PHP实现上传图片到数据库并显示输出的方法-php开发

这篇文章主要介绍了PHP实现上传图片到数据库并显示输出的方法,结合实例形式分析了php采用二进制形式存储图片及读取显示的相关操作技巧,需要的朋友可以参考下

本文实例讲述了PHP实现上传图片到数据库并显示输出的方法。分享给大家供大家参考，具体如下：

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 lang="en">
<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;

到浏览器查看已经上传的图片,看是否可以显示