Question

有人可以请帮忙，因为我不知所措。如何从MYSQL数据库中获取图像到我的网页上。 anyname.html

数据库名称是：upload 表名是：images

id Primary int（11）AUTO_INCREMENT
name varchar（100）

size int（11）

输入varchar（20）

content mediumblob

上传效果很好但是无法让它显示能够在html页面中使用它的图像。当我上传图片后，我得到了一个回复：对于存储的11号图像，但我看不到图像：这是我的代码，所以请，请有人帮忙，因为我已经多年没有结果了。

我的upload.php代码：

<?php

// Check for post data.
if ($_POST && !empty($_FILES)) {
$formOk = true;

//Assign Variables
$path = $_FILES['image']['tmp_name'];
$name = $_FILES['image']['name'];
$size = $_FILES['image']['size'];
$type = $_FILES['image']['type'];

if ($_FILES['image']['error'] || !is_uploaded_file($path)) {
    $formOk = false;
    echo "Error: Error in uploading file. Please try again.";
}

//check file extension
if ($formOk && !in_array($type, array('image/png', 'image/x-png', 'image/jpeg', 'image/pjpeg', 'image/jpg', 'image/gif'))) {
    $formOk = false;
    echo "Error: Unsupported file extension. Supported extensions are JPG / PNG.";
}
// check for file size.
if ($formOk && filesize($path) > 500000) {
    $formOk = false;
    echo "Error: File size must be less than 500 KB.";
}

if ($formOk) {
    // read file contents
    $content = file_get_contents($path);

    //connect to mysql database
    if ($conn = mysqli_connect('localhost', 'root', '', 'upload')) {
        $content = mysqli_real_escape_string($conn, $content);
        $sql = "insert into images (name, size, type, content) values ('{$name}', '{$size}', '{$type}', '{$content}')";

        if (mysqli_query($conn, $sql)) {
            $uploadOk = true;
            $imageId = mysqli_insert_id($conn);
        } else {
            echo "Error: Could not save the data to mysql database. Please try again.";
        }

        mysqli_close($conn);
    } else {
        echo "Error: Could not connect to mysql database. Please try   again.";
        }
    }
}
?>

<html>
<head>
    <title>Upload image to mysql database.</title>
    <style type="text/css">
        img{
            margin: .2em;
            border: 1px solid #555;
            padding: .2em;
            vertical-align: top;
        }
    </style>
</head>
<body>
    <?php if (!empty($uploadOk)): ?>
        <div>
            <h3>Your Image has been Uploaded:</h3>
        </div>
        <div>
            <img src="image.php?id=<?=$imageId ?>" width="300px" height="300px"></img>
            <strong><br /><br />Embed</strong>: <input size="30" value='<img src="image.php?id=<?=$imageId ?>">'>
        </div>

        <hr>
    <?php endif; ?>

    <form action="<?=$_SERVER['PHP_SELF']?>" method="post" enctype="multipart/form-data" >
      <div>
        <h3>Image Upload:</h3>
      </div>
      <div>
        <label>IMAGE SELECT<br /></label>
        <input type="hidden" name="MAX_FILE_SIZE" value="500000">
        <input type="file" name="image" />
        <br /><br />
        <input name="submit" type="submit" value="Upload Image">
      </div>
    </form>
</body>
</html>

Answer 1

这似乎是PHP display image BLOB from MySQL

问题的重复

希望它有所帮助。

Answer 2

如果将图像数据存储在数据库中，请使用下一个image.php：

<?php

  $id = $_GET['id'];

  $link = mysql_connect('localhost', 'root', '');
  mysql_select_db('upload');
  $sql = "SELECT content FROM images WHERE id=$id";
  $result = mysql_query("$sql");
  $row = mysql_fetch_assoc($result);
  mysql_close($link);

  header("Content-type: image/jpeg");
  echo $row['content'];
?>

如何在我的网页上显示存储在我的mysql数据库中的图像文件？

2 个答案: