PHP Level =初学者
我正试图写一个简单的程序，显示一个特定的数字的次数值。我尝试过使用post方法来做到这一点，但每次程序启动运行并选择提交按钮时，它都会显示php文件"timescalc.php"的整个代码。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。
下面是代码
<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>
您在php代码中有许多语法错误，但它们在这里得到纠正，您还需要像WAMP SERVER这样的web服务器来运行php代码。祝你编码好运!
<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>