获得PHP选择值没有SQL命令

-2

<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>

我想我的选择“类型”的值，并将其传递给我的文本框“yourType”。
而且每种类型的选项具有相应的费用：
对于类型1 = 250
类型2 = 90
类型3 = 90

这是我的PHP代码：

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

但我总获取$ varType的值时出现解析错误

来源

2017-02-04 R.Joy

也许是时候阅读手册了？访问数组是通过'[]' –

完成的。使用GET的正确方法是$ _GET ['type']。你不应该使用（）。 –

<?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>

来源

2017-02-04 11:24:09

获得PHP选择值没有SQL命令

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