Reputation: 59
I have a database which has a table called 'propImages' and there are two columns.- 'pid' and 'location'.
And i have data in the database where multiple images can contained by single pid. image contains database data
now i want to retrieve images from database according to given pid. there can be more than one image. All i know it there should be an iteration to retrieve images.
I want to display images in HTML .
can you please show me the way to do it in php?
Thanks in advance guys
Upvotes: 0
Views: 1367
Reputation: 409
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "databasename";
// Create connection
$con = mysqli_connect($servername, $username, $password, $dbname);
//create sql
$sql = "SELECT * FROM `propImages` where pid='$YOUR_PID'";
$result = mysqli_query($con, $sql);
$row = mysqli_num_rows($result);
//retrive data print here
if($row > 0){
while($col = mysqli_fetch_assoc($result))
{
echo $col['location'];
}
} else {
echo 'no result found.';
}
?>
wish it helps
Upvotes: 1
Reputation: 534
This may help you
<?php
include 'inc/database.php';
$conn = new mysqli($servername, $username, $password, $database);
$propid = $_GET['propid'];
$sql = "SELECT * FROM propImages WHERE propid='" . $propid . "';";
$result = $conn->query($sql);
if($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo "<img src=" . $row['image'] . ">";
}
}
else {
echo "No results";
}
?>
in the inc/database.php :
<?php
$servername = "localhost";
$username = "root";
$password = "";
$database = "database";
?>
To see how it works try visiting : file.php?propid=22
Upvotes: 1