上载的图像未保存在数据库中

Question

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> 

Answer 1

我想通了自己

<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>

Answer 2

1. 请在Prepare语句后绑定参数。 检查docs- http://php.net/manual/en/mysqli.prepare.php
2. 如果条件放在$ update（执行的返回值）而不是sql中
3. 关闭</form>标记。

4. 检查所有具有数据的变量。 使用var_dump() ， print_r()或echo()方法进行检查。

    <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> </form> </td> <?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); if( $sql2 = $conn->prepare("UPDATE stall SET file = ? WHERE stall_id = ?")) { $sql2->bindParam(1, $location, PDO::PARAM_STR); $sql2->bindParam(2, $userid, PDO::PARAM_INT); $update=$sql2->execute(); if($update){ echo ' <script> window.location = "stalls.php" </script>'; } }else{ echo 'sql prepare failed'; } } ?> 

Answer 3

从

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

至

$sql2 = $conn->prepare("UPDATE stall SET file = ? WHERE stall_id = ?");
$sql2->bindParam(1, $location, PDO::PARAM_STR);
$sql2->bindParam(2, $userid, PDO::PARAM_INT);
$sql2->execute();