PHP上传图片到数据库并显示的实例代码_PHP教程

PHP上传图片到数据库并显示的实例代码

2021-09-24 14:12藏色散人 PHP教程

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

				?

									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

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									<!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

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									<?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;