Question

我是php的新手，也是一个完全菜鸟。我使用以下代码将一些文件插入到数据库中。

<?php
include_once 'dbconfig.php';
if(isset($_POST['btn-upload']))
{    

    $file = rand(1000,100000)."-".$_FILES['file']['name'];
    $file_loc = $_FILES['file']['tmp_name'];
    $file_size = $_FILES['file']['size'];
    $file_type = $_FILES['file']['type'];
    $folder="uploads/";


    $new_size = $file_size/1024;  



    $new_file_name = strtolower($file);


    $final_file=str_replace(' ','-',$new_file_name);

    if(move_uploaded_file($file_loc,$folder.$final_file))
    {
        $sql = "INSERT INTO tbl_uploads(file,type,size) VALUES('$final_file','$file_type','$new_size')";
        //mysql_query($sql, $conn);
        ?>
        <script>
        alert('successfully uploaded');
        window.location.href='index.php?success';
        </script>
        <?php
    }
    else
    {
        ?>
        <script>
        alert('error while uploading file');
        window.location.href='index.php?fail';
        </script>
        <?php
    }
}
?>

尽管这会将文件保存在上载文件夹中，但它不会在数据库中插入任何内容。无论其上传的数据库保持为空多少次。它也没有显示任何错误。我的config.php文件是

<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "dbtuts";
$conn = new mysqli($dbhost, $dbuser, $dbpass);

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
echo "Connected successfully";
?>

HTML部分

<form action="upload.php" method="post" enctype="multipart/form-data">
    <input type="file" name="file" />
    <button type="submit" name="btn-upload">upload</button>
</form>

Answer 1

我认为，您的sql无法获取变量

您尝试这样写：

$sql = "INSERT INTO tbl_uploads(file,type,size) VALUES('$final_file','$file_type','$new_size')";

Answer 2

尝试一下 $ sql =“插入到tbl_uploads（file，type，size）VALUES（'$ final_file'，'$ file_type'，'$ new_size'）”； mysqli_query（$ conn，$ sql）;

Answer 3

未选择任何数据库

<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "";
$dbname = "dbtuts";
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname); // <--- here select db first

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
echo "Connected successfully";
?>

然后使用

<?php
include_once 'dbconfig.php';
if(isset($_POST['btn-upload']))
{    

    $file = rand(1000,100000)."-".$_FILES['file']['name'];
    $file_loc = $_FILES['file']['tmp_name'];
    $file_size = $_FILES['file']['size'];
    $file_type = $_FILES['file']['type'];
    $folder="uploads/";


    $new_size = $file_size/1024;  



    $new_file_name = strtolower($file);


    $final_file=str_replace(' ','-',$new_file_name);

    if(move_uploaded_file($file_loc,$folder.$final_file))
    {
        $sql = "INSERT INTO tbl_uploads(file,type,size) VALUES('$final_file','$file_type','$new_size')";
        if($conn->query($sql)) {
          echo "Success";
        } else {
           echo "Failed. Error: ".$conn->error;
        }
        ?>
        <script>
        alert('successfully uploaded');
        window.location.href='index.php?success';
        </script>
        <?php
    }
    else
    {
        ?>
        <script>
        alert('error while uploading file');
        window.location.href='index.php?fail';
        </script>
        <?php
    }
}
?>

PHP的SQL代码不插入数据库

3 个答案: