php – 使用mysqli获取行

它是一个简单的代码.我希望每次上传文件时都在数据库中保留一条记录.当每次上传任何文件时,我希望用户从数据库中查看他/她所有上传的文件.上传文件时保存的记录工作正常.但是在获取包含所有上载文件信息的表时会出现问题.

//connetion code 
$con = mysqli_connect("localhost", "root", "", "sss");

if (mysqli_connect_errno()) 
{
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

//file upload code
move_uploaded_file($_FILES["file"]["tmp_name"],"C:/xampp/htdocs/" . $_FILES["file"]["name"]);
mysqli_query($con, "INSERT INTO uploads ( filename, uploaded_on) VALUES ( '{$_FILES['file']['name']}', NOW());");
echo "Stored in: " . "C:/xampp/htdocs/" . $_FILES["file"]["name"];

//fetch rows 
$result =mysqli_query($con, "select * from uploads");

while ($row = mysqli_fetch_array($result))
{
    printf ("%s\n", $row);
}

mysqli_close($con);
}

我觉得编码有一些严重的问题.这是我第一次在mysqli工作,之前我习惯使用mysql进行编码.需要帮助才能知道实际问题和解决方案.

编辑:
它返回这个,

Notice: Array to string conversion in C:\xampp\htdocs\sss\upload_file.php on line 68
Array

Notice: Array to string conversion in C:\xampp\htdocs\sss\upload_file.php on line 68
Array

Notice: Array to string conversion in C:\xampp\htdocs\sss\upload_file.php on line 68
Array

但不幸的是,这段代码是巨大代码的一部分.所以这里第68行是$result = mysqli_query($con,“select * from uploads”)的行;开始.

最佳答案
$result =mysqli_query($con, "select * from uploads");
while ($row = mysqli_fetch_assoc($result))
   {
      printf ("%s\n", $row['column_name_1']);
      printf ("%s\n", $row['column_name_2']);
      printf ("%s\n", $row['column_name_3']);
   }
mysqli_close($con);

但是你应该认真考虑查看代码的安全性.

转载注明原文:php – 使用mysqli获取行 - 代码日志