Question

我正在尝试显示已存储在MySQL中的图像，但尚未获得成功。显然，回显表头（img）给了我类似的东西

此外，我希望能够在网站本身中添加图像，而不是使用phpmyadmin并在其中插入图像。

到目前为止，这是我在standing.php页面上使用的代码

<?php
    require_once('database.php');

    // Get all categories
    $query = 'SELECT * FROM categories
              ORDER BY categoryID';
    $statement = $db->prepare($query);
    $statement->execute();
    $teams = $statement->fetchAll();
    $statement->closeCursor();
?>
<!DOCTYPE html>
<html>

<!-- the head section -->
<head>
    <title>NBA</title>
    <link rel="stylesheet" type="text/css" href="css/index.css">
    <link rel="shortcut icon" type="image/png" href="images/favicon.ico"/>

</head>

<!-- the body section -->

<body>
    <main id="standingListMain">

    <h1 id="addCategoryh1">Team Standings</h1>
    <table id="standingListTable">
        <tr>
            <th>Team</th>
            <th>&nbsp;</th>
        </tr>
        <?php foreach ($teams as $team) : ?>
        <tr>
            <td><?php echo $team['categoryID']; ?></td>
            <td>
              <?php echo $team['categoryName']; ?>
              <?php echo $team['img']; ?>
            </td>
        </tr>
        <?php endforeach; ?>
    </table>
    <br>

    </main>
    <!-- <footer id="standingListFooter">
        <p>&copy; <?php echo date("Y"); ?> NBA</p>
    </footer> -->
</body>
</html>

基本上，用户可以在team_list.php页面上添加或删除球队，并在排名页面上查看

<?php
    require_once('../Model/database.php');

    // Get all categories
    $query = 'SELECT * FROM categories
              ORDER BY categoryID';
    $statement = $db->prepare($query);
    $statement->execute();
    $teams = $statement->fetchAll();
    $statement->closeCursor();
?>
<!DOCTYPE html>
<html>

<!-- the head section -->
<head>
    <title>NBA</title>
    <link rel="stylesheet" type="text/css" href="../css/index.css">
    <link rel="shortcut icon" type="image/png" href="images/favicon.ico"/>

</head>

<!-- the body section -->

<body>
    <main>
    <h1 id="addCategoryh1">Teams</h1>
    <table id="categoryListTable">
        <tr>
            <th>Name</th>
            <th>&nbsp;</th>
        </tr>
        <?php foreach ($teams as $team) : ?>
        <tr>
            <td><?php echo $team['categoryName']; ?></td>
            <td>
                <form action="delete_team.php" method="post"
                      id="delete_product_form">
                    <input type="hidden" name="team_id"
                           value="<?php echo $team['categoryID']; ?>">
                    <input id="deleteCategoryList" type="submit" value="Delete">
                </form>
            </td>
        </tr>
        <?php endforeach; ?>
    </table>
    <br>

    <h2 id="add_category_h2">Add Team</h2>
    <form action="add_team.php" method="post"
          id="add_category_form">

        <label>Name:</label>
        <input type="input" name="name">
        <input id="add_category_button" type="submit" value="Add">
    </form>
    <br>
    <p><a href="../index.php">View Team List</a></p>

    </main>
    <footer id="categoryListFooter">
        <p>&copy; <?php echo date("Y"); ?> NBA</p>
    </footer>
</body>
</html>

上面的代码是team_list.php页面，下面的代码是连接到数据库的代码add_team.php

<?php
// Get the team data
$name = filter_input(INPUT_POST, 'name');

// Validate inputs
if ($name == null) {
    $error = "Invalid team data. Check all fields and try again.";
    include('../Error/error.php');
} else {
    require_once('../Model/database.php');

    // Add the product to the database
    $query = 'INSERT INTO categories (categoryName)
              VALUES (:team_name)';
    $statement = $db->prepare($query);
    $statement->bindValue(':team_name', $name);
    $statement->execute();
    $statement->closeCursor();

    // Display the team List page
    include('team_list.php');
}
?>

上图显示了您可以在其中添加或删除团队的页面。

Answer 1

用于测试目的

首先，您需要知道图像是否确实存在。假设您的数据库中有一个ID为1的图像因此，创建另一个文件例如image.php

（请确保此代码运行良好。我尚未对其进行测试，但我知道它将为您工作）；

image.php

<?php 

 require_once('database.php');

    // Get all categories
    $query = "SELECT img FROM categories where categoryID=1";
    $statement = $db->prepare($query);
    $statement->execute();
    $num = $statement->rowCount();

if( $num ){

    $teams = $statement->fetchAll();

    // Ensure to specify header with content type,
    // you can do header("Content-type: image/jpg"); for jpg,
    // header("Content-type: image/gif"); for gif, etc.

    header("Content-type: image/png");

    //display the image file
    print $teams['img'];
    exit;
}else{
 //echo no image found with that Category Id.

}

?>

然后在您的 standing.php 中删除此代码

  <?php echo $team['img']; ?>

并替换为

<img src="image.php?id=1" />

让我知道您是否仍然需要帮助

从MYSQL插入和显示图像

1 个答案: