5

我正在尝试在 MySql 中显示上传到我的“商店”表的最后 5 张图片。我对 PHP 和数据库完全是个菜鸟，我已经阅读了很多关于如何做到这一点的文章，但没有运气。

我可以一次存储和显示一张图片，但我希望能够有一个画廊来显示最近上传的 5 张。

任何建议或帮助将不胜感激谢谢！

ps 我知道将图片存储到这样的数据库是不受欢迎的，但这个项目只是为了练习。

索引.php

<!DOCTYPE html>
<html>
<head>
<title>Project One</title>
</head>

<body>

<form action="index.php" method="POST" enctype="multipart/form-data">
    File:
    <input type="file" name="image"> <input type="submit" value="Upload">
<form>
<p />

<?php

//connect to database
(connect to server)
(select correct DB)

//file properties
$file = $_FILES['image']['tmp_name'];

if (!isset($file))
    echo "please select an image.";
else
  {
  $image = addslashes(file_get_contents($_FILES['image']['tmp_name']));
  $image_name = $_FILES['image']['name'];
  $image_size = getimagesize($_FILES['image']['tmp_name']); 

  if($image_size==FALSE)
    echo "That's not an image.";
  else
  {
    if (!$insert = mysql_query("INSERT INTO store VALUES ('', '$image_name', '$image')"))
        echo "Problem Uploading Image.";
    else
        {

        $lastid = mysql_insert_id();
        echo "Image uploaded. <p />Your image:<p /><img src=get.php?id=$lastid>";

        }
  }
  }

?>

<p />
<p />
<a href="http://WEBSITE.com/gallery.php"> Go to Gallery </a>
</body>

</html>

获取.php

<?php

   //connect to database
    (connect to server)
    (select correct DB)

$id = addslashes($_REQUEST['id']);

$image = mysql_query("SELECT * FROM store WHERE id=$id");
$image = mysql_fetch_assoc($image);
$image = $image['image'];

header("Content-type: image/jpeg");

echo $image;

?>
4

4 回答 4

7

这是我想做类似事情时使用的……很久以前！=P

$sql = "SELECT image FROM table WHERE cond ORDER BY xxxx DESC LIMIT 5";
$result = mysqli_query($db,$sql);
while($arraySomething = mysqli_fetch_array($result))
{
    echo "<img src='php/imgView.php?imgId=".$arraySomething."' />";
}
于 2012-11-03T23:57:32.797 回答
7

我尝试第一种方法，header('content-type: image/jpeg');但最终没有显示图像。在通过网站搜索了一些谷歌之后，我找到了可以将图像从数据库显示到我的页面的解决方案

尝试这个：

mysql_connect("localhost","root","")or die("Cannot connect to database"); //keep your db name
mysql_select_db("example_db") or die("Cannot select database");
$sql = "SELECT * FROM `article` where `id` = 56"; // manipulate id ok 
$sth = mysql_query($sql);
$result=mysql_fetch_array($sth);
// this is code to display 
echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image_file'] ).'"/>'
于 2014-05-12T08:08:05.780 回答
2 
mysql_connect("localhost","root","")or die("Cannot connect to database"); 

//keep your db name
mysql_select_db("example_db") or die("Cannot select database");

$sql = "SELECT * FROM `article` where `id` = 56"; 
// manipulate id ok 
$sth = mysql_query($sql);
$result=mysql_fetch_array($sth);
// this is code to display 

echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image_file'] ).'"/> width="xxxx" height="xxxx"';

还要添加高度和宽度

于 2015-10-08T06:16:53.710 回答
1

您也可以使用此功能

//Retrieve image from database and display it on html webpage    
function displayImageFromDatabase(){    
//use global keyword to declare conn inside a function    
global $conn;    
$sqlselectimageFromDb = "SELECT * FROM `imageuploadphpmysqlblob` ";    
$dataFromDb = mysqli_query($conn, $sqlselectimageFromDb);    
while ($row = mysqli_fetch_assoc($dataFromDb)) {    
echo '<img height="250px" width="250px" src=data:image;base64,'.$row['image'].'/>';    
}

像这样将其插入mysql数据库：

$image = $_FILES['imagefile']['tmp_name'];
$name = $_FILES['imagefile']['name'];
$image = base64_encode(file_get_contents(addslashes($image)));

参考：https ://mauricemutetingundi.blogspot.com/2019/04/how-to-upload-blob-image-to-mysql.html

于 2019-04-07T13:39:17.660 回答