PHP上传图片到数据库并显示的实例代码

更新时间：2020-04-14 16:42:29 作者：startmvc

PHP上传图片到数据库并显示1、创建数据表CREATETABLEccs_image(idint(4)unsignedNOTNULLauto_increment,descri

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