Question

这是我在数据库中插入和显示图像的代码（blob数据类型），Insertion每次都成功但我无法显示任何图像。而且我不知道每次都会出现什么问题。

<?php

    $hostname = "mysql.hostinger.in";
    $username = "u881956746_jk";
    $password = "jay@95373";
    $dbname = "u881956746_quote";

    $conn = mysqli_connect($hostname , $username , $password , $dbname);

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

    if(isset($_POST['submit']))
    {
        if(isset($_FILES['file']))
        {
            $count = count($_FILES['file']['name']);

            for($i = 0 ; $i < $count ; $i++)
            {
                $image = mysqli_real_escape_string($conn , $_FILES['file']['tmp_name'][$i]);
                $image = base64_encode($image);

                $sql = "INSERT INTO `u881956746_quote`.`quotes` (`Image`) VALUES ('$image');";

                if(mysqli_query($conn , $sql))
                {
                    echo "Successfully Inserted";
                }
                else
                {
                    echo "Not inserted Succesfully";
                }
            }
        }
        else
        {
            echo "Nothing to Submit";
        }
    }
    if(isset($_POST['Show']))
    {
        $sql = "SELECT * FROM `quotes`";
        $result = mysqli_query($conn , $sql);

        if(mysqli_num_rows($result) > 0)
        {
            while($row = mysqli_fetch_array($result))
            {
                echo '<img height = "300" width = "400" src="data:image/jpeg;base64,'.base64_encode( $row['Image'] ).'"/>';
                echo "<br />";
            }           
        }
    }

    mysqli_close($conn);
?>

<!DOCTYPE html>
<html>  
<body>  
    <form method = "POST" action = "practice2.php" enctype = "multipart/form-data">
    <input type = "file" name = "file[]" id = "image" multiple />
    <br />
    <br />
    <input type = "submit" name = "Show" value = "Show" />
    <br />
    <br />
    <input type = "submit" name = "submit" value = "Submit File" />
    </form>
</body>
</html>

Answer 1

从显示图片中移除base64_encode，因为它在插入期间已转换为base64

if(isset($_POST['Show']))
    {
        $sql = "SELECT * FROM `quotes`";
        $result = mysqli_query($conn , $sql);

        if(mysqli_num_rows($result) > 0)
        {
            while($row = mysqli_fetch_array($result))
            {
                echo '<img height = "300" width = "400" src="data:image/jpeg;base64,'.$row['Image'].'"/>';
                echo "<br />";
            }           
        }
    }

Answer 2

最后得到了输出：

<?php

$hostname = "mysql.hostinger.in";
$username = "u881956746_jk";
$password = "jay@95373";
$dbname = "u881956746_quote";

$conn = mysqli_connect($hostname , $username , $password , $dbname);

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

if(isset($_POST['submit']))
{
    if(isset($_FILES['file']))
    {
        $count = count($_FILES['file']['name']);

        for($i = 0 ; $i < $count ; $i++)
        {
            $image = addslashes(file_get_contents($_FILES['file']['tmp_name'][$i]));

            $sql = "INSERT INTO `u881956746_quote`.`quotes` (`Image`) VALUES ('$image');";

            if(mysqli_query($conn , $sql))
            {
                echo "Successfully Inserted";
            }
            else
            {
                echo "Not inserted Succesfully";
            }
        }
    }
    else
    {
        echo "Nothing to Submit";
    }
}
if(isset($_POST['Show']))
{
    $sql = "SELECT * FROM `quotes`";
    $result = mysqli_query($conn , $sql);

    if(mysqli_num_rows($result) > 0)
    {
        while($row = mysqli_fetch_array($result))
        {
            echo '<img height = "60" width = "80" src="data:image/jpeg;base64,'.base64_encode( $row['Image'] ).'"/>';
            echo "<br />";
        }           
    }
}

mysqli_close($conn);

?>

无法在PHP中显示来自MySQL数据库的图像

2 个答案: