PHP image is not updating in database

Question

I want to update my image in my database. But when I update it in my form the value in my database goes empty for the image.

Here is the PHP script I am using to update the image and other values. Good to notice is that only my image value goes to empty the rest of the values in my database are updating.

<?php 
    if(isset($_POST["submit"]) && isset($_GET['id']) && $_GET['id'] == "type") {

    include('config.php');

    $imgFile = $_FILES['user_image']['name'];
    $tmp_dir = $_FILES['user_image']['tmp_name'];
    $imgSize = $_FILES['user_image']['size'];

    if($imgFile)
    {
        $upload_dir = 'user_images/'; // upload directory   
        $imgExt = strtolower(pathinfo($imgFile,PATHINFO_EXTENSION)); // get image extension
        $valid_extensions = array('jpeg', 'jpg', 'png', 'gif'); // valid extensions
        $userpic = rand(1000,1000000).".".$imgExt;
        if(in_array($imgExt, $valid_extensions))
        {           
            if($imgSize < 5000000)
            {
                unlink($upload_dir.$edit_row['opzoekImage']);
                move_uploaded_file($tmp_dir,$upload_dir.$userpic);
            }
            else
            {
                $errMSG = "Sorry, your file is too large it should be less then 5MB";
            }
        }
        else
        {
            $errMSG = "Sorry, only JPG, JPEG, PNG & GIF files are allowed.";        
        }   
    }
    else
    {
        // if no image selected the old image remain as it is.
        $userpic = $edit_row['opzoekImage']; // old image from database
    }

    $sql2 = "UPDATE tblOpzoek SET opzoekName='".$_POST["typenaam"]."', opzoekValue='".$_POST["typewaarde"]."', opzoekImage='".$userpic."' WHERE opzoekId=".$_POST["typeid"]."";

    if($db->query($sql2) === TRUE) {
        header('Location: ' . $_SERVER['HTTP_REFERER']);
        } else {
        echo "<script type= 'text/javascript'>alert('Error: " . $sql2 . "<br>" . $db->error."');</script>";
        }
    }
?>

And here is the config.php code:

<?php
   define('DB_SERVER', 'localhost');
   define('DB_USERNAME', 'root');
   define('DB_PASSWORD', 'root');
   define('DB_DATABASE', 'offerteBVDO');
   $db = mysqli_connect(DB_SERVER,DB_USERNAME,DB_PASSWORD,DB_DATABASE);
?>

Update:

The SQL query is now outputting this: UPDATE tblOpzoek SET opzoekName='Greenline Veranda', opzoekValue='greenline_veranda', opzoekImage='12443.jpg' WHERE opzoekId='1'

But in the database the image does not have the value as in the statement before.

When i insert the SQL query in my phpmyadmin it is working, why is it not working in my PHP?

Check if `$userpic` is empty before doing an update. If empty, don't include the image field in the sql string, otherwise included it. — Karlo Kokkak, Apr 06 '18 at 09:27
print the query "echo $sql2 " and run the query directly to mysql and see any errors — Suresh Kamrushi, Apr 06 '18 at 09:28
Your script is wide open to [SQL Injection Attack](http://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php) Even [if you are escaping inputs, its not safe!](http://stackoverflow.com/questions/5741187/sql-injection-that-gets-around-mysql-real-escape-string) Use [prepared parameterized statements](http://php.net/manual/en/mysqli.quickstart.prepared-statements.php) — RiggsFolly, Apr 06 '18 at 09:33

score 1 · Accepted Answer · edited Apr 06 '18 at 09:38

1

Your update query seem error, please update with this one.

$sql2 = "UPDATE tblOpzoek SET opzoekName='".$_POST["typenaam"]."', 
                opzoekValue='".$_POST["typewaarde"]."', 
                opzoekImage='".$userpic."' 
        WHERE opzoekId='".$_POST["typeid"]."'";

edited Apr 06 '18 at 09:38

RiggsFolly

93,638
21
103
149

answered Apr 06 '18 at 09:37

Mukesh Jakhar

309
2
7

it still puts `opzoekImage` with an empty value – Jari Rengeling Apr 06 '18 at 09:41
then you have to check you html form, add form attribute enctype="multipart/form-data" otherwise image not upload – Mukesh Jakhar Apr 09 '18 at 03:29

PHP image is not updating in database

1 Answers1