Question

我最近开始使用PHP，但我无法弄清楚如何echo结果。任何帮助将不胜感激。

<html>
    <head>
    <style>
    input[type=submit]{
        background-color: #4CAF50;
        border: none;
        color: white;
        padding: 16px 32px;
        text-decoration: none;
        margin: 4px 2px;
        cursor: pointer;
    }
    input[type=text]{
        border: 3px solid black;
    }
    </style>
    </head>

    <body>
    <form action="fitness1.php" method="post">
    <b>weight:</b>  <input type="text" name="weight"><br>
    <br>
    <b>height:<b>   <input type="text" name="height"><br>
    <br>
    <input type="submit">
    </body>
</html>

这是PHP代码：

<html>
<body>
<?php

$WEight = $_Post['weight'];
$HEight = $_post['height'];

if (is_numeric($WEight)&& is_numeric($HEight)){

    $bmi= $WEight / $HEight * $HEight;
  }     

echo ("$bmi");


else {
    echo "please enter your weight and height";
}
?>
</body>
</html>

Answer 1

变量的情况在PHP中很重要。您正在寻找此：

$WEight = $_POST['weight'];
$HEight = $_POST['height'];

然后，这是我们将使用的逻辑：

if (is_numeric($WEight)&& is_numeric($HEight)) {
    $bmi = $WEight / $HEight * $HEight;
    echo $bmi;
} else {
    echo "please enter your weight and height";
}

您的echo需要位于if (...) { ... }区块内。

修改：

根据要求，我将如何改进它：

HTML：

<html> <head> <style> input[type=submit]{ background-color: #4CAF50; border: none; color: white; padding: 16px 32px; text-decoration: none; margin: 4px 2px; cursor: pointer; } input[type=text]{ border: 3px solid black; } </style> </head> <body> <form action="fitness1.php" method="post" /> <b>Weight:</b> <input type="text" name="weight" /><br /> <br /> <b>Height:<b> <input type="text" name="height" /><br /> <br /> <input type="submit" /> </body> </html>

PHP：

<html> <body> <?php $weightInKg = (int) $_POST['weight']; $heightInCm = (int) $_POST['height']; if ($heightInCm && $weightInKg) { $bmi = $weightInKg / pow($heightInCm, 2); echo "Your BMI is ${bmi}."; } else { echo "Please enter your height and weight as a number greater than 0."; } ?> </body> </html>

Answer 2

错误消息的主要问题是：

echo("$bmi"); // this is wrong you don't need brackets.

用

替换它

echo $bmi;

这是错误的原因但是如果建议将echo $ bmi移入内部，如果条件或初始化$ bmi，否则如果条件没有通过则可以得到未定义的变量错误。

尝试根据用户输入输出计算结果时出错

2 个答案: