Question

我有一个上传图片的按钮，每次上传图片时，图片都没有保存在数据库中，并且没有给出错误，

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

我想出了自己，

<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

准备好声明后请绑定参数。检查docs- http://php.net/manual/en/mysqli.prepare.php
in if condition put $ update（返回执行值）而不是sql
关闭</form>代码。

检查所有包含数据的变量。使用var_dump()，print_r()或echo()方法查看。

<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

改变

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

到

$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();

上传的图像不会保存在数据库中

3 个答案: