Question

这是计算器的表格

<form method='post' action='result.php' name='calc_form'>

<input type='text' name='input1' size='15'>

<select name='operation'>
<option value="plus">+</option>
<option value="minus">-</option>
<option value="multi">*</option>
<option value="invalid">**</option>
<option value="divide">/</option>
</select>  

<input type='text' name='input2' size='15'>
<input type='submit' value='go'>


</form>

这是带if ifif语句的php

<?php

define ('INVALID_INPUT', 'ERROR: invalid input');
define ('INVALID_OPERATOR', 'ERROR: invalid operator');


$result = null;

if 
(isset($_POST) and 
isset($_POST['input1']) and 
isset($_POST['input2']) and 
isset($_POST['operation']))
  {
  $input1 = $_POST['input1'];   
  $input2 = $_POST['input2'];
  $operation = $_POST['operation'];
     switch ($operation)
       {
       case 'plus':
         $result = $input1 + $input2;
         break;
       case 'minus':
         $result = $input1 - $input2;
         break;
       case 'multi':
         $result = $input1 * $input2;
         break;
       case 'divide':
         $result = $input1 / $input2;
         break;
       default: 
         $result = INVALID_OPERATOR;           
         break;
       }
    }  
elseif
(!isset($_POST) and 
isset($_POST['input1']) and 
isset($_POST['input2']) and 
isset($_POST['operation']))
{  
$result = INVALID_INPUT;
}

if ($result !== null)
{
echo <<<EOM


<h2>you calculate</h2> $input1 

<h2>and</h2>  $input2 

<h2>result is:</h2>

$result

EOM;
 }

＆GT;

我甚至尝试使用else else语句并且没有条件，但是当表单中没有添加任何内容时它不会显示INVALID INPUT。这有什么不对？

Answer 1

您可以使用else语句而不是else if语句。这样，如果您更改输入，则不必每次列出两次。

例如：

if (isset($_POST) and 
isset($_POST['input1']) and 
isset($_POST['input2']) and 
isset($_POST['operation'])) 
{
    // result = calculation
} else
{
    // result == code
}

试试这个 - 我在我的服务器上进行了测试，它确实有效！

Calculator.php中

<?php
error_reporting(E_ALL | E_STRICT);
ini_set('display_errors',1);;

define ('INVALID_INPUT', 'ERROR: invalid input');
define ('INVALID_OPERATOR', 'ERROR: invalid operator');
$input1 = $input2 = $operation = null;
if (isset($_POST['input1']) and  isset($_POST['input2']) and isset($_POST['operation']))
  {

    $input1 = $_POST['input1'];   
    $input2 = $_POST['input2'];
    $operation = $_POST['operation'];
    if(is_numeric($input1) and is_numeric($input2))
    {
       switch ($operation)
       {
       case 'plus':
         $result = $input1 + $input2;
         break;
       case 'minus':
         $result = $input1 - $input2;
         break;
       case 'multi':
         $result = $input1 * $input2;
         break;
       case 'divide':
         $result = $input1 / $input2;
         break;
       default: 
         $result = INVALID_OPERATOR;           
         break;
       }
    }
    else 
    {
        $input1 = '';
        $input2 = '';
        $operation = 'invalid';
        $result = INVALID_INPUT;
    }  
}

?>

<!DOCTYPE html>
<html>
<head>
    <meta charset="utf-8">
    <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
    <title></title>
    <meta name="description" content="">
    <meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body class="">

<form method='POST' action='calculator.php'>

<input type='text' name='input1' size='15' value="<?php echo $input1; ?>">

<select name='operation'>
<option <?php if($operation == 'plus') { echo 'selected';} ?> value="plus">+</option>
<option <?php if($operation == 'minus') { echo 'selected';} ?> value="minus">-</option>
<option <?php if($operation == 'multi') { echo 'selected';} ?> value="multi">*</option>
<option <?php if($operation == 'invalid') { echo 'selected';} ?> value="invalid">**</option>
<option <?php if($operation == 'divide') { echo 'selected';} ?> value="divide">/</option>
</select>  

<input type='text' name='input2' size='15' value="<?php echo $input2; ?>">
<span>=</span>
<input type='text' name='result' size='50' value="<?php echo $result; ?>">

<input type='submit' value='go'>


</body>
</html>

Answer 2

问题在于这段代码：

if 
(isset($_POST) and 
isset($_POST['input1']) and 
isset($_POST['input2']) and 
isset($_POST['operation']))

即使用户在输入字段中未输入任何内容，他们仍然会通过isset($_POST['input1'])，因为他们已经设置了＆＃39;一个空字符串。请尝试将其切换为此：

if 
(isset($_POST['input1']) && strlen($_POST['input1']) &&
 isset($_POST['input2']) && strlen($_POST['input2']) &&
 isset($_POST['operation']))

Answer 3

你的病情错了;这是不可能的：

!isset($_POST) and 
isset($_POST['input1']) and 
isset($_POST['input2']) and 
isset($_POST['operation'])

我怀疑你的意思

!isset($_POST) or 
!isset($_POST['input1']) or 
!isset($_POST['input2']) or 
!isset($_POST['operation'])

if elseif php简单的计算器

3 个答案: