Question

所以，我用PHP编写了一个计算器，它写了结果，并且计算器使用了多少时间。

我第一次收到错误，但是当我点击任何提交按钮时，程序运行正常。 首次出错时我该怎么办？

该计划：

<html>
<body>
    <div>
        <?php
            $using_calculator = $_POST['using_calculator'];
            if(isset($_POST['plus'])){
                    $x = $_POST['number_1'];
                    $y = $_POST['number_2'];
                    $z = $x+$y;
                    $using_calculator++;
            }else if(isset($_POST['minus'])){
                    $x = $_POST['number_1'];
                    $y = $_POST['number_2'];
                    $z = $x-$y;
                    $using_calculator++;
            }else if(isset($_POST['multiplication'])){
                    $x = $_POST['number_1'];
                    $y = $_POST['number_2'];
                    $z = $x*$y;
                    $using_calculator++;
            }else if(isset($_POST['division'])){
                    $x = $_POST['number_1'];
                    $y = $_POST['number_2'];
                    $z = $x/$y;
                    $using_calculator++;
            }
        ?>
        <form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="POST">
            <p>
                First number: <input type="number" name="number_1"><br/>
                Second number: <input type="number" name="number_2"><br/><br/>
                <input type = "submit" name="plus" value="Plus">
                <input type = "submit" name="minus" value="Minus">
                <input type = "submit" name="multiplication" value="Multiplicate">
                <input type = "submit" name="division" value="Divide">
                <input type = "hidden" name="using_calculator" value="<?php echo $using_calculator; ?>">            
            </p>
        </form>
        <br>Result: <?php echo $z; ?>
        <br>The calculator was used <?php echo $using_calculator; ?> times
    </div>
</body>

我必须使用隐藏的输入字段，但我可以使用函数和对象。那么最好的方法是什么？

Answer 1

因为第一次变量（在你的php代码中）不是isset。所以它会显示错误。例如


 $using_calculator = $_POST['using_calculator']; // first time $_POST['using_calculator']  is not isset

所以php会在第一时间输出错误信息。提交后，这些变量从POST初始化。因此，要避免此问题，请修改php代码，如下所示

 <?php

    if(isset($_POST['using_calculator']))
      {
                $using_calculator = $_POST['using_calculator'];
                if(isset($_POST['plus'])){
                        $x = $_POST['number_1'];
                        $y = $_POST['number_2'];
                        $z = $x+$y;
                        $using_calculator++;
                }else if(isset($_POST['minus'])){
                        $x = $_POST['number_1'];
                        $y = $_POST['number_2'];
                        $z = $x-$y;
                        $using_calculator++;
                }else if(isset($_POST['multiplication'])){
                        $x = $_POST['number_1'];
                        $y = $_POST['number_2'];
                        $z = $x*$y;
                        $using_calculator++;
                }else if(isset($_POST['division'])){
                        $x = $_POST['number_1'];
                        $y = $_POST['number_2'];
                        $z = $x/$y;
                        $using_calculator++;
                }
        }
        ?>

我在PHP中遇到一些错误，我该怎么办？

1 个答案: