简体   繁体   English

显示来自php脚本的回报

[英]Displaying return from php script

I have this script: 我有这个脚本:

/** Key */
private $k = 't0kJKx27zm4ronxqXe';

/** Private salt */
private $p = 't0kez47ARpOxFnoVNJ';

private $host = 'http://url.com';

public function __construct() 
{
   ;
}


public function generateUrl($iniciator = NULL)
{
    $token = hash_hmac('sha1', date('Y-m-d') . $this->k, $this->p);

    if($iniciator !== NULL)
    {
        if(!is_string($iniciator))
        {
            throw new \Exception('Iniciator must be string!');
        }
    }

    return $this->host . 'a=' . $this->a . '&k=' . $token . '&u=' . $iniciator;

}
}

How do i echo/print... the string in return? 我如何回显/打印...返回的字符串?

return $this->host . 返回$ this-> host。 'a=' . 'a ='。 $this->a . $ this-> a。 '&k=' . '&k =' $token . $ token。 '&u=' . '&u =' $iniciator; $ initiator;

Your code is not complete it misses first line, something as: 您的代码不完整,缺少第一行,如下所示:

class Test {

Then you would run your code with: 然后,您将使用以下代码运行代码:

$someVar = new Test;
echo $someVar->generateUrl();

If you want to print from your class you can simply write echo instead of return: 如果要从您的班级打印,则可以简单地编写echo而不是return:

echo $this->host . 'a=' . $this->a . '&k=' . $token . '&u=' . $iniciator;

And then you would run your code with: 然后,您将使用以下代码运行代码:

$someVar = new Test;
$someVar->generateUrl(); // no need for echo because it's already in class

Warning: you have not defined variable $a in return. 警告:您尚未定义变量$ a作为回报。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM