Question

我在访问我在sql表add_deal中保存为Long BLOB类型的图像时遇到问题，我面临的问题是我无法访问图像并在浏览器中显示它。当我运行程序时，我看到一个破碎的页面。我试过放置“header（'Content-type：image / jpeg'）;”上面的回声反过来给出一个错误，指出“因为它包含错误而无法显示图像”我将图像保存为：

$image = addslashes($_FILES['image']['tmp_name']);
        $name=addslashes($_FILES['image']['name']);
        $image_size=getimagesize($_FILES['image']['tmp_name']);
        $deal_type = $_POST['dealtype'];
        $description = $_POST['description'];
        $quantity = $_POST['quantity'];
        $validity = $_POST['validity'];
        $price=$_POST['cost'];

    $qry="INSERT INTO `add_deal`(`dealtype`, `description`, `quantity`, `validity`, `price`, `image`, `name`) VALUES ('$deal_type','$description','$quantity','$validity','$price','$image','$name')";
    $result=mysqli_query($db,$qry);
    if(! $result ) {
          die("Query FAILED. " . mysqli_error($db));
        }
    else
    {
        echo "Your Deal has been placed in your Gallery";
        echo "<br/>Go to <a href='dealhome.php'>home</a> page to see whats in your gallery</b>";
    }

并将其作为：

访问

    if(! $result ) {
          die("Query FAILED. " . mysqli_error($db));
        }
            else
    {

    if ($result->num_rows > 0) {
        // output data of each row
        //echo "<table>";
    while($row = mysqli_fetch_row($result)) {
    header('Content-type: image/jpeg');
    echo $row['image'];

    }
    }
    //echo "</table>";
    }

使用PHP从数据库访问图像

0 个答案: