Question

这是我编写的代码，但它无效。它假设向用户询问一个奇数并检查它是否正确。

代码：

$guess = $_POST['guess'];
$submit = $_POST['submit'];

if(isset($submit)){
        if ($guess/2 %0){
                echo "the number you are guessing is not odd";
        }elseif{

         ($guess<$rand)
                echo "Your Guess is Smaller than the secret number";
        }elseif{
                ($guess>$rand)
                         echo "Your Guess is bigger than the secret number";
        }else{
                          ($guess==$rand)
        echo "you guessed currectly. now try anouthe number!";}





else
        header("Location: index.php");
                exit();}

?>

Answer 1

你能试试吗？

您放置了＆＃39;（）＆＃39;在你的elseif错了。

<?php
$rand = rand(1, 99);

$guess  = $_POST['guess'];
$submit = $_POST['submit'];

if(isset($submit))
{
    if($guess / 2 % 0)
    {
        echo "the number you are guessing is not odd";
    }
    elseif($guess < $rand)
    {
        echo "Your Guess is Smaller than the secret number";
    }
    elseif($guess > $rand)
    {
        echo "Your Guess is bigger than the secret number";
    }
    elseif($guess == $rand)
    {
        echo "you guessed currectly. now try anouthe number!";
    }
}
else
{
    header("Location: index.php");
    exit();
}

?>

我尚未测试此代码，因此我需要您的反馈。 修改：您已确认此操作有效。

我想向您提供有关elseif的手册： http://php.net/manual/en/control-structures.elseif.php

请考虑更简单/更清晰的编码。就个人而言，我喜欢使用＆＃39;：＆＃39;而不是＆＃39; {}＆＃39;，当您使用与HTML混合的PHP时，代码更少，更容易阅读：

<?php
$rand = rand(1, 99);

$guess  = $_POST['guess'];
$submit = $_POST['submit'];

if(isset($submit)):
    if($guess / 2 % 0):
        echo "the number you are guessing is not odd";
    elseif($guess < $rand):
        echo "Your Guess is Smaller than the secret number";
    elseif($guess > $rand):
        echo "Your Guess is bigger than the secret number";
    elseif($guess == $rand):
        echo "you guessed currectly. now try anouthe number!";
else:
    header("Location: index.php");
    exit();
endif;
?>

不要忘记查看$_POST数据。

同样适用于array，但这是旁注：

$arr = array(1 => 'hi', 2 => 'hello'); // old
$arr = [1 => 'hi', 2 => 'hello']; // new

Answer 2

这不是php中if-else构造的正确语法。

elseif部分需要在它之后（在开始大括号之前）有一个条件，而else根本不期待一个条件。

if ($guess/2 %0){
        echo "the number you are guessing is not odd";
} elseif ($guess<$rand) {
    // ....
} else {
    echo "you guessed currectly. now try anouthe number!";
}

当然，在你的其他人之前，你必须确保if和elseif匹配所有＆＃34;错误＆＃34;例。

PHP语法错误 - 使用elseif

2 个答案: