从表中获取Blob图片，并使用php sqlite3将其显示 [英] Fetch blob image from table and display it using php sqlite3
本文介绍了从表中获取Blob图片，并使用php sqlite3将其显示的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我知道这个问题已经被问过很多次了，但是我无法使用其中任何一个来解决. 我是sqlite的新手，无法理解我在做什么错.
I know this question has been asked many times but I couldnot solve this using any of them. I am new to sqlite and cannot understand what I am doing wrong.
我正在尝试创建个人资料查看页面.我可以从sqlite数据库中获取所有详细信息，但无法显示个人资料图片.
I am trying to make a profile view page. I am able to fetch all details from my sqlite database but i am not able to display my profile picture.
**username|landline|mobile|email|profilepicture**
john |xxxxxxxx|xxxxxx|x@x.x|blob
我尝试过的东西
$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>
但是当我放header('Content-Type: image/png');
推荐答案
创建一个image.php:
Create an 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'];
在profile.php中:
In profile.php:
<img src='image.php?id=<?php echo $row['id'];?>'/>
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