PHP上传图片到数据库并显示
一、创立数据表
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
二、用于上传图片到效劳器的页面 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"> 形容: <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>
三、解决图片上传的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 "申请失败,请重试";
注:图片是以二进制blob方式存进数据库的,像这样
四、显示图片的php getimage.php
<?php $id =2;// $_GET['id']; 为简约,间接将id写上了,失常应该是经过用户填入的id猎取的 $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;
五、到阅读器查看曾经上传的图片,看能否能够显示
是不成绩的,证实图片曾经以二进制的方式存储到数据库了
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