PHP上传图片到数据库并显示的实例代码

这篇文章主要介绍了PHP上传图片到数据库并显示的实例代码,有兴趣的朋友们参考下。

PHP上传图片到数据库并显示

1、创建数据表

  1. CREATE TABLE ccs_image (
  2. id int(4) unsigned NOT NULL auto_increment,
  3. description varchar(250) default NULL,
  4. bin_data longblob,
  5. filename varchar(50) default NULL,
  6. filesize varchar(50) default NULL,
  7. filetype varchar(50) default NULL,
  8. PRIMARY KEY (id)
  9. )engine=myisam DEFAULT charset=utf8

2、用于上传图片到服务器的页面 upimage.html

  1. <!doctype html>
  2. <html>
  3. <head>
  4. <meta charset="UTF-8">
  5. <meta name="viewport"
  6. content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
  7. <meta http-equiv="X-UA-Compatible" content="ie=edge">
  8. <style type="text/css">
  9. *{margin: 1%}
  10. </style>
  11. <title>Document</title>
  12. </head>
  13. <body>
  14. <form method="post" action="upimage.php" enctype="multipart/form-data">
  15. 描述:
  16. <input type="text" name="form_description" size="40">
  17. <input type="hidden" name="MAX_FILE_SIZE" value="1000000"> <br>
  18. 上传文件到数据库:
  19. <input type="file" name="form_data" size="40"><br>
  20. <input type="submit" name="submit" value="submit">
  21. </form>
  22. </body>
  23. </html>

3、处理图片上传的php upimage.php

  1. <?php
  2. if (isset($_POST['submit'])) {
  3. $form_description = $_POST['form_description'];
  4. $form_data_name = $_FILES['form_data']['name'];
  5. $form_data_size = $_FILES['form_data']['size'];
  6. $form_data_type = $_FILES['form_data']['type'];
  7. $form_data = $_FILES['form_data']['tmp_name'];
  8. $dsn = 'mysql:dbname=test;host=localhost';
  9. $pdo = new PDO($dsn, 'root', 'root');
  10. $data = addslashes(fread(fopen($form_data, "r"), filesize($form_data)));
  11. //echo "mysqlPicture=".$data;
  12. $result = $pdo->query("INSERT INTO ccs_image (description,bin_data,filename,filesize,filetype)
  13. VALUES ('$form_description','$data','$form_data_name','$form_data_size','$form_data_type')");
  14. if ($result) {
  15. echo "图片已存储到数据库";
  16. } else {
  17. 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

  1. <?php
  2. $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
  3. $dsn ='mysql:dbname=test;host=localhost';
  4. $pdo = new PDO($dsn,'root','root');
  5. $query = "select bin_data,filetype from ccs_image where ;
  6. $result = $pdo->query($query);
  7. $result = $result->fetchAll(2);
  8. // var_dump($result);
  9. $data = $result[0]['bin_data'];
  10. $type = $result[0]['filetype'];
  11. Header( "Content-type: $type");
  12. echo $data;

5、到浏览器查看已经上传的图片,看是否可以显示