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：

注意：未定义的索引：conv 注意：未定义索引：温度2

当 PHP 脚本位于另一个文件中时，一切正常。 有谁知道我做错了什么？

Answer 1

您必须确认您已发送页面并且 $_POST 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 2

在处理表单之前不会设置变量 ($type = $_POST["conv"];)。做

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

Answer 3

这是我的答案... 首先，最好验证它是否已提交？？如果调用了提交按钮，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

您的 PHP 代码将在您加载页面时每次运行，而不仅仅是当有人按下提交时。这意味着它会在外面寻找$_POST['conv'] 和$_POST['temperature2']，但由于表单尚未发布，因此什么也找不到。

您需要命名您的提交按钮，然后用 if 包围您的所有 PHP 处理，如下所示：

<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内容时查看该 PHP 代码。将 @ 放在 @$_POST['mysubmit'] 之前，这样您就不会遇到与之前在这个新数组键上遇到的相同的错误。