Question

我试图创建一个

的代码

＆＃13;

<html>
<body>
	<form method = "post"> 
		<?php
	    	    Number : <input type = "text" name = "prime">;
	    	    <input type = "submit" value ="Submit">;
	    	    $num = $_REQUEST["prime"];
	    	    $flag = 0;
	    	    for($i = 2; $i <= $num/2; $i++)
	    	    {
			if( $num % $i == 0)
			{
		    	    $flag = 1;
		            break;
		        }
	    	    }
	    	    if($flag == 0)
			echo "$num is a prime number";
	    	    else
			echo "$num is not a prime number";
		?>
	</form>
</body>
</html>

＆＃13;

每当我尝试运行它时，我都会收到错误消息 解析错误：语法错误，意外＆＃39;＆lt;＆＃;;期待文件结束第5行的C：\ xampp \ htdocs \ pc.php

任何帮助将不胜感激

Answer 1

你的代码错了。第5行没有任何说明。更正的代码：

<html>
<body>
<?php
    if($_SERVER['REQUEST_METHOD']=='post'){
        $num = $_POST["prime"];
        $flag = 0;
        for($i = 2; $i <= $num/2; $i++)
        {
            if( $num % $i == 0)
            {
                $flag = 1;
                break;
            }
        }
        if($flag == 0)
            echo $num." is a prime number";
        else
            echo $num." is not a prime number";
    }
?>
    <form method = "post" action="#">
        Number : <input type = "text" name = "prime">
        <input type = "submit" value ="Submit">
    </form>
</body>
</html>

这将检查当前的请求方法。如果是锅

Answer 2

您可以按照以下程序解决问题

 <html>
<body>
    <?php
        if(isset($_POST['prime']))
        {
            $num = $_POST["prime"];
            $flag = 0;
            for($i = 2; $i <= $num/2; $i++)
            {
                if( $num % $i == 0)
                {
                    $flag = 1;
                    break;
                }
            }
                if($flag == 0)
                    echo "$num is a prime number";
                else
                    echo "$num is not a prime number";
        }


    ?>
    <form method = "post"> 
        <input type = "text" name="prime">
        <input type = "submit" value ="Submit">

    </form>
</body>
</html>

我想写一个PHP脚本来查找给定的数字是否为素数

2 个答案: