if statment issues

Jason_Batten

This code below works but it requires a mysql table to have the data to list in the selection box. See how it works then view the second code which does not work.

$result = mysql_query('SELECT field FROM table');
// get list of fields to place in option selection box
while($row = mysql_fetch_array($result))
{
   $result2 = mysql_query('SELECT field2 FROM table2 WHERE id="'.$_GET['id'].'"');
// get the selected field(2) from the result with matching id
while($row2 = mysql_fetch_array($result2))
{
   // if the selected field(2) matches field in list make it selected
   if($row2['field2'] == "{$row['field']}")
   {
      $selected = 'selected';
   }else{
      $selected = '';
   }
   echo '<option '.$selected.' value="'.$row['field'].'">'.$row['field'].'</option>';
   }
}
}

Now the second code.

$result = mysql_query('SELECT field FROM table WHERE id="'.$_GET['id'].'"');
// get the selected field from the result with matching id
while($row = mysql_fetch_array($result))
{
   // if the selected field matches my option in list make it selected
   if($row['field'] == 'my_option1')
   {
      $selected = 'selected';
   }
   elseif($row['field'] == 'my_option2')
   {
      $selected = 'selected';
   }else{
      $selected = '';
   }
   echo '<option '.$selected.' value="my_option1">My Option One</option>';
   echo '<option '.$selected.' value="my_option2">My Option Two</option>';
}

To the best of my knowledge it should work but it doesn't therefore I am wrong so can someone please help me. Should I be using a switch thingo instead? 😕

I mean the first way is best for 10 results but if you only have 2 it seems too much work to have a mysql table for 2 results :bemused:

Weedpacket

Originally posted by NetNerd85

   echo '<option '.$selected.' value="my_option1">My Option One</option>';
   echo '<option '.$selected.' value="my_option2">My Option Two</option>';

[/b]

The problem here is that there is nothing to distinguish between one $selected and the other. Either both are selected if either option is matched, or neither are.

$selected1 = $selected2 = '';
if($row['field'] == 'my_option1')
{
  $selected1 = 'selected';
}
elseif($row['field'] == 'my_option2')
{
  $selected2 = 'selected';
}
echo '<option '.$selected1.' value="my_option1">My Option One</option>';
echo '<option '.$selected2.' value="my_option2">My Option Two</option>';

Jason_Batten

Oh, cheers weed 😃

I thought that the variable would be ignored because the the if statment was "wrong" but I see I was wrong :bemused:

Thanks for the help, you've solved two problems for me 😃

Jason_Batten

Alright so why does the next code work?

$id = $_GET['id'];
$query = 'SELECT * FROM table';
if($result = mysql_query($query))
{
   while($row = mysql_fetch_array($result))
   {
      if($row['property_id'] == "$id")
      {
         $var = $row['field'];
      }else{
         $var = 'not correct';
      }
      echo "$var";
   }
}
else
{
   echo '<p>Failed to connect to the database.</p>';
}

It displays results with both the result shown and the 'not correct' text.

Weedpacket

It will echo something for every $row, however many rows there are. Is this not the desired behaviour?

Oh, and do you mean $result or $result3?

Jason_Batten

Sorry, yeah I meant $result not $result3

Compared the above post, to my original posts second example. They are the same, that I can see yet they do not work the same.

The first example was used to make a chosen field selected, the above example was used to display an image if a field had an image if not it would display nothing.

:bemused: They work differently, I don't understand why.