Question

我试图使用php获取所选单选按钮的值。但是，当我尝试回显收音机的值时，无论我在提交表单时选择了哪个单选按钮，它都会返回'on'。

HTML：

<article>
    <form id="quiz" method="post" action="question3.php">
        <fieldset id="question">
            <legend>Question 2</legend>
            <p>Where was CoffeeScripts first stable build announced? <br />
            <label for="github">GitHub</label>
            <input type="radio" name="ans2" id="github" value="GitHub"/> <br />
            <label for="hackernews">Hacker News</label>
            <input type="radio" name="ans2" id="hackernews" value="Hacker News"/> <br />
            <label for="coffeescript">CoffeeScript.com</label>
            <input type="radio" name="ans2" id="coffeescript" value="CoffeeScript"/> <br />
            <label for="dropbox">DropBox</label>
            <input type="radio" name="ans2" id="dropbox" value="Dropbox"/> <br />
        </fieldset>
        <p><input type="submit" value="Submit Answer" id="subans" /></p>
    </form>
</article>

然后在流程页面上：

<?php

if (isset($_POST['ans2'])) {
    $ans = $_POST['ans2'];
    echo "<p>$ans</p>";
} 
else {
    echo "Nothing was selected.";
}

?>

就像我说的那样，这只是向页面输出“on”。

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Answer 1

你可以尝试这个吗，在PHP块中添加了isset（$ _ POST ['subans']）

        <article>
            <form id="quiz" method="post" action="question3.php">
                <fieldset id="question">
                    <legend>Question 2</legend>
                    <p>Where was CoffeeScripts first stable build announced? <br />
                    <label for="github">GitHub</label>
                    <input type="radio" name="ans2" id="github" value="GitHub"/> <br />
                    <label for="hackernews">Hacker News</label>
                    <input type="radio" name="ans2" id="hackernews" value="Hacker News"/> <br />
                    <label for="coffeescript">CoffeeScript.com</label>
                    <input type="radio" name="ans2" id="coffeescript" value="CoffeeScript"/> <br />
                    <label for="dropbox">DropBox</label>
                    <input type="radio" name="ans2" id="dropbox" value="Dropbox"/> <br />
                </fieldset>
                <p><input type="submit" value="Submit Answer" id="subans" name="subans" /></p>
            </form>
        </article>

在php页面/部分中，添加了isset（$ _ POST ['subans']）。

        <?php 

            if(isset($_POST['subans'])){ // added the submit button action check

                    if (isset($_POST['ans2'])) {
                        $ans = $_POST['ans2'];
                        echo "<p>$ans</p>";
                    } 
                    else {
                        echo "Nothing was selected.";
                    }
            }

        ?>

在PHP中获取无线电选择不是返回值

1 个答案: