使用 mysqli 获取行

问题描述

这是一个简单的代码.每次上传文件时，我都想在数据库中保留一条记录.每次上传任何文件时，我都希望用户从数据库中以串行方式查看他/她的所有上传文件.上传文件时的记录保存工作正常.但是在获取包含所有上传文件信息的表时出现问题.

its a simple code. I want to keep a record in the database each time when a file is being uploaded. And when each time any file is being uploaded I want the user to see he/her's all uploaded files in a serial from the database. The record keeping while uploading file works fine. But the problem appears while fetching the table containing all the uploaded file information.

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

if (mysqli_connect_errno()) 
{
    printf("Connect failed: %s
", 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
", $row);
}

mysqli_close($con);
}

我能感觉到编码存在一些严重的问题.这是我第一次在 mysqli 中工作，之前我曾经使用 mysql 进行编码.需要帮助以了解实际问题和解决方案.

I can feel that there are some serious problem in coding. This is my first time working in mysqli, before I used to code using mysql. need help to know the actual problem and the solution.

已它返回这个，

Notice: Array to string conversion in C:xampphtdocssssupload_file.php on line 68
Array

Notice: Array to string conversion in C:xampphtdocssssupload_file.php on line 68
Array

Notice: Array to string conversion in C:xampphtdocssssupload_file.php on line 68
Array

但不幸的是，这段代码是庞大代码的一部分.所以这里的第 68 行是 $result = mysqli_query($con, "select * from uploads"); 开始的那一行.

but unfortunately this code is a part of a huge code. so here line 68 is the line where the $result = mysqli_query($con, "select * from uploads"); starts.

推荐答案

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

但是您应该认真考虑查看代码的安全性.

But you should seriously consider looking at the security of your code.

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