无法使 html 表单与 PHP 一起使用

Question

我这辈子都无法让这个简单的表格发挥作用...

怎么了？

    <form action="" method="post">
    <input type="text" name="kupongkode" placeholder=" Kupongkode?"> <input
        type="submit" value="&#x2713;" id="kupongkodeKnapp">
</form>

<?php
if (isset ( $_POST ['kupongkodeKnapp'] )) {
    if ($_POST ['kupongkode'] == "TEST")
        echo "Godkjent!";

    else
        echo "Ikke godkjent";
}

?>

Answer 1

提交按钮的名称必须与您检查时相同 isset

你应该通过 this link.

Answer 2

在提交表单时，您总是希望通过 "name" 属性引用表单元素。前端 (JavaScript) 和后端（PHP 或其他）就是这种情况.

<?php
if (isset ( $_POST ['kupongkode'] )) {
if ($_POST ['kupongkode'] == "TEST")
    echo "Godkjent!";
else
    echo "Ikke godkjent";
}

Answer 3

您提交的名称属性丢失

    <form action="index.php" method="post">
        <input type="text" name="kupongkode" placeholder=" Kupongkode?">
        <input type="submit" value="&#x2713;" name="kupongkodeKnapp" id="kupongkodeKnapp">
    </form>

    <?php
    if (isset ( $_POST ['kupongkodeKnapp'] )) {
        if ($_POST ['kupongkode'] == "TEST")
            echo "Godkjent!";
        else
            echo "Ikke godkjent";
    }
    ?>

Answer 4

你弄错了，你应该用name代替id:

<form action="" method="post">
    <input type="text" name="kupongkode" placeholder=" Kupongkode?"> <input
        type="submit" value="&#x2713;" name="kupongkodeKnapp">
</form>

<?php
if (isset ( $_POST ['kupongkodeKnapp'] )) {
    if ($_POST ['kupongkode'] == "TEST")
        echo "Godkjent!";

    else
        echo "Ikke godkjent";
}

?>

Answer 5

这样试试：

<form action="" method="post">
<input type="text" name="kupongkode" placeholder=" Kupongkode?"> 
<input type="submit" value="&#x2713;" id="kupongkodeKnapp">
 </form>
 <?php
 if (isset($_POST)){
if ($_POST ['kupongkode'] == "TEST")
    echo "Godkjent!";

else
    echo "Ikke godkjent";
}

?>

无法使 html 表单与 PHP 一起使用

Can't get html form to work with PHP

html

php

forms