我在PHP中遇到一些错误,我该怎么办


I get some error in PHP, what can I do?

所以,我用PHP做了一个计算器,它会写结果以及使用计算器的时间。

第一次我收到一些错误,但当我点击任何提交按钮时,程序都能完美地工作如何处理首次出现的错误

程序:

<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。那么什么是最好的方法呢?

这是因为第一时间变量(在php代码中)不是isset。所以它会显示错误。例如

$using_calculator=$_POST['using_calcalculator'];//第一次未设置$_POST['using_calculator']

因此php将第一次输出错误消息。提交后,这些变量将从POST中初始化。因此,为了避免这个问题,请修改php代码,如下所示

<?phpif(isset($_POST['using_calculator']){$using_calculator=$_POST['using_calcalculator'];if(isset($_POST['plus'])){$x=$_POST['number_1'];$y=$_POST['number_2'];$z=$x+$y;$using_calculator++;}否则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++;}否则if(isset($_POST['division']){$x=$_POST['number_1'];$y=$_POST['number_2'];$z=$x/$y;$using_calculator++;}}?>