PHP上传图片到数据库并显示的实例代码

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我是靠谱客的博主魁梧皮带，最近开发中收集的这篇文章主要介绍PHP上传图片到数据库并显示的实例代码，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

PHP上传图片到数据库并显示

1、创建数据表

CREATE TABLE ccs_image (

 id int(4) unsigned NOT NULL auto_increment,

 description varchar(250) default NULL,

 bin_data longblob,

 filename varchar(50) default NULL,

 filesize varchar(50) default NULL,

 filetype varchar(50) default NULL,

 PRIMARY KEY (id)

)engine=myisam DEFAULT charset=utf8

2、用于上传图片到服务器的页面 upimage.html

<!doctype html>

<html>

<head>

  <meta charset="UTF-8">

  <meta name="viewport"

     content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">

  <meta http-equiv="X-UA-Compatible" content="ie=edge">

  <style type="text/css">

    *{margin: 1%}

  </style>

  <title>Document</title>

</head>

<body>

<form method="post" action="upimage.php" enctype="multipart/form-data">

  描述:

  <input type="text" name="form_description" size="40">

  <input type="hidden" name="MAX_FILE_SIZE" value="1000000"> <br>

  上传文件到数据库:

  <input type="file" name="form_data" size="40"><br>

  <input type="submit" name="submit" value="submit">

</form>

</body>

</html>

3、处理图片上传的php upimage.php

<?php

if (isset($_POST['submit'])) {

  $form_description = $_POST['form_description'];

  $form_data_name = $_FILES['form_data']['name'];

  $form_data_size = $_FILES['form_data']['size'];

  $form_data_type = $_FILES['form_data']['type'];

  $form_data = $_FILES['form_data']['tmp_name'];

   

  $dsn = 'mysql:dbname=test;host=localhost';

  $pdo = new PDO($dsn, 'root', 'root');

  $data = addslashes(fread(fopen($form_data, "r"), filesize($form_data)));

  //echo "mysqlPicture=".$data;

   

  $result = $pdo->query("INSERT INTO ccs_image (description,bin_data,filename,filesize,filetype)

         VALUES ('$form_description','$data','$form_data_name','$form_data_size','$form_data_type')");

  if ($result) {

    echo "图片已存储到数据库";

  } else {

    echo "请求失败,请重试";

注:图片是以二进制blob形式存进数据库的,像这样

4、显示图片的php getimage.php

<?php

 

  $id =2;// $_GET['id']; 为简洁,直接将id写上了,正常应该是通过用户填入的id获取的

  $dsn ='mysql:dbname=test;host=localhost';

  $pdo = new PDO($dsn,'root','root');

  $query = "select bin_data,filetype from ccs_image where id=2";

  $result = $pdo->query($query);

  $result = $result->fetchAll(2);

//  var_dump($result);

  $data = $result[0]['bin_data'];

  $type = $result[0]['filetype'];

  Header( "Content-type: $type");

  echo $data;

5、到浏览器查看已经上传的图片,看是否可以显示