获取PHP选择值没有sql命令

Question

Here's my HTML code:

<form method = "get" action = "PHPFile.php">
    Type:<select name = "type">
        <option value = "1">1</option>
        <option value = "1">2</option>
        <option value = "1">3</option>
    </select>
    Your Type:<input type = "text" name = "yourType" value = "$varType;">
    Fee:<input type = "text" name = "fee" value = "$varFee;">
</form>

I want to get the value of my select "type" and pass it to my textbox "yourType".
And each type options has corresponding fees:
For Type 1 = 250
Type 2 = 90
Type 3 = 90

Here is my Php code:

<?php
    $varType = $_GET("type");
    if($varType == 1){
      $varFee = 250;
    }
    else {
      $varFee = 90;
    }
?>

But I always get an parse error on getting the value of $varType

Answer 1

<?php
$varType  = null;
if(isset($_GET['type'])) {
    $varType = $_GET['type'];
    if($varType == 1){
      $varFee = 250;
    }
    else {
      $varFee = 90;
    }
}
?>

<form method = "get" action = "PHPFile.php">
    Type:<select name = "type" ONCHANGE="if(this.value!='-1') location = this.options[this.selectedIndex].value; else return;">
        <option value = "-1">---</option>
        <option value = "?type=1">1</option>
        <option value = "?type=2">2</option>
        <option value = "?type=3">3</option>
    </select>
    Your Type:<input type = "text" name = "yourType" value = "<?php echo $varType; ?>">
    Fee:<input type = "text" name = "fee" value = "<?php echo $varFee; ?>">
</form>