Updating table using PHP, HTML and MySQL

Question

I am trying to update a table in my database with PHP.

I used a loop to get all of the data from the table. Then, adding an "Approve" button in front of each row where the user can click if they want to approve the request, which would change the status to Approved.

Now I'm stuck, I can't find a way to update the Status field and I am not good with JavaScript.

I thought about making the button name dynamic, but I couldn't.

Output image

This is the loop code:

while($row = mysqli_fetch_array($result))
  {

echo "<tr bgcolor='#cccccc'>";
echo "<td>" . $row['nursename'] . "</td>";
echo "<td>" . $row['dr'] . "</td>";
echo "<td>" . $row['natureofreq'] . "</td>";
echo "<td>" . $row['clinic'] . "</td>";
echo "<td>" . $row['clinicfloor'] . "</td>";
echo "<td>" . $row['qitem'] . "</td>";
echo "<td>" . $row['moreinfo'] . "</td>";
echo "<td>" . $row['user_check'] . "</td>";
echo "<td>" . $row['id'] . "</td>";
echo "<td bgcolor='#009FE3'>" . $row['rstatus'] . "</td>";
echo '<form name="submit" method="post" action="">';
echo "<td>" . '<button id="approve" name="approve"> Approve</button>' . "</td>";
echo '</form>';
echo "</tr>";
}

This is the update code, which is not working:

$id=$row['id'];
if(isset($_POST['approve']))
{
  $allowed = mysqli_query($n," UPDATE newrequest SET rstatus = 'Approved' WHERE id = '$id' ");
}

I would appreciate any help.

output

Try $id = $_POST['id'];... if I understand what you’re trying to do, I guess you need to fetch whatever you send in the form... but $row is not set in the second code. Also, actually send it, because I think you are not. In other words, I’m assuming your update code is somewhere else because it should be... after all. — Zeke
– Zeke, Commented Nov 19, 2017 at 1:12
Can you please update your loop code with a row for database id? — hans-könig
– hans-könig, Commented Nov 19, 2017 at 1:20

hans-könig · Accepted Answer · 2017-11-19 02:18:51Z

2

     $url_of_current_page = $_SERVER['REQUEST_URI'];

     if(isset($url_of_current_page)){

         $_SESSION['url_of_current_page'] = $url_of_current_page;

     }

     $approve_row = '';

     while($row = mysqli_fetch_array($result)){

         $approve_row .= "<tr bgcolor='#cccccc'>

                             <td>" . $row['nursename'] . "</td>

                             <td>" . $row['dr'] . "</td>

                             <td>" . $row['natureofreq'] . "</td>

                             <td>" . $row['clinic'] . "</td>

                             <td>" . $row['clinicfloor'] . "</td>

                             <td>" . $row['qitem'] . "</td>

                             <td>" . $row['moreinfo'] . "</td>

                             <td>" . $row['user_check'] . "</td>

                             <td bgcolor='#009FE3'>" . $row['rstatus'] . "</td>

                             <td>
                                 <a href='action.php?id=" . $row['id'] . "'>
                                     <button>Approve</button>
                                 </a>
                             </td>

                         </tr>";

    }


     if(isset($approve_row)){

          echo $approve_row;

     }

And the action code. If your action page is not action.php, then tell me the name of the page.

     if(isset($_GET['id'])){

         $id = $_GET['id'];

         mysqli_query($n," UPDATE newrequest SET rstatus = 'Approved' WHERE id = '$id' ");

         if(isset($_SESSION['url_of_current_page'])){

              header( "Location: ". $_SESSION['url_of_current_page'] );
              exit();

         }
     }

answered Nov 19, 2017 at 2:18

hans-könig

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2 Comments

Sara M. Over a year ago

It is working! thank you. Can you please give a breif explaination of what you did?

hans-könig Over a year ago

Fantastic. The 1st code save the name of the current page in a session variable so that it can be used at the action page. At the action page, the exit() is saying that return to the Approve buttons page after the update is done. Then on the while loop code $approve_row .=" "; returns all rows from the database table. But $approve_row must 1st be initialized(i.e $approve_row = ' ';). I used but link tags around the buttons because it is simpler that way and less coding.

Seyi Daniels · Accepted Answer · 2017-11-19 01:38:01Z

0

You do not seem to be sending any data to the database yet from the code u dropped

answered Nov 19, 2017 at 1:38

Seyi Daniels

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1 Comment

Sara M. Over a year ago

No, there is data in the table but it is sent from a form in another code. The thing I am trying to do here is to change the Status to Approved with a button..

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