How to store images in mysql database using php

Question

How can i store and display the images in a MySQL database. Till now i have only written the code to get the images from the user and store them in a folder, the code that i wrote till now is: HTML FILE

<input type="file" name="imageUpload" id="imageUpload">

PHP FILE

    $target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["imageUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);


if (move_uploaded_file($_FILES["imageUpload"]["tmp_name"], $target_file)) {
    echo "The file ". basename( $_FILES["imageUpload"]["name"]). " has been uploaded.";
} else {
    echo "Sorry, there was an error uploading your file.";}

Solution 1

I found the answer, For those who are looking for the same thing here is how I did it. You should not consider uploading images to the database instead you can store the name of the uploaded file in your database and then retrieve the file name and use it where ever you want to display the image.

HTML CODE

<input type="file" name="imageUpload" id="imageUpload">

PHP CODE

if(isset($_POST['submit'])) {

    //Process the image that is uploaded by the user

    $target_dir = "uploads/";
    $target_file = $target_dir . basename($_FILES["imageUpload"]["name"]);
    $uploadOk = 1;
    $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);

    if (move_uploaded_file($_FILES["imageUpload"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["imageUpload"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }

    $image=basename( $_FILES["imageUpload"]["name"],".jpg"); // used to store the filename in a variable

    //storind the data in your database
    $query= "INSERT INTO items VALUES ('$id','$title','$description','$price','$value','$contact','$image')";
    mysql_query($query);

    require('heading.php');
    echo "Your add has been submited, you will be redirected to your account page in 3 seconds....";
    header( "Refresh:3; url=account.php", true, 303);
}

CODE TO DISPLAY THE IMAGE

while($row = mysql_fetch_row($result)) {
    echo "<tr>";
    echo "<td><img src='uploads/$row[6].jpg' height='150px' width='300px'></td>";
    echo "</tr>\n";
}

Solution 2

if(isset($_POST['form1']))
{
    try
    {


        $user=$_POST['username'];

        $pass=$_POST['password'];
        $email=$_POST['email'];
        $roll=$_POST['roll'];
        $class=$_POST['class'];



        if(empty($user)) throw new Exception("Name can not empty");
        if(empty($pass)) throw new Exception("Password can not empty");
        if(empty($email)) throw new Exception("Email can not empty");
        if(empty($roll)) throw new Exception("Roll can not empty");
        if(empty($class)) throw new Exception("Class can not empty");


        $statement=$db->prepare("show table status like 'tbl_std_info'");
        $statement->execute();
        $result=$statement->fetchAll();
        foreach($result as $row)
        $new_id=$row[10];


        $up_file=$_FILES["image"]["name"];

        $file_basename=substr($up_file, 0 , strripos($up_file, "."));
        $file_ext=substr($up_file, strripos($up_file, ".")); 
        $f1="$new_id".$file_ext;

        if(($file_ext!=".png")&&($file_ext!=".jpg")&&($file_ext!=".jpeg")&&($file_ext!=".gif"))
        {
            throw new Exception("Only jpg, png, jpeg or gif Logo are allow to upload / Empty Logo Field");
        }
        move_uploaded_file($_FILES["image"]["tmp_name"],"../std_photo/".$f1);


        $statement=$db->prepare("insert into tbl_std_info (username,image,password,email,roll,class) value (?,?,?,?,?,?)");

        $statement->execute(array($user,$f1,$pass,$email,$roll,$class));


        $success="Registration Successfully Completed";

        echo $success;
    }
    catch(Exception $e)
    {
        $msg=$e->getMessage();
    }
}

Comments

[Strawberry]

A user may have valid reasons for uploading images to a database, and it can be performative. It's just not very popular, and so not as well 'supported' (i.e. a less extensive knowledge base)

[Boann]

Riddled with security holes.

[Clinton]

WARNING: The above example code is wide open for SQL injection vulnerabilities.

[user5226582]

While the code which would solve the problem is welcome, please add some explanation. (Also a good idea to make examples as small as possible)

[Clinton]

@Oy3 You are necro'ing a 5 year old comment. Learning about SQL injection is an exercise every developer should do, and there are many examples online. I will not provide a better method in this case on my 5 year old warning....