php - 上载的图像未保存在数据库中

我有一个上传图片的按钮，每次上传图片时，图片都不会保存在数据库中，也不会出现错误，
这是我的代码：

<td>
  <form method="POST" enctype="multipart/form-data">
    <input type="file" name="file" required>
        <button type="submit" name="files" class="btn btn-primary btn-xs">
            Submit
        </button>
        <?php
            if(isset($_POST['files']))
              {
                $userid = $row['stall_id'];
                $a = $_FILES['file']['name'];
                $ab = $_FILES['file']['tmp_name'];
                $location = "".$a;
                move_uploaded_file($ab, "../pictures/".$location);

                $sql2 = $conn->prepare("UPDATE stall SET file = ? WHERE stall_id = ?");
                $sql2->execute(array($location,$userid));

            if($sql){
                echo '
                <script>
                    window.location = "stalls.php"
                </script>';
                    }
                    }
                ?>
</td>

最佳答案

我自己想出来了，

<td>
<form method="POST" enctype="multipart/form-data">
    <input type="file" name="file" required>
    <input type="hidden" name="stall_id" value="<?php echo $value['stall_id']?>">
    <button type="submit" name="files" class="btn btn-primary btn-xs">
    Submit
    </button>
    <?php
    if(isset($_POST['files']))
    {
    $userid = $_POST['stall_id'];
    $a = $_FILES['file']['name'];
    $ab = $_FILES['file']['tmp_name'];
    $location = "".$a;
    move_uploaded_file($ab, "../pictures/".$location);

    $sql2 = $conn->prepare("UPDATE stall SET file = ? WHERE stall_id = ?");
    $sql2->execute(array($location,$userid));

    if($sql2){
    echo '
    <script>
    window.location = "stalls.php"
    </script>';
    }
    }
    ?>
    </td>

关于php - 上载的图像未保存在数据库中，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/41455487/