Retrieving image from php mysql database using specific id inserted by user - php

Question

I created a page for Inserting Image with some data into database in php/mysql .
Table Structure
name Varchar(20) |
rollno Varchar(50) PRIMARY KEY |
address Varchar(100) |
image LONGBLOB
but i don't know how to retrieve it using roll no. i.e. when user insert roll no in a text box and click on submit it should display all the details of that roll no including image.
I am Using below code for image displaying.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "galleryupload";
// Create connection
$conn = new mysqli($servername, $username, $password,$dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM `vision` where `rollno` = VCI/2012-13/C/03"; // manipulate id ok
$sth = mysqli_query($conn,$sql);
$result=mysqli_fetch_array($sth);
echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image_file'] ).'"/>'
?>

Answer 1

Actually, you do not have to use Javascript for this (Unless you want to retrieve and display the image without reloading the page).
You can do this by having the form and the script responsible for retrieving image in the same php file. Firstly, display the form (it will be displayed even after it was submitted, and image shown):
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>">
<input type="text" name="rollno">
<input type="submit" name="Search">
</form>
Then, let's check if a POST request was received, lookup and display the image. I am using Jay S. code for DB connection.
<?php
if (isset($_POST['rollno'])) {
$rollno = htmlspecialchars($_POST['rollno']);
$db = new PDO('mysql:host=localhost;dbname=testdb;', 'username', 'password');
$stmt= $db->prepare("SELECT * FROM table WHERE rollno = :rollno");
$stmt->execute(array(":rollno", $rollno));
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
$row = $rows[0];
//display info about image
?>
<div class="row-info">
<span>Name: </span>
<span><?php echo $row['name'] ?></span>
</div>
<div class="row-info">
<span>Rollno: </span>
<span><?php echo $row['rollno'] ?></span>
</div>
<div class="row-info">
<span>Address: </span>
<span><?php echo $row['address'] ?></span>
</div>
<div class="image">
<img src="<?php echo $row['image'] ?>"/>
</div>
<?php
}
Of course, you should add your own markup to display the information.

Answer 2

Ok, I am assuming this textbox is on and HTML page, so you would need to use Javascript to get the information from the server, where the MySQL database is. You will need a php page on the server that the Javascript can call. For this answer, i'll just name it getRecord.php. Also, to make the JavaScript alot easier, I recommend using jQuery. I will be assuming you know ho to use that in this answer.
Javascript
var rollno = $("#inputbox").val();
$.get("getRecord.php",{
data: {rollno : rollno},
success: function(data){
data = JSON.parse(data);
//Display the data how you wish
}
);
getRecord.php
<?
$rollno = $_GET['rollno'];
//For the Database retrieval part, I'm going to assume you are using PDO, but if you're using mysqli, just translate this
$db = new PDO('mysql:host=localhost;dbname=testdb;charset=utf8', 'username', 'password');
$stmt= $db->prepare("SELECT * FROM table WHERE rollno = :rollno");
$stmt->execute(array(":rollno", $rollno));
$rows = $stmt->fetchAll(PDO::FETCH_ASSOC);
echo json_encode($rows[0]);
//Because this should only return one row, im doing [0]. After echo, this will be sent back to the Javascript where the success function will execute
?>

Answer 3

// this is the code of get.php page...
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "galleryupload";
// Create connection
$conn = new mysqli($servername, $username, $password,$dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM `vision` where `rollno` = VCI/2012-13/C/03"; // manipulate id ok
$sth = mysqli_query($conn,$sql);
$result=mysqli_fetch_array($sth);
echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image_file'] ).'"/>'
?>