IF statement inside WHILE - PHP, MySQL and HTML dropdown select

Question

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:

id_user (primary key)
income_input
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!

MightyPork · Accepted Answer · 2013-08-04 16:36:03Z

1

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>

edited Aug 4, 2013 at 16:36

answered Aug 4, 2013 at 16:27

MightyPork

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3 Comments

Adrian Over a year ago

This worked perfectly. One thing tho, I tried to use 'fetch_assoc' but if the "current logged-in user" had no previous data, the form wasn't displaying at all. I used 'mysql_fetch_array' ($row = mysql_fetch_array($result);) and now it's working properly. Still running some test but this was a real help. Thanks again!

MightyPork Over a year ago

Since you want an associative array, assoc is the right choice - but I guess you should first use mysql_num_rows to check if anything was selected at all, and only if it was, fetch it.

Adrian Over a year ago

Ohh alright, that makes sense now. Thank you for the clarification.

Elpy · Accepted Answer · 2013-08-04 16:31:11Z

0

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>

answered Aug 4, 2013 at 16:31

Elpy

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1 Comment

MightyPork Over a year ago

Your while loop is pretty useless here.

Rohan Kumar · Accepted Answer · 2013-08-04 16:31:54Z

0

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>

answered Aug 4, 2013 at 16:31

Rohan Kumar

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