How to display image from database using php [duplicate] - php

Question

This question already has answers here:
How to retrieve images from MySQL database and display in an html tag
(4 answers)
Closed 1 year ago.
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;
}

Answer 1

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'] ).'"/>';

Answer 2

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">';
?>

Answer 3

<?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'].'">';
}
?>

Answer 4

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.

Answer 5

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.'" />';

Answer 6

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.

Answer 7

Simply replace
print $image;
with
echo '<img src=".$image." >';

Answer 8

$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'].'" />';
}

Answer 9

<?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

Answer 10

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.