如何在数据库中散列上传的文件名 - How to hash the uploaded file name in db?

How to hash the uploaded file name in db?

本文关键字：文件名数据库 | 更新日期: 2023-09-27

我有一个"个人资料图片"系统，每个注册用户都可以更改他们的个人资料图片，以便显示图片。上传的过程是上传名为pictures的文件夹中的图像，然后将文件名写入名为"picture"的数据库行，这很有效，一切都很好，但我需要知道如何哈希上传的图片的名称，所以如果两个图片不同但图片名称相同的人会上传一个文件，就会出现问题。。。

这是我的提交表格：

<form action="" method="post" enctype="multipart/form-data">
    <input type="file" name="file">
    <input type="submit" name="submit">
</form>

我有这样的代码上传文件到文件夹和数据库行：

<?php
    if(isset($_POST['submit'])){
        move_uploaded_file($_FILES['file']['tmp_name'],"pictures/".$user_id.$_FILES['file']['name']);
        $con = mysqli_connect("host","name","pass","db");
        $q = mysqli_query($con,"UPDATE members SET picture = '".$_FILES['file']['name']."' WHERE username = '".$_SESSION['username']."'");
    }
?>

这是在网站上显示图片的代码：

<?php   
    $con = mysqli_connect("host","name","pass","db");
    $q = mysqli_query($con,"SELECT * FROM members WHERE username = '".$_SESSION['username']."'");
    while($row = mysqli_fetch_assoc($q)){
        if($row['picture'] == ""){
            echo " <img src='pictures/public_icon.png' width='25' height='25'>"; *//This is default pic//*
        } else {
            echo " <img width='25' height='25' src='pictures/".$row['picture']."' alt='Profile Pic'>"; *//And this is pic from db row//*
        }                                    
    }
?>

我想知道如何在上传的文件夹和数据库行中散列这张照片的名称。。。我能做热的吗？我会用数据库中的"id"answers"date"行来处理它，但我不知道如何处理..：/

将其用于上传：

<?php
    if(isset($_POST['submit'])){
      /*
      * create a unique file name with the current time and ip address of 
      * user, since it is highly unlikely to have the same file uploaded 
      *form the same ip address at the same time
      */
      $unique_file_name = time(). $_SERVER['REMOTE_ADDR']. $_FILES['file']['name']; 
         //run query only if file upload and transfer succeeded
         if( move_uploaded_file($_FILES['file']['tmp_name'],"pictures/".$user_id. $unique_file_name) ){ 
            $con = mysqli_connect("host","name","pass","db");
            $q = mysqli_query($con,"UPDATE members SET picture = '".$_FILES['file']['name']."' WHERE username = '".$_SESSION['username']."'");
        }
    }
?>

你可以试试这个

<?php
    if(isset($_POST['submit'])){
            $filename = $user_id.$_FILES['file']['name'];
            move_uploaded_file($_FILES['file']['tmp_name'],"pictures/".$filename);
            $con = mysqli_connect("host","name","pass","db");
            $q = mysqli_query($con,"UPDATE members SET picture = '".$filename."' WHERE username = '".$_SESSION['username']."'");
    } ?>