保留php if语句中的select选项

Question

I am attempting to create a single PHP page that opens a salesman or customer window based off a hidden input field in a form being posted. The tricky part for me is I also need two select option dropdowns that appear based on the same hidden input field being posted but ALSO retain their value after submit.

I know how to retain the values of a form like so..

<form action="" method="post">
    <select name="newyear">
        <option <?php if ($sqlyear == '2015cust') { ?>selected="true" <?php }; ?>value="2015cust">2015</option>
        <option <?php if ($sqlyear == '2016cust') { ?>selected="true" <?php }; ?>value="2016cust">2016</option>
    </select>
</form>

But when I try and implement this method inside my PHP if statements it does not retain the selected option.

if ($open == 'salesman_window'){
    echo "This is salesman dropdown menu";
    echo "$sqlyear";
    echo "<form action='' method='post'>";
    echo "<input type='hidden' name='newyear_trigger' value=''>";
    echo "<input type='hidden' name='window' value='salesman_window'>";
    echo "<select name='newyear'>";
    echo "<option if ($sqlyear == '2015users') { selected='true'  };  value='2015users'> 2015</option>";
    echo "<option if ($sqlyear == '2016users') { selected='true'  };  value='2016users'>2016</option>";
    echo "</select>";
    echo "<button class='documentation_button'>Submit</button>";
    echo "</form>";

}

if ($open == 'customer_window'){
    echo "this is customer dropdown menu";
    echo "$sqlyear";
    echo "<form action='' method='post'>";
    echo "<input type='hidden' name='newyear_trigger' value=''>";
    echo "<input type='hidden' name='window' value='customer_window'>";
    echo "<select name='newyear'>";
    echo "<option if ($sqlyear == '2015cust') { selected='true'  }; value='2015cust'>2015</option>";
    echo "<option if ($sqlyear == '2016cust') { selected='true'  }; value='2016cust'>2016</option>";
    echo "</select>";
    echo "<button class='documentation_button'>Submit</button>";
    echo "</form>";
}
?>

I have tried breaking up the option with multiple echo ""; and tried a few ".." of these in there as well but either get syntax errors or it just doesn't retain the selection. Everything works in regards to showing the correct dropdown menu based on what is posted but I just cannot get the select option to retain. The $sqlyear is for sure getting the correct value each time the dropdown option is selected so I know it isn't that either. Can anyone help?

Answer 1

A couple of options:

1. Use a ternary operator in your echo:

   echo "<option" . (($sqlyear == '2015users') ? " selected='true'" : '') 
              . " value='2015users'>2015</option>";
   echo "<option" . (($sqlyear == '2016users') ? " selected='true'" : '') 
              . " value='2016users'>2016</option>";
   

---

1. Keep using the if, but without opening/closing PHP tags as much:

   <option <?php if ($sqlyear == '2015users') echo " selected='true'"; ?> value='2015users'>
       2015</option>
   <option <?php if ($sqlyear == '2016users') echo " selected='true'"; ?> value='2016users'>
       2016</option>