php下将图片以二进制存入Phpadmin数据库中并显示的实现代码
时间: 2024-05-22 08:16:00 浏览: 9
以下是将图片以二进制存入Phpadmin数据库中并显示的实现代码:
1. 存入数据库
// 连接数据库
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// 读取图片文件
$image_path = "image.jpg";
$image_data = file_get_contents($image_path);
// 将图片数据存入数据库
$sql = "INSERT INTO images (image_data) VALUES ('$image_data')";
if ($conn->query($sql) === TRUE) {
echo "Image saved to database successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
2. 显示图片
// 连接数据库
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// 从数据库读取图片数据
$sql = "SELECT image_data FROM images WHERE image_id = 1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// 输出图片数据
$row = $result->fetch_assoc();
header("Content-Type: image/jpeg");
echo $row["image_data"];
} else {
echo "Image not found";
}
$conn->close();
注意:以上代码仅供参考,实际应用中需要根据具体情况进行修改和优化。同时,将图片存入数据库会增加数据库的负担和数据存储量,建议仅在特定情况下使用,如需要保护图片版权等。
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