这里我们从dropDownMenu中选择性别，然后输入身高。这个高度需要转换成英寸。由于某些原因，最终结果不会与当前代码一起显示。我相信问题可能在于我试图呼应$result变量的方式。
背景：女性的理想体重是将身高（英寸）乘以3.5，再减去108。一个人的理想体重是用他的身高（英寸）乘以4再减去128。
我需要找到一种简单的方法来显示结果，无论是回声还是显示在文本框中。有人有什么想法吗？
<?php
if(isset($_POST['submit'])) {
$gender = isset($_POST['gender']) ? $_POST['gender']: 0;
$height = (int)$_POST['height'];
switch ($gender) {
case 0:
$result = ($height * 3.5) - 108;
break;
case 1:
$result = ($height * 4) - 128;
break;
default:
$result = 0;
} echo "Ideal Weight: ". $result .'Unit';
}
?>
<html>
<div align="center">
<body>
<form name="form" method="post" action="<?php echo $PHP_SELF;?>">
Select Your Gender: <select name="gender">
<option value=""></option>
<option value="1">Male</option>
<option value="0">Female</option>
</select>
<br><br>
Enter Your Height: <input type="number" name="height" placeholder="unit inches">
<br><br>
<input type="submit" name="submit" value="Calculate Your Ideal Weight"/>
</form>
</body>
</div>
</html>
case值与选项值不相同，关闭开关后回显。尝试：
<?php
if(isset($_POST['submit'])) {
$gender = isset($_POST['gender']) ? $_POST['gender']: 0;
$height = (int)$_POST['height'];
switch ($gender) {
case 0:
$result = ($height * 3.5) - 108;
break;
case 1:
$result = ($height * 4) - 128;
break;
default:
$result = 0;
}
echo "Ideal Weight: ". $result .'Unit';
}
?>
<html>
<div align="center">
<body>
<form name="form" method="post" action="<?php echo $PHP_SELF;?>">
Select Your Gender: <select name="gender">
<option value=""></option>
<option value="1">Male</option>
<option value="0">Female</option>
</select>
<br><br>
Enter Your Height: <input type="number" name="height" placeholder="unit inches">
<br><br>
<input type="submit" name="submit" value="Calculate Your Ideal Weight"/>
</form>
</body>
</div>
</html>