PHP实现上传图片到数据库并显示输出的方法

2024-05-04 22:43:10

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本文实例讲述了PHP实现上传图片到数据库并显示输出的方法。分享给大家供大家参考，具体如下：

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 lang="en"><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

<?phpif (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;

到浏览器查看已经上传的图片,看是否可以显示

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