Question

PHP Level = beginner

我正在尝试编写一个显示特定次数的简单程序放入输入框时的值。我试图使用post方法来执行此操作，但每次程序运行并选择提交按钮时，它会显示php文件'timescalc.php'的完整代码。我想知道我做错了什么，虽然我知道 if语句的计算可能是错误的。

继承代码

<DOCTYPE html>
<html>
<head>
<title>Number Times</title>
</head>

<body>

 <h1>Number Times Table Calculator</h1>

<form method="post" action="timescalc.php">
 Enter Number : <input type="text" name="number"> <br>
 <input type="submit" value="submit"/>
</form>

<?php

$number = $_POST['number'];

if ($number == 2, $number ++2)
{
echo $number . ;
}

else if ($number == 3, $number ++3)
{
echo $number . ;
}

else if ($number == 4, $number ++4)
{
echo $number . ;
}

else if ($number == 5, $number ++5)
{
echo $number . ;
}

else 
{   
echo "pick numbers from 2  to 5 only" ;
}

?>

</body>

</html>

Answer 1

您的php代码中存在大量语法错误，但在此处进行了更正，您还需要像WAMP SERVER这样的网络服务器来运行php代码。祝你好运！

<DOCTYPE html>
<html>
<head>
<title>Number Times</title>
</head>
<body>

 <h1>Number Times Table Calculator</h1>

<form method="post" action="test.php">
Enter Number : <input type="text" name="number"> <br>
<input type="submit" value="submit"/>
</form>

<?php

$number = $_POST['number'];

if ($number == 2)
{
$number = $number * 2;
echo $number ."." ;
}

else if ($number == 3)
{
$number = $number * 3;
echo $number ."." ;
}

else if ($number == 4)
{
$number = $number * 4;
echo $number ."." ;
}

else if ($number == 5)
{
$number = $number * 5
echo $number ."." ;
}

else 
{   
echo "pick numbers from 2  to 5 only" ;
}

?>

</body>
</html>

php Number Times程序

1 个答案: