Php drop down menu. If value exists in mysql, it is the selected value in drop down menu

Question

Here is code without mysql

foreach ( $transaction_type as $i1=>$opt1 ) : 
echo '<option value="' .htmlspecialchars($opt1) .'"' .( 
(
htmlspecialchars($opt1) == htmlspecialchars($_POST['transaction_type'][$i])
)
? 'selected' : "" ) .'>'.htmlspecialchars($opt1) .'</option>';
endforeach;

Trying to modify. If in mysql certain field exists value, the value from mysql is the selected value. Else, echo the same as in code above.

Update

Here is code that at least for me is much more understandable. All works.

foreach ( $transaction_type as $i1=>$opt1 ) {
?>

<option value="<?php echo htmlspecialchars($opt1);?>"
<?php 
if (
($opt1 == $_POST['transaction_type'][$i]) OR ($opt1 == $result[VatReturnTransactionType01])
}
{
echo 'selected';
}
?> >

<?php echo htmlspecialchars($opt1); ?>

</option>

<?php
}
?>

AmazingDreams · Accepted Answer · 2013-08-03 18:33:36Z

1

Very difficult to read.. there might be an extra bracket or a bracket missing but ...

foreach ( $transaction_type as $i => $opt1 ) : 
    $option = '<option value="' .htmlspecialchars($opt1) .'"';

   if(($result[VatReturnTransactionType01] != '' AND 
       htmlspecialchars($opt1) == htmlspecialchars($result[VatReturnTransactionType01]))
       OR htmlspecialchars($opt1) == htmlspecialchars($_POST['transaction_type'][$i]))
   {
       $option .= 'selected';
   }

   $option .= '>'.htmlspecialchars($opt1) .'</option>';
endforeach;

edited Aug 3, 2013 at 18:33

answered Aug 3, 2013 at 17:57

AmazingDreams

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7 Comments

user2465936 Over a year ago

Get blank page. Yes, difficult. Possibly will try to make something simpler.

AmazingDreams Over a year ago

I missed a bracket at the end of the second line of the if statement ;)

user2465936 Over a year ago

Unfortunately can not get it to work. Checked brackets. The first after if matches the last after [$i])) Second after if matches VatReturnTransactionType01]). Removed last brackets from [VatReturnTransactionType01])) and [$i]))

AmazingDreams Over a year ago

Ok, so the syntax is correct but is the logic correct? Double check that.

user2465936 Over a year ago

Tried, but can not get to work... for my brains not understandable:)

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