Drop down - get value from database set option to selected

PHP
1

Hi,

I am trying to create a drop down menu that will select a value that is stored in the database - right now the code creats a dropdown (with nothing selected) - hope someone can help.

in the database, the values are stored as
–null-
Option1
Option2
Option3

my code is


$instruction = $_GET['instruction'];

<?php 
<select id="instruction" name="instruction">
<option value="" <?php if (!empty($instruction) && $instruction == '' ) echo 'selected = "selected"'; ?>></option>      
<option value="Option1" <?php if (!empty($instruction) && $instruction == 'Option1')  echo 'selected = "selected"'; ?>>Option1</option>
<option value="Option2" <?php if (!empty($instruction) && $instruction == 'Option2')  echo 'selected = "selected"'; ?>>Option2</option>
<option value="Option3" <?php if (!empty($instruction) && $instruction == 'Option3')  echo 'selected = "selected"'; ?>>Option3</option>
</select>
<?
2

Hi missingcolor,

Your code works fine when the php delimiters are positioned correctly


<?php $instruction = $_GET['instruction']; ?>

<select id="instruction" name="instruction">
<option value="" <?php if (!empty($instruction) && $instruction == '' ) echo 'selected = "selected"'; ?>></option>
<option value="Option1" <?php if (!empty($instruction) && $instruction == 'Option1')  echo 'selected = "selected"'; ?>>Option1</option>
<option value="Option2" <?php if (!empty($instruction) && $instruction == 'Option2')  echo 'selected = "selected"'; ?>>Option2</option>
<option value="Option3" <?php if (!empty($instruction) && $instruction == 'Option3')  echo 'selected = "selected"'; ?>>Option3</option>
</select>
3

i got it to work by using this instead

<option value=“Option3” <?php if ($row['instruction] == ‘Option3’) echo ‘selected = “selected”’; ?>>Option3</option>

1 Like