PHP上传图片到数据库并显示

藏色散人转载: 2023-04-08 06:28:024377浏览

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