if elseif php简单计算器
这是计算器的表格if elseif php简单计算器,php,forms,if-statement,calculator,Php,Forms,If Statement,Calculator,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> <
<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>
+
-
*
**
/
这是带有if-elseif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>
$result
EOM;
}
我甚至试过只使用else语句,不带条件,但当表单中没有添加任何内容时,它不会显示无效输入。这里怎么了?你的情况不对;这是不可能的:
!isset($_POST) and
isset($_POST['input1']) and
isset($_POST['input2']) and
isset($_POST['operation'])
我猜你是说
!isset($_POST) or
!isset($_POST['input1']) or
!isset($_POST['input2']) or
!isset($_POST['operation'])
你的情况不对;这是不可能的:
!isset($_POST) and
isset($_POST['input1']) and
isset($_POST['input2']) and
isset($_POST['operation'])
我猜你是说
!isset($_POST) or
!isset($_POST['input1']) or
!isset($_POST['input2']) or
!isset($_POST['operation'])
您可以使用else语句而不是else-if语句。这样,如果您更改输入,您就不必将每个输入都列出两次
例如:
试试这个-我在我的服务器上测试了这个,效果很好
计算器.php
<?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>
value=“plus”>+
value=“multi”>*
value=“divide”>/
您可以使用else语句而不是else-if语句。这样,如果您更改输入,您就不必将每个输入都列出两次
例如:
试试这个-我在我的服务器上测试了这个,效果很好
计算器.php
<?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 $result; ?>">
<input type='submit' value='go'>
</body>
</html>
value=“plus”>+
value=“multi”>*
value=“divide”>/
问题在于这段代码:
if
(isset($_POST) and
isset($_POST['input1']) and
isset($_POST['input2']) and
isset($_POST['operation']))
即使用户在输入字段中不输入任何内容,他们仍然会传递isset($\u POST['input1'])
,因为他们被“设置”为空字符串。请尝试将其切换到以下位置:
if
(isset($_POST['input1']) && strlen($_POST['input1']) &&
isset($_POST['input2']) && strlen($_POST['input2']) &&
isset($_POST['operation']))
问题在于这段代码:
if
(isset($_POST) and
isset($_POST['input1']) and
isset($_POST['input2']) and
isset($_POST['operation']))
即使用户在输入字段中不输入任何内容,他们仍然会传递isset($\u POST['input1'])
,因为他们被“设置”为空字符串。请尝试将其切换到以下位置:
if
(isset($_POST['input1']) && strlen($_POST['input1']) &&
isset($_POST['input2']) && strlen($_POST['input2']) &&
isset($_POST['operation']))
我试过这个,但是当表单为空而不是无效输入时,结果显示为零。我试过这个,但是当表单为空而不是无效输入时,结果显示为零。结果是一样的,结果显示为零而不是无效输入。这是不同的标记,但可以正常工作。只需添加$input1=$input2=$operation=$result=null;代码行中的result变量。我忘了将方法从get更改为post。结果是一样的,它显示的是零而不是无效的input。这是不同的标记,但它可以正常工作。只需添加$input1=$input2=$operation=$result=null;代码行中的结果变量。我忘了将方法从get更改为post。是的,这很好用。为了避免php注意到$input1=$input2=$result=null,这段代码中缺少了一段平静的代码;是的,这个很好用。为了避免php注意到$input1=$input2=$result=null,这段代码中缺少了一段平静的代码;