Print selected attribute on select option with PHP and MySQL

Question

I'm printing a select list from my database.

Here's a part of my code:

<form action="" method="post">
    <div class="form-item">
        <select name="specialties">
        <?php
        $query = "SELECT * FROM specialties";
        $result = mysqli_query($connection, $query);

        echo '<option value="All">' . $filterTitle . '</option>';
        while($row = mysqli_fetch_assoc($result)) {
            $id = $row['id'];
            $title = $row['title_'. $lang];

            if($_POST['specialties'] == $id) {
                $selected = 'selected';
            }

            echo '<option value="' . $id . ' "' . $selected . '>' . $title . '</option>';
        }
        ?>  
        </select>
    </div>

    <div class="form-item">
        <input class="form-submit" type="submit" name="submit" value="<?php echo $filterSubmit; ?>">
    </div>
</form>

In this form when I choose some of the options, it filters the content. That's working properly and I don't need to post that part of the code.

First when I choose some option from the select list and after submit, it filter the content, but the select option shows the All value, instead of the selected option. After that I've tried with the code above and now when I submit the form, the select value shows the last option and all the options has selected attribute.

How can I change that logic, so when I select let's say the second option, after submitting the form, the selected attribute will appear only on the second option, so the text will be the second option?

CONCLUSION: Thank you guys for your answers. You're all correct. I gave you all +1 and I choose the Hassaan's answer, because it's very well explained.

This is because you set the $selected value once but you never unset it. So after your if statement succeeds once all the subsequent options will also be set as selected. Try adding a else { $selected = ''; } Or something to the same effect. — apriede
– apriede, Commented May 9, 2016 at 10:53
Yes, I've checked the answers from the other guys before and now I know the missing gap. Thanks anyway. :) — user2519032
– user2519032, Commented May 9, 2016 at 11:08

Hassaan · Accepted Answer · 2016-05-09 11:03:56Z

1

Currently you are checking/comparing values and generating variable. You need to use else condition because you must change/reset the value of generated variable. See the example below.

if($_POST['specialties'] == $id) {
    $selected = 'selected';
}
else
    $selected = '';

edited May 9, 2016 at 11:03

answered May 9, 2016 at 10:55

Hassaan

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Alex Andrei · Accepted Answer · 2016-05-09 10:54:08Z

1

Add an else clause to reset the selected attribute to an empty string.

if($_POST['specialties'] == $id) {
    $selected = 'selected';
}
else{
    $selected = '';
}

answered May 9, 2016 at 10:54

Alex Andrei

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Comments

Gokul Shinde · Accepted Answer · 2016-05-09 10:54:48Z

1

You need to make empty selected variable every time like below

while($row = mysqli_fetch_assoc($result)) {
    $selected = '';
  $id = $row['id'];
  $title = $row['title_'. $lang];

  if($_POST['specialties'] == $id) {
      $selected = 'selected';
  }

  echo '<option value="' . $id . ' "' . $selected . '>' . $title . '</option>';
}

answered May 9, 2016 at 10:54

Gokul Shinde

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