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PHP implements method of uploading images to database and displaying output

不言Original: 2018-05-31 15:11:415847browse

This article mainly introduces the method of uploading pictures to the database and displaying the output in PHP. It analyzes the related operation skills of using PHP to store pictures in binary form and read and display them in the form of examples. Friends in need can refer to the following

The example of this article describes the method of uploading images to the database and displaying the output in PHP. Share it with everyone for your reference, the details are as follows:

1. Create a data table

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. A page for uploading images to the server 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. PHP that handles image uploads upimage.php

<?php
if (isset($_POST[&#39;submit&#39;])) {
  $form_description = $_POST[&#39;form_description&#39;];
  $form_data_name = $_FILES[&#39;form_data&#39;][&#39;name&#39;];
  $form_data_size = $_FILES[&#39;form_data&#39;][&#39;size&#39;];
  $form_data_type = $_FILES[&#39;form_data&#39;][&#39;type&#39;];
  $form_data = $_FILES[&#39;form_data&#39;][&#39;tmp_name&#39;];
  $dsn = &#39;mysql:dbname=test;host=localhost&#39;;
  $pdo = new PDO($dsn, &#39;root&#39;, &#39;root&#39;);
  $data = addslashes(fread(fopen($form_data, "r"), filesize($form_data)));
  //echo "mysqlPicture=".$data;
  $result = $pdo->query("INSERT INTO ccs_image (description,bin_data,filename,filesize,filetype)
         VALUES (&#39;$form_description&#39;,&#39;$data&#39;,&#39;$form_data_name&#39;,&#39;$form_data_size&#39;,&#39;$form_data_type&#39;)");
  if ($result) {
    echo "图片已存储到数据库";
  } else {
    echo "请求失败,请重试";

Note: The picture is stored in the database in the form of a binary blob, like this

4. Display the php getimage.php

<?php
  $id =2;// $_GET[&#39;id&#39;]; 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=&#39;mysql:dbname=test;host=localhost&#39;;
  $pdo=new PDO($dsn,&#39;root&#39;,&#39;root&#39;);
  $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][&#39;bin_data&#39;];
  $type = $result[0][&#39;filetype&#39;];
  Header( "Content-type: $type");
  echo $data;

Go to the browser to view the uploaded image and see if it can be displayed.

is no problem, which proves that the image has been stored in binary form. Reached the database

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