WHILE中的IF语句 - PHP，MySQL和HTML下拉选择

I'll try to be short with this one. Most probably I'm missing something obvious so any input would be greatly appreciated.

SQL structure/columns:

1. id_user (primary key)
2. income_input
3. income_select

PHP Code:

<?php
$con = mysql_connect("localhost","dbname","pwd");
if (!$con)
  {
  die('Could not connect to the database: ' . mysql_error());
  }

mysql_select_db("dbname", $con);

global $user;            // Drupal 7 global
$userid =  $user->uid;   // Drupal 7 currently logged-in user id

$result = mysql_query("SELECT * FROM my_table WHERE id_user='$userid'");

if($result === FALSE) {
    die(mysql_error());
}

if ($user->uid) {
?>

<form>
  <label for="edit-1">Label</label>
  <input id="edit-1" maxlength="60" size="60" type="text" value="<?php while($row = mysql_fetch_array($result)){ echo $row['income_input']; } ?>">
  <select id="select-edit-1">
     <option value="daily" <?php while($row = mysql_fetch_array($result)){ $income_select = $row['income_select']; if ($income_select == 'daily') echo 'selected="selected"'; }?>>Daily</option>
     <option value="weekly" <?php while($row = mysql_fetch_array($result)){ $income_select = $row['income_select']; if ($income_select == 'weekly') echo 'selected="selected"'; }?>>Weekly</option>
  </select> 
</form>

<?php   
  }
  mysql_close($con);
?>

The input value (income_select) is fetched properly from the database, but the dropdown (income_select) won't return the selected state. If I include the entire form into the WHILE statement it's working fine and I can echo quickly the DB values and the select dropdown "selected" state is working fine too. I would opt for this method but I have to keep the form as it is and only populate it with data from the DB if there is any entry for the currently logged-in user.

I hope the above one makes sense and finally thank you in advance for your help!

You do not need a while if you want to fetch just one row.

First, fetch the row and get what option should be selected.

<?php 

// better use 'fetch_assoc' here
$row = mysql_fetch_assoc($result) or die(mysql_error());

$income_select = $row['income_select'];

$daily = ($income_select=='daily');
$weekly = ($income_select=='weekly');

?>

And in form, use it like this:

<option value="daily" <?php if($daily) echo 'selected'; ?>>Daily</option>
<option value="weekly" <?php if($weekly) echo 'selected'; ?>>Weekly</option>

If your PHP settings and version allow it, you can even use this:

<option value="daily" <?= $daily ? 'selected' : '' ?>>Daily</option>
<option value="weekly" <?= $weekly ? 'selected' : '' ?>>Weekly</option>

Change the code like this so it is readable and more efficient.

<?php
$income_select = "weekly"; //default value
$row = mysql_fetch_array($result);
$income_select = $row['income_select'];

echo $income_select; // echo so you can see the value and debug why it doesnt work because the code looks ok at first sight
?>

<select id="select-edit-1">
 <option value="daily" <?php if ($income_select == 'daily') echo 'selected="selected"'; }?>>Daily</option>
 <option value="weekly" <?php if ($income_select == 'weekly') echo 'selected="selected"'; }?>>Weekly</option>

You should try it like,

<select id="select-edit-1">
    <?php
       $row = mysql_fetch_array($result);
       $income_select = $row['income_select'];// will return daily or weekly
       $dailySel='';$weeklySel='';
       if ($income_select == 'daily') 
       {
          $dailySel='selected="selected"';
       }   
       else
          $weeklySel='selected="selected"';
    ?>
    <option value="daily" <?php echo $dailySel;?>Daily</option>
    <option value="weekly" <?php echo $weeklySel; ?>Weekly</option>
 </select>