我找到了答案，对于那些在这里寻找同样事情的人，我就是这样做的.您不应该考虑将图像上传到数据库，而是可以将上传文件的名称存储在数据库中，然后检索文件名并在您想要显示图像的任何地方使用它.

I found the answer, For those who are looking for the same thing here is how I did it. You should not consider uploading images to the database instead you can store the name of the uploaded file in your database and then retrieve the file name and use it where ever you want to display the image.

HTML 代码

<input type="file" name="imageUpload" id="imageUpload">

PHP 代码

if(isset($_POST['submit'])) {

    //Process the image that is uploaded by the user

    $target_dir = "uploads/";
    $target_file = $target_dir . basename($_FILES["imageUpload"]["name"]);
    $uploadOk = 1;
    $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);

    if (move_uploaded_file($_FILES["imageUpload"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["imageUpload"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }

    $image=basename( $_FILES["imageUpload"]["name"],".jpg"); // used to store the filename in a variable

    //storind the data in your database
    $query= "INSERT INTO items VALUES ('$id','$title','$description','$price','$value','$contact','$image')";
    mysql_query($query);

    require('heading.php');
    echo "Your add has been submited, you will be redirected to your account page in 3 seconds....";
    header( "Refresh:3; url=account.php", true, 303);
}

显示图像的代码

while($row = mysql_fetch_row($result)) {
    echo "<tr>";
    echo "<td><img src='uploads/$row[6].jpg' height='150px' width='300px'></td>";
    echo "</tr>
";
}