[RESOLVED] Sticky Drop Down

ten120963

I want to write something that has 8 drop down boxes on a form. I would have the code all
in one, so PHP would be at the top, and html at the bottom. PHP code would run:

if ($_SERVER['REQUEST_METHOD'] == 'POST') { // handle the form

If a user makes choices in the 8 drop down boxes that are the same as choices made by
someone else in a database, I want to generate an error, and keep the choices in the drop
down boxes (sticky) so they can be modified.

Does anyone know how this can be accomplished?

bradgrafelman

If you're displaying the form again because an error occurred, simply check to see if the user previously provided a value and, if so, output the appropriate HTML (e.g. selected="selected") for the appropriate <option> element.

ten120963

This is the code that i'm using that does not work:

print "<tr><td>Name:</td><td><select name=\"$name\">";

$query = "SELECT name FROM users ORDER BY name ASC";

if ($r = mysql_query($query, $dbc)) { // run the query

  // retrieve every record and add to a dropdown box
  while ($row = mysql_fetch_array($r)) {
    if (isset($_POST['name']) &&
    $_POST['name'] == $row['name']) {
      print "<option selected=\"selected\" value=\" {$row['name']} \"> {$row['name']} </option>";      
    } else {
      print "<option value=\" {$row['name']} \"> {$row['name']} </option>";  
    }
  }  

} else { // query did not run
  print '<p style="color: red;">Could not retrieve the data because:<br />' . mysql_error($dbc) 
  . '.</p><p>The query being run was: ' . $query . '</p>';
} // end of query if

bradgrafelman

When posting PHP code, please use the board's [noparse]

..

[/noparse] bbcode tags as they make your code much easier to read and analyze.

What is the value of $name? Also, note that your <option> elements have spaces on either side of $row['name'], yet you're not adding those spaces in the if() comparison. Either add the spaces in the latter or get rid of them in the former.

ten120963

The spaces were the cause of the problem. Once I removed them and ran a test, the first Drop Down retained
the choice. Thank you so much for fixing my problem!

One more thing. Is it possible to use a variable in the $_POST ?

For example, I have:

$choices = array ('choice1', 'choice2', 'choice3', 'choice4', 'choice5', 'choice6', 'choice7', 'choice8');

foreach ($choices as $key => $value) {

then later I want to use $value in the $POST, but I cant use $POST['$value'] as it does not work.

This is to avoid the same block of code for each of 8 Drop Down boxes.

bradgrafelman

ten120963;11010967 wrote:
I cant use $_POST['$value'] as it does not work.

It doesn't work for the same reason why this code:

$color = 'blue';
echo '$color';

will output a dollar sign followed by the word color, whereas this code:

$color = 'blue';
echo $color;

will output the word blue.

ten120963

Wow. What a simple fix. I was trying lots of things and i'm sure I tried single and double quotes, but not the one you provided.
Thank you so much for helping out a beginner.