Question

我正在尝试使用表单和PHP并使用SWITCH案例制作温度转换器。我相信我已经接近解决方案，但是当我按下“提交”按钮时，我的页面变成空白，并且看不到echo语句。这是我的代码：

<?php

if(isset($_POST['convertTemp']))

{

switch('$convertTemp'){

    case 'FtoC':
        $newTemp = (($tEmp - 32)* (5/9));
        break;

    case 'FtoK':
       $newTemp = (($tEmp - 32)* (5/9) + 273.15);
        break;

    case 'KtoF':
       $newTemp = (($tEmp - 273.15)* (9/5) + 32);
        break;

     case 'KtoC':
       $newTemp = ($tEmp - 273.15);
        break;

    case 'CtoK':
       $newTemp = ($tEmp + 273.15);
        break;

    case 'CtoF':
       $newTemp = (($tEmp * 9/5) + 32);
        break;      
  echo " <h2 align='center'>The initial temperature was" . $tEmp . "and the converted temperature is:" . $newTemp . "/h2>";    

}
}else{
    echo'




<html>
<body>

<h1 align="center">Convert a Temperature</h1> 

    <form align="center" method="POST">

        Enter the tempurature you wish to convert:<input type="number" name="tEmp">

        <h2>Convert temperature from: </h2>
        <input type="radio" name="convertTemp" value="FtoC"> Farenheit to Celcius <br>
        <input type="radio" name="convertTemp" value="FtoK"> Farenheit to Kelvin <br>
        <input type="radio" name="convertTemp" value="KtoF"> Kelvin to Farenheit <br>
        <input type="radio" name="convertTemp" value="KtoC"> Kelvin to Celcius <br>
        <input type="radio" name="convertTemp" value="CtoK"> Celcius to Kelvin <br>
        <input type="radio" name="convertTemp" value="CtoF"> Celcius to Farenheit <br>

    <input type="submit" value="Convert Tempurature!">

    </form>



    </body>
    ';

    }

?>

Answer 1

这里有几个问题。您未看到的echo在最后一个switch之后的break语句中，需要在切换之后（即switch的右括号之后）出现声明）。但是，即使这样做，也不会设置变量（未设置$tEmp，您需要先从$_POST获取其值，与$convertTemp相同）。

<?php
  if(isset($_POST['convertTemp']) && isset($_POST['tEmp'])) {
    $convertTemp = $_POST['convertTemp'];
    $tEmp = $_POST['tEmp'];
    switch($convertTemp){
        case 'FtoC':
            $newTemp = (($tEmp - 32)* (5/9));
            break;
        case 'FtoK':
           $newTemp = (($tEmp - 32)* (5/9) + 273.15);
            break;
        case 'KtoF':
           $newTemp = (($tEmp - 273.15)* (9/5) + 32);
            break;
         case 'KtoC':
           $newTemp = ($tEmp - 273.15);
            break;
        case 'CtoK':
           $newTemp = ($tEmp + 273.15);
            break;
        case 'CtoF':
           $newTemp = (($tEmp * 9/5) + 32);
            break;      
    }
    echo "<h2 align='center'>The initial temperature was " . $tEmp . " and the converted temperature is: " . $newTemp . "</h2>";
  }
  else {
    echo'
      <html>
        <body>
          <h1 align="center">Convert a Temperature</h1>
          <form align="center" method="POST">
            Enter the tempurature you wish to convert:<input type="number" name="tEmp">

            <h2>Convert temperature from: </h2>

            <input type="radio" name="convertTemp" value="FtoC"> Farenheit to Celcius <br>
            <input type="radio" name="convertTemp" value="FtoK"> Farenheit to Kelvin <br>
            <input type="radio" name="convertTemp" value="KtoF"> Kelvin to Farenheit <br>
            <input type="radio" name="convertTemp" value="KtoC"> Kelvin to Celcius <br>
            <input type="radio" name="convertTemp" value="CtoK"> Celcius to Kelvin <br>
            <input type="radio" name="convertTemp" value="CtoF"> Celcius to Farenheit <br>

            <input type="submit" value="Convert Tempurature!">
          </form>
        </body>
      </html>
    ';
  }
?>

提交后，使用SWITCH和POST的PHP温度转换器表单空白

1 个答案: