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.";}

I wouldnt't recommend storing image in the database, what your doing currently is a better approach. All you need to do is store the path to the image in the database — David Jones
– David Jones, Commented Nov 5, 2014 at 12:58
Thanks for your help, i figured it out. Now i am just storing the name of the uploaded file in my database and retrieving that name to open the image where ever i want it. — Vipul Behl
– Vipul Behl, Commented Nov 5, 2014 at 13:25

rap-2-h · Accepted Answer · 2014-11-05 14:13:28Z

11

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";
}

edited Nov 5, 2014 at 14:13

rap-2-h

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answered Nov 5, 2014 at 13:33

Vipul Behl

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3 Comments

Strawberry Over a year ago

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)

Clinton Over a year ago

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

Clinton Over a year ago

@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....

m7913d · Accepted Answer · 2017-04-28 08:22:57Z

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();
    }
}

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)

Regolith · Accepted Answer · 2019-08-16 05:39:55Z

-1

insert image zh

-while we insert image in database using insert query

$Image = $_FILES['Image']['name'];
    if(!$Image)
    {
      $Image="";
    }
    else
    {
      $file_path = 'upload/';
      $file_path = $file_path . basename( $_FILES['Image']['name']);    
      if(move_uploaded_file($_FILES['Image']['tmp_name'], $file_path)) 
     { 
}
}

edited Aug 16, 2019 at 5:39

Regolith

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answered Aug 16, 2019 at 4:52

zalak patel

11 bronze badge

Comments

rahul patel · Accepted Answer · 2015-03-10 14:49:43Z

<!-- 
//THIS PROGRAM WILL UPLOAD IMAGE AND WILL RETRIVE FROM DATABASE. UNSING BLOB
(IF YOU HAVE ANY QUERY CONTACT:[email protected])


CREATE TABLE  `images` (
  `id` int(100) NOT NULL AUTO_INCREMENT,
  `name` varchar(100) NOT NULL,
  `image` longblob NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB ;

-->
<!-- this form is user to store images-->
<form action="index.php" method="post"  enctype="multipart/form-data">
Enter the Image Name:<input type="text" name="image_name" id="" /><br />

<input name="image" id="image" accept="image/JPEG" type="file"><br /><br />
<input type="submit" value="submit" name="submit" />
</form>
<br /><br />
<!-- this form is user to display all the images-->
<form action="index.php" method="post"  enctype="multipart/form-data">
Retrive all the images:
<input type="submit" value="submit" name="retrive" />
</form>



<?php
//THIS IS INDEX.PHP PAGE
//connect to database.db name is images
        mysql_connect("", "", "") OR DIE (mysql_error());
        mysql_select_db ("") OR DIE ("Unable to select db".mysql_error());
//to retrive send the page to another page
if(isset($_POST['retrive']))
{
    header("location:search.php");

}

//to upload
if(isset($_POST['submit']))
{
if(isset($_FILES['image'])) {
        $name=$_POST['image_name'];
        $email=$_POST['mail'];
        $fp=addslashes(file_get_contents($_FILES['image']['tmp_name'])); //will store the image to fp
        }
                // our sql query
                $sql = "INSERT INTO images VALUES('null', '{$name}','{$fp}');";
                            mysql_query($sql) or die("Error in Query insert: " . mysql_error());
} 
?>



<?php
//SEARCH.PHP PAGE
    //connect to database.db name = images
         mysql_connect("localhost", "root", "") OR DIE (mysql_error());
        mysql_select_db ("image") OR DIE ("Unable to select db".mysql_error());
//display all the image present in the database

        $msg="";
        $sql="select * from images";
        if(mysql_query($sql))
        {
            $res=mysql_query($sql);
            while($row=mysql_fetch_array($res))
            {
                    $id=$row['id'];
                    $name=$row['name'];
                    $image=$row['image'];

                  $msg.= '<a href="search.php?id='.$id.'"><img src="data:image/jpeg;base64,'.base64_encode($row['image']). ' " />   </a>';

            }
        }
        else
            $msg.="Query failed";
?>
<div>
<?php
echo $msg;
?>

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