Update value in database using PHP

Question

This is how the table users looks like:

---------------------------------------------------
|  ID  |  user  |             avatar              |
|  3   |  ane22 |     /img/default-avatar.png     |
|  4   |  cuz33 |     /img/default-avatar.png     |

Upload button:

<form method="POST" 
              action="" 
              enctype="multipart/form-data"> 
            <input type="file" 
                   name="uploadfile" 
                   value="" /> 
  
            <div> 
                <input type="submit" name="action" value="Upload"> 
            </div> 
        </form>

This is how I am trying to update the profile picture for the current user:

} else if ($_POST['action'] == 'Upload') {
    //action for delete
    error_reporting(0); 
      $msg = ""; 
  $name = $_SESSION["user"];/* user */
  
  // If upload button is clicked ... 
  if (isset($_POST['upload'])) { 
  
    $filename = $_FILES["uploadfile"]["name"]; 
    $tempname = $_FILES["uploadfile"]["tmp_name"];     
        $folder = "image/".$filename; 
  
        // Get all the submitted data from the form 
        $sql = "UPDATE users SET avatar='" . $filename . "' WHERE user='" . $name . "'"; 
  
        // Execute query 
        mysqli_query($link, $sql); 
          
        // Now let's move the uploaded image into the folder: image 
        if (move_uploaded_file($tempname, $folder))  { 
            $msg = "Image uploaded successfully"; 
        }else{ 
            $msg = "Failed to upload image"; 
      } 
  } 
  $result = mysqli_query($link, "SELECT * FROM users");

When I press the upload button, nothing happens. And nothing changes in my table also. Any tips?

Are you sure isset($_POST['upload']) is correct? Have you checked what $_POST contains. — Nigel Ren
– Nigel Ren, Commented Jul 30, 2020 at 8:51
Your query is vulnerable to SQL injections, consider using prepared statements — Cid
– Cid, Commented Jul 30, 2020 at 8:54
You can try uploading the image first and then update to database. Something like below: if (move_uploaded_file($tempname, $folder)) { $msg = "Image uploaded successfully"; mysqli_query($link, $sql); }else{ $msg = "Failed to upload image"; } — AbuHuraira Lakdawala
– AbuHuraira Lakdawala, Commented Jul 30, 2020 at 8:57

Egon · Accepted Answer · 2020-07-30 09:39:04Z

0

Man you have no action on your form, how do you want it to submit something if the form doesn't do anything... And also why would you name a variable $id if you store a name in it... That's why "WHERE user = $id" works, because it's not an id it's a name... Before asking a question in stackoverflow pls try to find your own mistake, or if your error doesn't already exist on it

answered Jul 30, 2020 at 9:39

Egon

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14 Comments

Erik Auranaune Over a year ago

Man I don't get any error message, thats why. Doesn't really matter what I name the $id, because the value is still the name.

Egon Over a year ago

of course you don't get any error message because if the form has no action it won't load anything. PHP doesn't take it as an error, bcs it might be done on purpose, but if you want to make your form upload something, it has to get a link to something

Erik Auranaune Over a year ago

What do you mean no action? Do you not see the button?

Egon Over a year ago

the button has as action a submit, it's not the action of the form, it's acting like an onclick listener, it just submit to your form when you click on it, so it is supposed to act as the "sender" of the form, but if the form has no receiver, it can't send anything

Egon Over a year ago

You are trying to send data, without saying where/who should get it, that's the problem. You are supposed to declare it on the attribute "action=" of you form

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