从表中获取 blob 图像并使用 php sqlite3 显示它
[英]Fetch blob image from table and display it using php sqlite3
问题描述
I know this question has been asked many times but I couldnot solve this using any of them.
我知道这个问题已被问过很多次，但我无法使用其中任何一个来解决这个问题。 I am new to sqlite and cannot understand what I am doing wrong.我是 sqlite 的新手，无法理解我做错了什么。
WHAT I AM TRYING我在尝试什么
I am trying to make a profile view page.我正在尝试制作个人资料视图页面。 I am able to fetch all details from my sqlite database but i am not able to display my profile picture.我可以从我的 sqlite 数据库中获取所有详细信息，但我无法显示我的个人资料图片。
TABLE STRUCTURE表结构
**username|landline|mobile|email|profilepicture**
john |xxxxxxxx|xxxxxx|x@x.x|blob
WHAT I TRIED我的尝试
$sql = "SELECT * FROM profile";
$query = $db->query($sql);
while($row = $query->fetchArray(SQLITE3_ASSOC) ){
echo "NAME = ". $row['user_name'] . "<br/>";
echo "LANDLINE = ". $row['user_landline'] ."<br/>";
echo "MOBILE = ". $row['user_mobile'] ."<br/>";
echo "EMAIL = ".$row['user_email'] ."<br/>";
header('Content-Type: image/png');
echo $row['user_profile_picture'];
}
<html>
<img src='profile.php?imgid=<?php echo $row['user_profile_picture'];?>'/>
</html>
But the image dosenot show and also the rest of the data dosenot display when i put header('Content-Type: image/png');但是当我放置header('Content-Type: image/png');时，图像不显示，其余数据也不显示header('Content-Type: image/png');
1 个解决方案
解决方案1
2 已采纳 2013-06-06 06:29:06
Create an image.php:创建一个 image.php：
<?php
$sql = "SELECT user_profile_picture FROM profile WHERE id = " . $_GET['id'];
$query = $db->query($sql);
$row = $query->fetchArray(SQLITE3_ASSOC);
header('Content-Type: image/png');
echo $row['user_profile_picture'];
In profile.php:在 profile.php 中：
<img src='image.php?id=<?php echo $row['id'];?>'/>
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