的圖片上傳到數據庫

這是上傳表單，使我將圖像上傳到數據庫的圖片上傳到數據庫

<html> 
<head> 
<title>File Uploading Form</title> 
</head> 

<body> 
<h3>File Upload:</h3> 
Select a file to upload: <br /> 

<form action="file_uploader.php" method="post" enctype="multipart/form-data"> 
<input type="file" name="file" size="50" /> 
<br /> 
<input type="submit" value="Upload File" /> 
</form> 
</body> 
</html>

這是我有一個在一個名爲file_uploader.php代碼的代碼。當試圖完成這個我得到的錯誤無法複製文件！

<?php 
if($_FILES['file']['name'] != "") 
{ 
    copy($_FILES['file']['name'], "databasehostdetails") or 
      die("Could not copy file!"); 
} 
else 
{ 
    die("No file specified!"); 
} 
?> 
<html> 
<head> 
<title>Uploading Complete</title> 
</head> 

<body> 
<h2>Uploaded File Info:</h2> 
<ul> 
<li>Sent file: <?php echo $_FILES['file']['name']; ?> 
<li>File size: <?php echo $_FILES['file']['size']; ?> bytes 
<li>File type: <?php echo $_FILES['file']['type']; ?> 
</ul> 
</body> 
</html>

來源

2013-04-27 user2326279

粘貼PHP錯誤日誌中變得越來越快回答 – vaichidrewar 2013-04-27 19:05:01

你應該應對文件到服務器，然後就在數據庫的路徑可能會有所幫助。 – Sumoanand 2013-04-27 19:07:14

爲什麼不把它存儲在服務器上，只是在db中存儲文件名/路徑？ – member8888 2013-04-27 19:05:23

您需要使用tmp_name。另外，什麼是databasehostdetails？第二個參數是目標（您要將文件複製到的位置）。

copy($_FILES['file']['tmp_name'], DESTINATION_PATH);

來源

2013-04-27 19:09:13 SeanWM

這是一個簡單的方法來做到這一點。我給你我的腳本的index.html 但要小心：

不推薦，因爲它是昂貴的數據庫來存儲文件
修復任何SQL注入vurnelabilities。我自己不知道該怎麼做。你可以進行研究並修復它。
還檢查了這裏Storing Images in DB - Yea or Nay?

然而代碼工作，如果它不是一個大規模應用，並沒有SQL注入攻擊可能發生相當不錯。

<html> 
<body> 
<form method="post" enctype="multipart/form-data" action="doupload.php"> 
<input type="hidden" name="MAX_FILE_SIZE" value="2000000"> 
<input name="userfile" type="file" id="userfile"> 
<input name="upload" type="submit" id="upload" value=" Upload "> 
</form> 

</body> 
</html>

doupload.php

<?php   
include("config.php"); 

$fileName = $_FILES['userfile']['name']; 
$tmpName = $_FILES['userfile']['tmp_name']; 
$fileSize = $_FILES['userfile']['size']; 
$fileType = $_FILES['userfile']['type']; 
$fp = fopen($tmpName, 'r'); 
$content = fread($fp, filesize($tmpName)); 
$content = addslashes($content); 
fclose($fp); 

if(!get_magic_quotes_gpc()) 
{ 
$fileName = addslashes($fileName); 
} 

$query = "INSERT INTO upload set name='".$fileName."', size='".$fileSize."', type='".$fileType."', content='".$content."'";   
mysql_query($query) ;  
?>

getuploaded.php

<?php 
// select records from database if exists to display 
include("config.php"); 
$query1 = "SELECT id, name FROM upload"; 
$result1 = mysql_query($query1) or die('Error, query failed'); 


if(mysql_num_rows($result1)>0) 
{ 
while(list($id, $name) = mysql_fetch_array($result1)) 
{ 
?> 
<a href="download.php?id=<?php echo $id;?>"><?php echo $name;?></a> <br> 

<?php 
} 
} 
?>

的download.php

<?php 
//header("Content-type: $type"); 
include("config.php"); 
$id = $_GET['id']; 
$query = "SELECT name, type, size, content " . 
"FROM upload WHERE id = '$id'"; 

$result = mysql_query($query) or die('Error, query failed'); 
list($name, $type, $size, $content) = mysql_fetch_array($result); 

header("Content-length: $size"); 
header("Content-type: $type"); 
header("Content-Disposition: attachment; filename=$name"); 
echo $content; 

?>

的config.php

<?php 
$dbhost = 'localhost'; 
$dbuser = 'root'; 
$dbpass = ''; 
$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die('Error connecting to mysql'); 
$dbname = 'uploadfile'; 
mysql_select_db($dbname); 
?>

Createing表

CREATE TABLE upload (
id INT NOT NULL AUTO_INCREMENT, 
name VARCHAR(30) NOT NULL, 
type VARCHAR(30) NOT NULL, 
size INT NOT NULL, 
content MEDIUMBLOB NOT NULL, 
PRIMARY KEY(id) 
);

來源

2013-04-27 19:20:02 themis

的圖片上傳到數據庫

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