仅在会话不存在时才加载session_start()吗?

问题描述

是否有一个简短的代码技巧来检查会话是否已开始，如果没有，则加载一个会话?当前，无论是否进行检查，如果我插入会话开始，我都会收到错误消息会话已开始...".

is there a short code trick to check if a session has started and if not then load one? Currently I receive an error "session already started..." if I put in a session start regardless of checking.

推荐答案

isset通常是检查当前是否定义了预定义变量的正确方法:

isset is generally the proper way to check if predefined variables are currently defined:

如果您使用的是5.4之前的php版本，

您通常可以通过下面的代码来摆脱它的诱惑，但是要注意，如果由于某种原因禁用了会话，session_start()仍然会被调用，这可能导致错误和不必要的调试，因为$ _SESSION数组不允许创建.

you can usually get away with and its tempting just doing the code below, but be aware that if for some reason sessions are disabled, session_start() will still be called which could end up leading to errors and needless debugging since the $_SESSION array isn't allowed to be created.

if(!isset($_SESSION)){session_start();}

如果您要考虑检查会话是否由于某种原因而被禁用，则可以阻止该错误消息，设置一个测试会话变量，然后验证它是否已设置.如果不是，则可以在响应中编写禁用会话的代码.您需要在脚本开始时或在脚本开始时执行所有这些操作.因此，以一个示例为例(根据错误处理设置的不同，其工作方式也不同):

if you want to account for checking to see if sessions are disabled for some reason, you can supress the error message, set a test session variable, and then verify it was set. If its not, you can code in a reaction for disabled sessions. You'll want to do all this at or near the start of your script. So as an example(works differently depending on error handling setting):

if(!isset($_SESSION)){session_start();}
$_SESSION['valid'] = 'valid';
if($_SESSION['valid'] != 'valid')
{
   //handle disabled sessions
}

但是，如果您使用的是5.4或更高版本的php ，

您可以使用 session_status()函数，这是一个更好的选择，因为它可以解决会话问题被禁用，并检查会话是否已经存在.

You can use the session_status() function, a better option as it accounts for sessions being disabled and checks if a session already exists.

if (session_status() === PHP_SESSION_NONE){session_start();}

请注意， PHP_SESSION_NONE 是PHP设置的常量，因此不需要用引号引起来.它的计算结果为整数1，但它是专门为消除幻数而设置的常数，最好针对该常数进行测试.

NOTE that PHP_SESSION_NONE is a constant set by PHP and therefore does not need to be wrapped in quotes. It evaluates to integer 1, but as its a constant that was specifically set to eliminate the need for magic numbers, best to test against the constant.

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