獲得PHP選擇值沒有SQL命令

Question

這裏是我的HTML代碼：

<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> 

我想我的選擇「類型」的值，並將其傳遞給我的文本框「yourType」。
而且每種類型的選項具有相應的費用：
對於類型1 = 250
類型2 = 90
類型3 = 90

這是我的PHP代碼：

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

但我總獲取$ varType的值時出現解析錯誤

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>