因此,我正在构建一个简单的程序,它要求用户输入一个数字,然后将其张贴并生成几个随机的相似数字。
然后,程序要求用户选择正确的变量,然后由程序进行验证。
我遇到了一些问题,因为在if/else
语句中没有更好的词,所以我认为变量是不可用的。
我觉得我犯了一个非常简单/愚蠢的错误。
<?php
$numb = $_GET["number"];
switch ($numb) {
case 1:
echo "1x<br>";
$ans = 1; $n1 = rand(($ans - 5), ($ans + 5)); $n2 = rand(($ans - 5), ($ans + 5)); $n3 = 1; $n4 = rand(($ans - 5), ($ans + 5));
break;
case 2:
echo "2";
break;
case 3:
echo "3";
break;
} echo $ans;
if(isset($_POST['submit']))
{
$rb = $_POST['radio'];
if($rb == $ans){echo "test";}
else{echo "fail";}
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<input type="radio" name="radio" value="<?php echo $n1; ?>"><?php echo $n1 ?>
<input type="radio" name="radio" value="<?php echo $n2; ?>"><?php echo $n2 ?>
<input type="radio" name="radio" value="<?php echo $n3; ?>"><?php echo $n3 ?>
<input type="radio" name="radio" value="<?php echo $n4; ?>"><?php echo $n4 ?>
<input type="submit" name="submit" value="submit" />
</form>
发布于 2017-09-29 10:05:43
在switch/case中使用变量之前,必须先定义变量。
// Check if get is used
if (isset($_GET['number'])){
$numb = $_GET["number"];
// predefine Variables here
$ans = '';
$n1 = '';
$n2 = '';
$n3 = '';
$n4 = '';
switch ($numb)
{
case 1:
echo "1x<br>";
$ans = 1;
$n1 = rand(($ans - 5), ($ans + 5));
$n2 = rand(($ans - 5), ($ans + 5));
$n3 = 1;
$n4 = rand(($ans - 5), ($ans + 5));
break;
case 2:
echo "2";
break;
case 3:
echo "3";
break;
}
echo $ans;
if (isset($_POST['submit']))
{
$rb = $_POST['radio'];
if ($rb == $ans)
{
echo "test";
}
else
{
echo "fail";
}
}
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<input type="radio" name="radio" value="<?php echo $n1; ?>"><?php echo $n1 ?>
<input type="radio" name="radio" value="<?php echo $n2; ?>"><?php echo $n2 ?>
<input type="radio" name="radio" value="<?php echo $n3; ?>"><?php echo $n3 ?>
<input type="radio" name="radio" value="<?php echo $n4; ?>"><?php echo $n4 ?>
<input type="submit" name="submit" value="submit"/>
</form>
<?php
} else {
echo "Number not set in get!";
}
https://stackoverflow.com/questions/46480672
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