”并选择“在新选项卡中打开图像”，我可以看到图像的路径和图像的链接Parent Directory 我正在将图像的路径上传到数据库并将图像存储到文件夹（ images/profilepic ）。

有人可以帮帮我吗？非常感谢！

这是我的 insert.php 代码：

<?php
$server = "localhost";
$user = "root";
$pass = "";
$dbname = "employees";

// Create connection
$conn = mysqli_connect($server, $user, $pass, $dbname);

// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$profile_file = basename($_FILES['profileimg']['name']);
$profile_path = "images/profilepic/$profile_file";
$profile_file = mysqli_real_escape_string($conn, $profile_file);

move_uploaded_file($_FILES['profileimg']['tmp_name'], $profile_path);


$sql = "INSERT INTO addemployees (profileimg)
        VALUES ('$profile_file')";

if (mysqli_query($conn, $sql)) {
  header("location: employees.php");

} else {
    echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

//Close the connection
mysqli_close($conn);

?>

这就是我展示它的地方：

<table id="dtBasicExample" class="table table-bordered table-hover text-center">
                             <thead>
                               <tr class="naslov">
                                   <th class="th-sm">ID:</th>
                                   <th class="th-sm">Profile Picture:</th>
                                   <th class="th-sm">First Name:</th>
                                   <th class="th-sm">Last Name:</th>
                                   <th class="th-sm">DOB:</th>
                                   <th class="th-sm">EMBG:</th>
                                   <th class="th-sm">Work Position:</th>
                                   <th class="th-sm">Address:</th>
                                   <th class="th-sm">Contract:</th>
                                   <th class="th-sm" colspan="2">Опции:</th>
                               </tr>
                            </thead>

                            <tbody id="myTable">

                          <?php

                              $conn = mysqli_connect("localhost", "root", "", "employees");

                            if (!$conn) {
                            die("Connection failed: " . mysqli_connect_error());
                            }

                              $sql = "SELECT * from addemployees";
                              $result = $conn-> query($sql);

                              if ($result-> num_rows > 0) {
                             while ($row = $result-> fetch_assoc()) {




                               echo "<tr>
                                      <td>".$row['id']."</td>
                                      <td><img src='images/profilepic/'". $row['profileimg'] ."' border=0 class='tile' id='profileimg'></td>
                                      <td>".$row['fname']."</td>
                                      <td>".$row['lname']."</td>
                                      <td>".$row['dob']."</td>
                                      <td>".$row['embg']."</td>
                                      <td>".$row['workposition']."</td>
                                      <td>".$row['address']."</td>
</tr>";


                          }

                            echo "</table>";
                          }
                            else {
                            echo "0 results";
                          }

                        $conn-> close();

                              ?>

                              </tbody>

Answer 1

删除<img src中变量之前的单引号：

<td><img src='images/profilepic/". $row['profileimg'] ."'
                                ^-- remove the ' here   ^-- but leave this one

按照你现在的方式，你的变量在src 属性之外：

// if $row['profileimg'] is image.jpg:
<img src='images/profilepic/'". $row['profileimg'] ."' border=0>

// will become:
<img src='images/profilepic/'image.jpg' border=0>
//                          ^-- end of the src='' argument