問題描述

這是一個簡單的代碼.每次上傳文件時，我都想在數據庫中保留一條記錄.每次上傳任何文件時，我都希望用戶從數據庫中以串行方式查看他/她的所有上傳文件.上傳文件時的記錄保存工作正常.但是在獲取包含所有上傳文件信息的表時出現問題.

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);
}

我能感覺到編碼存在一些嚴重的問題.這是我第一次在 mysqli 中工作，之前我曾經使用 mysql 進行編碼.需要幫助以了解實際問題和解決方案.

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.

已它返回這個，

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，這段代碼是龐大代碼的一部分.所以這里的第 68 行是 $result = mysqli_query($con, "select * from uploads"); 開始的那一行.

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.

推薦答案

$result =mysqli_query($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);

但是您應該認真考慮查看代碼的安全性.

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

這篇關于使用 mysqli 獲取行的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！