PHP 在回显中提交按钮

Question

我正在尝试构建一个零食机，您可以在其中选择您的零食，获取价格，然后点击按钮支付。

喜欢：价格是 0,60 欧元 0.05（按钮）0.10（按钮）.... 如果您按下 0.05 按钮，价格将降至 0,55€

为什么我在单击按钮后没有收到 "test" 回显？

这是我的代码

<?php
if(isset($_GET['mars']))
{
  $mars = "0,60";
  echo "Bitte Zahlen Sie noch <input type=\"button\" value=\"$mars\"> Euro<br>";
  echo "<input type=\"submit\" value=\"0,05\" name=\"fcent\">";
  if(isset($_GET['fcent']))
  {
   echo "test";
  }
}

?>

Answer 1

First, there appears to be no form-tag in your code. Without a form tag, it would be a miracle that pressing that button actually submits it via PHP. In other words, you need to wrap your form elements in a <form></form> Tag.

Secondly, the nested if: if(isset($_GET['fcent'])) is unreachable because when you press the fcent button; the $_GET['mars'] is no more in scope and since your code explicitly demands to be run when $_GET['mars'] is SET, nothing would happen. The Snippet below takes this 2 Points into account and you can fine-tune it even further to meet your needs...

NOTE: You have to be sure that your URL reads something similar to this: http://localhost/index.php?mars=some-value

<?php

    $mars       = "0,60";
    $payForm    = "<form name='whatever' method='get' action=''><br>";
    $payForm   .= "Bitte Zahlen Sie noch <input type=\"button\" value=\"$mars\"> Euro<br>";
    $payForm   .= "<input type=\"submit\" value=\"0,05\" name=\"fcent\">";
    $payForm   .="</form>";
    if(isset($_GET['mars'])){
        echo $payForm;
    }
    if(isset($_GET['fcent'])){
        echo $payForm;
        echo "test";
    }

PHP 在回显中提交按钮

PHP submit button in echo

php

echo

isset