Question

我正在学习php基础知识，并且在练习这项练习时遇到了困难请帮忙吗？我提前感谢您的帮助丹尼尔。

请注意，我只允许使用 if ... elseif ... else-statement

进行练习

编写一个PHP脚本，根据表单中输入的信息打印如下所示的语句。男性0-55岁：“你是一个鼎盛时期的男人！” 超过55岁的男性：“你是个聪明人！” 0-55岁的女性：“你最美丽的女人！” 超过55岁的女性：“你的年龄看起来很年轻！”。

发送信息的表单如下所示：

<form action="printinfo.php" method="get">
Choose your gender: <input type="radio" value="male" name="gender" checked>male
<input type="radio" value="female" name="gender">female
<br>
Write your age:
<select name="age">
<option value=1 selected>0-55</option>   
<option value=2>Over 55</option>
</select>
<br>
<input type="submit" value="Send">               
</form>

示例输出

You’re a man in his prime!

我的脚本是：

<?php
$male=$_GET["male"];
$female=$_GET["female"];
if($male > 55){
echo "You’re a man in his prime!";

}else if ($male <= 55){

echo "You are a wise man!";

}else if ($female <= 55){

echo "You are a damsel at her most beautiful!";

}else if ($female > 55){

echo "You look young for your age!";

?>

Answer 1

您以错误的方式访问全局GET变量：

$gender = $_GET['gender'];
$age = $_GET['age'];

if($gender == 'male' && $age == '0-55')
    echo "You’re a man in his prime!";

现在将其改为代码的其余部分。

Answer 2

<?php
    $gender = $_GET["gender"];
    $age=$_GET["age"];
    if($gender == 'male' && $age == 2)
    {
    echo "You’re a man in his prime!";

    }
    else if ($gender == 'male' && $age == 1)
    {

    echo "You are a wise man!";

    }
    else if ($gender == 'female' && $age == 1)
    {

    echo "You are a damsel at her most beautiful!";

    }
    else if ($gender  == 'female' && $age == 2)
    {

    echo "You look young for your age!";
    }
?>

HTML：

<html>
<form action="printinfo.php" method="get">
Choose your gender: <input type="radio" value="male" name="gender" checked>male
<input type="radio" value="female" name="gender">female
<br>
Write your age:
<select name="age">
<option value=1 selected>0-55</option>   
<option value=2>Over 55</option>
</select>
<br>
<input type="submit" value="Send">               
</form>
</html>

你忘了使用年龄单选按钮.. 而你是否if语句不正确使用这样。在你的If语句中你没有使用这个'人'的年龄，你可以看到，代码永远不会起作用，因为'55'和你之间没有选择。你的男性和女性很奇怪，他们没有价值。原因，

<input type="radio" value="male" name="gender"

您需要使用'name =“gender”'而不是'value =“男/女”'

Answer 3

<?php
    if(isset($_GET['gender'])){
    $gender=$_GET['gender'];
    }

    if($gender = 'male' && $_GET['age'] > 55){
    echo "You’re a man in his prime!";

    }else if ($gender = 'male' && $_GET['age'] <= 55){

    echo "You are a wise man!";

    }else if ($gender = 'female' && $_GET['age'] <= 55){

    echo "You are a damsel at her most beautiful!";

    }else if ($gender = 'female' && $_GET['age'] > 55){
    echo "You look young for your age!";
    }
    ?>

Answer 4

我希望你能对这项任务的分析有所帮助（也为我笨拙的英语道歉）。

首先，您希望在PHP脚本中获取的变量具有您为“name”标记指定的名称“input”

现在获取值时可以先检查变量if (isset($_GET['gender']))是否存在然后使用应用程序说明切换

执行所有操作

以下是示例代码：

    <?php

//new variable for age
$age_human = $_GET['age'];

// for example
//if (!isset($age_human, $_GET['gender'])) die "Empty value";

// male or female
switch($_GET['gender'])
{
    case "male": echo "male ";
             if($age_human > 55)
             {
            echo "You’re a man in his prime!";
             }
                     else if ($age_human <= 55)
             {
            echo "You are a wise man!";
                     }
                     else if ($age_human <= 55)
                     {
            echo "You are a damsel at her most beautiful!";
             }
                     else if($age_human > 55)
                     {
                        echo "You look young for your age!";
                     }
                 break;

    case "female": echo "female ";
             if($age_human > 55)
             {
            echo "You’re a man in his prime!";
             }
                     else if ($age_human <= 55)
             {
            echo "You are a wise man!";
                     }
                     else if ($age_human <= 55)
                     {
            echo "You are a damsel at her most beautiful!";
             }
                     else if($age_human > 55)
                     {
                        echo "You look young for your age!";
                     }
                     break;
}

?>

php简介

4 个答案: