PHP 自引用脚本

Question

我正在尝试使用以下代码在 HTML 表单中嵌入一个自引用 PHP 脚本：

未定义的索引：conv

<form action = "<?php $_SERVER['PHP_SELF'] ?>" method = "post">
    <input type = "number" id = "temp2" name = "temperature2" placeholder = "28">
    <label for = "temp2"> degrees </label>

    <select>
        <option name = "conv" value = "f"> Fahrenheit </option> 
        <option name = "conv" value = "c"> Celsius </option>
    </select>   

    <input type = "submit" value = "equals">

    <?php
        $type = $_POST["conv"];
        $tmp = $_POST["temperature2"];
        if ($type == "f") {
            $newTmp = (9/5 * $tmp) + 32;
            echo $newTmp . " degrees Celsius.";
        }
        elseif ($type == "c") {
            $newTmp = (5 * ($tmp - 32)) / 9;
            echo $newTmp . " degrees Fahrenheit."; 
        }
    ?>

</form>

我收到以下消息：

Notice: Undefined index: conv
Notice: Undefined index: temperature2

当 PHP 脚本在另一个文件中时，一切正常。 有谁知道我做错了什么？

Answer 1

变量（ $type = $_POST["conv"]; ）直到表单被处理后才被设置。 做

if (!empty($_POST["conv"])) {
$type = $_POST["conv"];
}

Answer 2

您必须验证您发送的页面和 $_POST 是否存在。 并更正选择元素

<form action = "<?php $_SERVER['PHP_SELF'] ?>" method = "post">
    <input type = "number" id = "temp2" name = "temperature2" placeholder = "28">
    <label for = "temp2"> degrees </label>

<select name = "conv">
    <option  value = "f"> Fahrenheit </option> 
    <option  value = "c"> Celsius </option>
</select>   

    <input type = "submit" value = "equals">

    <?php

        if(isset($_POST["temperature2"])) {        

        $type = $_POST["conv"];
        $tmp = $_POST["temperature2"];
        if ($type == "f") {
            $newTmp = (9/5 * $tmp) + 32;
            echo $newTmp . " degrees Celsius.";
        }
        elseif ($type == "c") {
            $newTmp = (5 * ($tmp - 32)) / 9;
            echo $newTmp . " degrees Fahrenheit."; 
        }
}
    ?>

</form>

Answer 3

这是我的答案......首先，最好验证一下，它是否已提交？？，如果调用提交按钮，则代码继续休息。 否则你会得到错误。 此外，在您单击提交按钮之前，不会显示结果和变量。

 <form action = "<?php $_SERVER['PHP_SELF'] ?>" method = "post"> <input type = "number" id = "temp2" name = "temperature2" placeholder = "28"> <label for = "temp2"> degrees </label> <select name = "conv"> <option value = "f"> Fahrenheit </option> <option value = "c"> Celsius </option> </select> <input type = "submit" name="submit" value = "equals"> <?php if(isset($_POST["submit"])) { $type = $_POST["conv"]; $tmp = $_POST["temperature2"]; if ($type == "f") { $newTmp = (9/5 * $tmp) + 32; echo $newTmp . " degrees Celsius."; } elseif ($type == "c") { $newTmp = (5 * ($tmp - 32)) / 9; echo $newTmp . " degrees Fahrenheit."; } } ?> </form>

Answer 4

每次加载页面时，您的 PHP 代码都会运行，而不仅仅是在有人按下提交时。 这意味着它在那里寻找$_POST['conv']和$_POST['temperature2']但没有找到任何东西，因为表单尚未发布。

你需要命名你的提交按钮，然后用这样的if包围你的所有 PHP 处理：

<input type = "submit" name="mysubmit" value = "equals">

<?php
if (@$_POST['mysubmit']) {
    $type = $_POST["conv"];
    $tmp = $_POST["temperature2"];
    if ($type == "f") {
        $newTmp = (9/5 * $tmp) + 32;
        echo $newTmp . " degrees Celsius.";
    }
    elseif ($type == "c") {
        $newTmp = (5 * ($tmp - 32)) / 9;
        echo $newTmp . " degrees Fahrenheit."; 
    }
}
?>

现在，它只会在有人实际提交某些内容时查看该 PHP 代码。 将@ 放在@$_POST['mysubmit'] 之前，这样您就不会遇到之前在这个新数组键上遇到的错误。