Ktm*_*k13 5 html php mysql blob image
我试图在MySql中显示上传到我的"商店"表的最后5张图片.我是PHP和数据库的完整菜鸟,我已经阅读了很多关于如何做到这一点,但没有运气.
我可以一次存储和显示一个图片,但我希望能够有一个画廊来显示最后5个上传的图片.
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礼貌的,但这个项目只是为了练习.
的index.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>
get.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;
?>
这就是我想做类似的事情时使用的......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."' />";
}
小智 6
我尝试了第一种方法header('content-type: image/jpeg');但最终没有显示图像.经过一些谷歌浏览网站后,我找到了解决方案,我可以将数据从数据库显示到我的页面
试试这个:
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'] ).'"/>'
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