PHP将信息上传到数据库

我试图上传一些信息到数据库中的项目，并上传到一个文件夹中的图像。我没有收到任何错误，但是sql查询没有运行，图像也没有上传

    <?php

    include'connect.php';

    if( isset($_POST['item_name']) && isset($_POST['description']) && isset($_POST['price']) && isset($_POST['quantity']) && isset($_POST['manufacturer']) &&     isset($_POST['pic']) )
    {
        $name = $_FILES['pic']['name'];
    $tmp_name = $_FILES['pic']['tmp_name'];

    if(!empty($_POST['item_name']) && !empty($_POST['description']) &&  !empty($_POST['price']) && !empty($_POST['quantity']) && !empty($_POST['manufacturer']) &&  !empty($name) )
    {   
        $item_name = $_POST['item_name'];
        $description = $_POST['description'];
        $price = $_POST['price'];
        $quantity = $_POST['quantity'];
        $manufacturer = $_POST['manufacturer'];

        $location = 'Upload/';
        move_uploaded_file($tmp_name, $location.$name);

        $query = "INSERT INTO `item` VALUES ('$item_name', '$description','$price','$quantity','$name','$manufacturer')";

        if($query_run = mysql_query($query))
        {   echo 'Item successfully added.';    }

        else
        {   echo 'Query Failure';   }


    }
}

?>

<form  action = 'additems.php' method = 'POST' enctype = 'multipart/form-data' >
    Name: <input type = 'text' name = 'item_name' ><br><br>
    Description: <input type = 'text' name = 'description' ><br><br>
    Price: <input type = 'text' name = 'price' ><br><br>
    Quantity: <input type = 'text' name = 'quantity' ><br><br>
    Manufacturer: <select name = 'manufacturer' >
                  <option value = '1'> Manufacturer 1 </option>
                  <option value = '2'> Manufacturer 2 </option>
                  <option value = '3'> Manufacturer 3 </option>
                  <option value = '4'> Manufacturer 4 </option>
                  </select>
                  <br><br>
    Picture: <input type = 'file' name = 'pic' > <br><br>

    <input type = 'submit' value = 'Submit' >

</form>

更改

 isset($_POST['pic'])

到

它至少输出了一些东西吗？”从您的echo中“查询失败”，即？是的，输出了echo错误。如前所述：发布您的mysql错误消息是的，成功了。现在我的“查询失败”错误出现了。只需要修正我的问题。我是指我的echo语句中的问题。
 isset($_FILE['pic']['name'])