因为你没有正确描述你的问题或没有提供任何代码。
所以，我以我的方式提出了你的问题并发布了我的答案。
按照步骤操作。

Since, you didn't described your question properly or didn't provided any code. So, i assumed your question in my way and posted my answer. Follow the step.

1）用于上传图片

1) For Uploading Image

<form method='POST' action='UploadImage.php' enctype="multipart/form-data">
    <input type='file' name='UploadImage'>
    <input type='submit' value="submit">
</form>

假设您的项目文件夹名称为：MyProject（所有项目文件都存在），
在MyProject文件夹中创建一个文件夹，即MyUploadImages
现在，

Suppose your Project Folder Name Is : MyProject (Where all project files are present), Make one folder inside "MyProject" Folder namely "MyUploadImages" Now,

UploadImage.php

<?php
include('connect.php'); // Do Database Connection in this file (create a file namely connect.php inside MyProject Folder)
extract($_POST);

$UploadedFileName=$_FILES['UploadImage']['name'];
if($UploadedFileName!='')
{
  $upload_directory = "MyUploadImages/"; //This is the folder which you created just now
  $TargetPath=time().$UploadedFileName;
  if(move_uploaded_file($_FILES['files']['tmp_name'], $upload_directory.$TargetPath)){    
    $QueryInsertFile="INSERT INTO TableName SET ImageColumnName='$TargetPath'"; 
    // Write Mysql Query Here to insert this $QueryInsertFile   .                   
  }
}
?>

现在，在数据库表中，您可以找到ImageColumnName，将图像路径设置为MyUploadImages / 1417Flower。 jpg

Now, In your Database Table, you can find ImageColumnName that image path is set as MyUploadImages/1417Flower.jpg

2）从数据库中恢复图像

2) Retreiving Image from database

AnyPage.php

<?
$Query="SELECT * FROM TableName";
// Write mysql query to fetch $Query

store that ImageColumnName value to any variable say $MyPhoto.
?>

<img src="<?echo $MyPhoto;?>">