php 如何使用php显示数据库中的图像
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How to display image from database using php
提问by Xavi
I am trying to display an image coming from the database and I was not able to display the image .but its showing like this user-1.jpgPlease see my code can one guide me how to display the image.
我试图显示来自数据库的图像，但我无法显示图像。但它的显示方式是这样的user-1.jpg请看我的代码可以指导我如何显示图像。
$sqlimage = "SELECT image FROM userdetail where `id` = $id1";
$imageresult1 = mysql_query($sqlimage);
while($rows = mysql_fetch_assoc($imageresult1))
{
$image = $rows['image'];
print $image;
}
回答by Charles
Displaying an image from MySql Db.
显示来自 MySql Db 的图像。
$db = mysqli_connect("localhost","root","","DbName");
$sql = "SELECT * FROM products WHERE id = $id";
$sth = $db->query($sql);
$result=mysqli_fetch_array($sth);
echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image'] ).'"/>';
回答by MANZAR HUSSAIN
If you want to show images from database then first you have to insert the images in database then you can show that image on page. Follow the below code to show or display or fetch the image.
如果要显示数据库中的图像，则首先必须将图像插入数据库中，然后才能在页面上显示该图像。按照以下代码显示或显示或获取图像。
Here I am showing image and name from database.
在这里，我显示了数据库中的图像和名称。
Note:I am only storing the pathof image in database but actual image i am storing in photofolder.
注意：我只将图像的路径存储在数据库中，但实际图像存储在照片文件夹中。
PHP Complete Code with design: show-code.php
PHP 完整代码设计：show-code.php
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, intial-scale=1.0"/>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<title>Show Image in PHP - Campuslife</title>
<style>
body{background-color: #f2f2f2; color: #333;}
.main{box-shadow: 0 .125rem .25rem rgba(0,0,0,.075)!important; margin-top: 10px;}
h3{background-color: #4294D1; color: #f7f7f7; padding: 15px; border-radius: 4px; box-shadow: 0 1px 6px rgba(57,73,76,0.35);}
.img-box{margin-top: 20px;}
.img-block{float: left; margin-right: 5px; text-align: center;}
p{margin-top: 0;}
img{width: 375px; min-height: 250px; margin-bottom: 10px; box-shadow: 0 .125rem .25rem rgba(0,0,0,.075)!important; border:6px solid #f7f7f7;}
</style>
</head>
<body>
<!-------------------Main Content------------------------------>
<div class="container main">
<h3>Showing Images from database</h3>
<div class="img-box">
<?php
$host ="localhost";
$uname = "root";
$pwd = '123456';
$db_name = "master";
$file_path = 'photo/';
$result = mysqli_connect($host,$uname,$pwd) or die("Could not connect to database." .mysqli_error());
mysqli_select_db($result,$db_name) or die("Could not select the databse." .mysqli_error());
$image_query = mysqli_query($result,"select img_name,img_path from image_table");
while($rows = mysqli_fetch_array($image_query))
{
$img_name = $rows['img_name'];
$img_src = $rows['img_path'];
?>
<div class="img-block">
<img src="<?php echo $img_src; ?>" alt="" title="<?php echo $img_name; ?>" width="300" height="200" class="img-responsive" />
<p><strong><?php echo $img_name; ?></strong></p>
</div>
<?php
}
?>
</div>
</div>
</body>
</body>
</html>
If you found any mistake then you can directly follow the tutorial which is i found from where. You can see the live tutorial step by step on this website.
如果您发现任何错误，那么您可以直接按照我从哪里找到的教程进行操作。您可以在本网站上逐步查看实时教程。
I hope may be you like my answer.
我希望你会喜欢我的回答。
Thank You
谢谢你
https://www.campuslife.co.in/Php/how-to-show-image-from-database-using-php-mysql.php
https://www.campuslife.co.in/Php/how-to-show-image-from-database-using-php-mysql.php
Output
输出
[Showing Images from Database][1]
[显示数据库中的图像][1]
回答by Yo Ne S
For example if you use this code , you can load image from db (mysql) and display it in php5 ;)
例如，如果您使用此代码，则可以从 db (mysql) 加载图像并将其显示在 php5 中；)
<?php
$con =mysql_connect("localhost", "root" , "");
$sdb= mysql_select_db("my_database",$con);
$sql = "SELECT * FROM `news` WHERE 1";
$mq = mysql_query($sql) or die ("not working query");
$row = mysql_fetch_array($mq) or die("line 44 not working");
$s=$row['photo'];
echo $row['photo'];
echo '<img src="'.$s.'" alt="HTML5 Icon" style="width:128px;height:128px">';
?>
回答by Paulo Rodrigues
$sqlimage = "SELECT image FROM userdetail where `id` = $id1";
$imageresult1 = mysqli_query($link, $sqlimage);
while($rows=mysqli_fetch_assoc($imageresult1))
{
echo "<img src = 'Image/".$row['image'].'" />';
}
回答by Muhammad Raheel
You need to do this to display image
您需要这样做才能显示图像
$sqlimage = "SELECT image FROM userdetail where `id` = $id1";
$imageresult1 = mysql_query($sqlimage);
while($rows=mysql_fetch_assoc($imageresult1))
{
$image = $rows['image'];
echo "<img src='$image' >";
echo "<br>";
}
You need to use html img tag.
您需要使用 html img 标签。
回答by Lizwi
<?php
$connection =mysql_connect("localhost", "root" , "");
$sqlimage = "SELECT * FROM userdetail where `id` = '".$id1."'";
$imageresult1 = mysql_query($sqlimage,$connection);
while($rows = mysql_fetch_assoc($imageresult1))
{
echo'<img height="300" width="300" src="data:image;base64,'.$rows['image'].'">';
}
?>
回答by saurabh
<?php
$conn = mysql_connect ("localhost:3306","root","");
$db = mysql_select_db ("database_name", $conn);
if(!$db) {
echo mysql_error();
}
$q = "SELECT image FROM table_name where id=4";
$r = mysql_query ("$q",$conn);
if($r) {
while($row = mysql_fetch_array($r)) {
header ("Content-type: image/jpeg");
echo $row ["image"];
}
}else{
echo mysql_error();
}
?>
sometimes problem may occures because of port number of mysql server is incoreect to avoid it just write port number with host name like this "localhost:3306"
in case if you have installed two mysql servers on same system then write port according to that
in order to display any data from database please make sure following steps
1.proper connection with sql
2.select database
3.write query
4.write correct table name inside the query
5.and last is traverse through data
回答by Ali Shahbaz
put this code to your php page.
将此代码放入您的php页面。
$sql = "SELECT * FROM userdetail";
$result = mysqli_query("connection ", $sql);
while ($row = mysqli_fetch_array($result,MYSQLI_BOTH)) {
echo "<img src='images/".$row['image']."'>";
echo "<p>".$row['text']. "</p>";
}
i hope this is work.
我希望这是工作。
回答by Mehar
Simply replace
只需更换
print $image;
with
和
echo '<img src=".$image." >';
回答by user2656474
instead of print $image;you should go for print "<img src=<?$image;?>>"
而不是print $image;你应该去print "<img src=<?$image;?>>"
and note that $image should contain the path of your image.
并注意 $image 应包含图像的路径。
So, If you are only storing the name of your image in database then instead of that you have to store the full path of your image in the database like /root/user/Documents/image.jpeg.
因此，如果您仅将图像名称存储在数据库中，那么您必须将图像的完整路径存储在数据库中，例如 /root/user/Documents/image.jpeg。
