时间：2021-07-01 10:21:17 帮助过：151人阅读
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
是没有问题的,证明图片已经以二进制的形式存储到数据库了
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