Question

我有一个html表单来从用户那里获取一个数字进行计算。给出第一个数字。第二个是由用户输入的。结果发布到同一页面。但结果答案现在是第一个数字，用户又输入第二个数字。第一个代码适用于第一次计算。但是下一个仍然使用给定的数字，而不是结果。我相信我应该使用if / else来测试是否输入了提交。但第一个答案是没有通过下一轮传递$ first = $ total（我得到未定义的变量总数）。

没有else语句：

 <head> 
 <title>Calculate</title> 
 </head> 
 <body> 

 <?php 
 $first = 2;

 if(isset($_POST['Calculate'])) 
     {
      $second = (!empty($_POST['second']) ? $_POST['second'] : null);
      $total = $first+$second; 
      print "<h2>Results</h2>"; 
      print "The total of the two numbers: $first + $second = $total <p>"; 
      $first=$total;
     }

 ?> 



 <h2>Calculate</h2> 
 <p><?php echo "First Number: $first"; ?></p> 
 <br>
 <form action = "index.php" method = "POST"> 
  Second Number: <input type = "text" name = "second"><br>
 <input type = "submit" name = "Calculate"/> 
 </form> 
 </body> 
 </html>

使用else语句：

 <head> 
 <title>Calculate</title> 
 </head> 
 <body> 

 <?php 


 if(isset($_POST['Calculate'])) 
      {
      $first=$total;
      $second = (!empty($_POST['second']) ? $_POST['second'] : null);
      $total = $first+$second; 
      print "<h2>Results</h2>"; 
      print "The total of the two numbers: $first + $second = $total <p>"; 
      }
 else {
       $first = 2;
       $second = (!empty($_POST['second']) ? $_POST['second'] : null);
       $total = $first+$second;
       }
 ?> 



 <h2>Calculate</h2> 
 <p><?php echo "First Number: $first"; ?></p> 
 <br>
 <form action = "index.php" method = "POST"> 
  Second Number: <input type = "text" name = "second"><br>
 <input type = "submit" name = "Calculate"/> 
 </form> 
 </body> 
 </html>

我将表单更改为

<form action = "index.php" method = "POST"> 
 <input type="hidden" name="first" value="$first" />

  Second Number: <input type = "text" name = "second"><br>
 <input type = "submit" name = "Calculate"/> 
 </form> 
 </body> 
 </html>

但得到了相同的结果。

Answer 1

你永远不会费心传递这两个值，只显示结果然后提示输入新值。您需要将“previous”值嵌入为隐藏的表单字段：

<form ...>
   <input type="hidden" name="first" value="$first" />
   <input type="text" name="second" />

Answer 2

首先，您不需要在条件内设置$second和$total的值，如果两者都相同的话。其次，您将三元组赋值括在括号中，以便将值设置为布尔值，而不是$_POST数组的值。所以你有：

$second = (!empty($_POST['second']) ? $_POST['second'] : null);

应该是：

$second = !empty($_POST['second']) ? $_POST['second'] : null;

这是一种更清洁的方法：

$first = isset($_POST['Calculate']) ? $total : 2;
$second = !empty($_POST['second']) ? $_POST['second'] : 0;
$total = $first + $second;

if(isset($_POST['Calculate'])) 
{
      print "<h2>Results</h2>"; 
      print "The total of the two numbers: $first + $second = $total <p>"; 
}

使用先前计算结果的PHP表单计算

2 个答案: