输出 HTML 用户输入*使用 HTML 代码* [重复]

Question

我正在尝试输出用户输入网站的信息。这是我的页面第一部分的代码，显示所填写的信息（此代码有效）：

<form action = "comment_process.php" method = "post">
    Name: 
    <input type = "text" name = "name"/> <br>

    Do you like this page?
    <input type = "radio" name = "answer" value = "Yes"> Yes 
    <input type = "radio" name = "answer" value = "No"> No <br>

    Comment:
    <textarea rows="3" cols="50" name = "comment"/> </textarea> <br>

    Rating 
    <select name = "rating"> <br>
        <option value = "0"> 1 </option>
        <option value = "1"> 2 </option>
        <option value = "2"> 3 </option>
        <option value = "3"> 4 </option>
        <option value = "4"> 5 </option> 

    <input type = "submit" value = "go"/>
</form>

但是，处理输入数据的页面不起作用。我的代码如下：

<?php

    $v = $_POST['firstname'];
    $x = $_POST['comment'];
    $y = $_POST['answer'];
    $z = $_POST['rating'];

    echo "Hello $v! <br>"

    if ($y == "Yes") {

        echo "I am happy that you like the page :)"

    }
    else {

        echo "I am sorry that you do not like the page :("

    }

    echo "Thanks for your comment [$x]";
    echo "You rated our page as Rating $z";

?>

我正在尝试根据用户输入的数据输出屏幕右侧小文本框中的内容：

我想知道是否有人可以帮助我！

Answer 1

你有一些错误

这对我有用：

<form action = "comment_process.php" method = "post">
    Name: 
    <input type = "text" name = "name"/> <br>

    Do you like this page?
    <input type = "radio" name = "answer" value = "Yes"> Yes 
    <input type = "radio" name = "answer" value = "No"> No <br>

    Comment:
    <textarea rows="3" cols="50" name = "comment"/> </textarea> <br>

    Rating 
    <select name = "rating"> <br>
        <option value = "0"> 1 </option>
        <option value = "1"> 2 </option>
        <option value = "2"> 3 </option>
        <option value = "3"> 4 </option>
        <option value = "4"> 5 </option> 

    <input type = "submit" value = "go"/>
</form>


<?php

    $v = $_POST['firstname'];
    $x = $_POST['comment'];
    $y = $_POST['answer'];
    $z = $_POST['rating'];

    echo "Hello $v";

    if ($y == "Yes") {

        echo "I am happy that you like the page :)";

    }
    elseif ($y == "No") {

        echo "I am sorry that you do not like the page :(";

    }

    echo "Thanks for your comment [$x]";
    echo "You rated our page as Rating $z";

?>

Answer 2

检查您的输入命名。它是您的 HTML 中的 name 和您的 php 中的 firstname。 还要确保在 php 中的每一行之后使用分号，否则会出错。 echo "Hello $v! <br>";

这不是错误，但可能会令人困惑，您在下拉列表中的选项具有与用户预期 <option value = "0"> 1 </option> 不同的值。这意味着他们在评分 1 时被告知评分为 0。也意味着您的 if 将在 sorry 选项中包含评分 3 或更低。

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编辑：使用您编辑的代码，您的 php 中的第一行 echo 将由于没有分号而导致错误。