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.jpg Please see my code can one guide me how to display the image.
$sqlimage = "SELECT image FROM userdetail where `id` = $id1";
$imageresult1 = mysql_query($sqlimage);
while($rows = mysql_fetch_assoc($imageresult1))
{
$image = $rows['image'];
print $image;
}
put this code to your php page.
$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.
$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'].'" />';
}
put you $image in img tag of html
try this
echo '<img src="your_path_to_image/'.$image.'" />';
instead of
print $image;
your_path_to_image would be absolute path of your image folder like eg: /home/son/public_html/images/ or as your folder structure on server.
Update 2 :
if your image is resides in the same folder where this page file is exists
you can user this
echo '<img src="'.$image.'" />';
You need to do this to display image
$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.
instead of print $image; you should go for print "<img src=<?$image;?>>"
and note that $image should contain the path of your image.
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.
Simply replace
print $image;
with
echo '<img src=".$image." >';
For example if you use this code , you can load image from db (mysql) and display it in php5 ;)
<?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">';
?>
<?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'].'">';
}
?>
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.
Here I am showing image and name from database.
Note: I am only storing the path of image in database but actual image i am storing in photo folder.
PHP Complete Code with design: 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.
I hope may be you like my answer.
Thank You
https://www.campuslife.co.in/Php/how-to-show-image-from-database-using-php-mysql.php
Output
[Showing Images from Database][1]
<?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
Displaying an image from MySql Db.
$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'] ).'"/>';
Source: Stackoverflow.com