Question

 <div style="text-align: left;">
    <form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?>">
    Name:<input type="text" name="name"><br>    
    Age:<input type="text" name="age">  <br>
    Job:<input type="text" name="job">  <br>
    <input type="submit" name="">
     </form>
    <?php
    $name=$_POST['name'];
    $age=$_POST['age'];
    $job=$_POST['job'];


    function rabby($name,$age,$job){

    if($age>=18){
    throw new exception('Your Name Is: $name<br>Age:$age<br> Nowadays You are working as: $job<br>Alooha!!Your are Adult :D');

    }
    return true;

    }
    try{

        rabby($name,$age,$job);
        echo "Sorry $name ,You are not not enable to see this page";
    }

    catch(exception $e){
        echo "Message:" .$e->getMessage();
    }
    ?>

在这里-------- 输出是：消息：您的姓名是：$ name 年龄：$年龄如今你正在努力：$ job Alooha !!你是成年人：D

     But i want to see--- 
    Message:  Your Name Is: rabby //when i will fillup html form
     Age:  22
     Nowadays You are working as:   student
     Alooha!!Your are Adult :D

如果我添加的年龄小于18岁，则抛出：抱歉拉比，你没有看到这个页面

请解决此问题。

Answer 1

它应该是：

throw new exception('Your Name Is: '.$name.'<br>Age:'.$age.'<br> Nowadays You are working as: '.$job.'<br>Alooha!!Your are Adult :D');

或

throw new exception("Your Name Is: $name<br>Age:$age<br> Nowadays You are working as: $job<br>Alooha!!Your are Adult :D");

在单引号中，您需要连接字符串以打印PHP变量。

Answer 2

首先，您需要了解单引号（''）和双引号（“”）之间的区别。单引号内的变量将按原样打印，而不是打印变量值。但是当它在double qoute中使用时，它将首先被执行并打印该值。

请参阅此链接以了解更多信息：What is the difference between single-quoted and double-quoted strings in PHP?。

为什么变量打印抛出新异常而不是打印变量值？

2 个答案: