uuid_create要求通过引用传递参数。
uuid_create(&$foo);问题是,这将产生错误:
Message: Call-time pass-by-reference has been deprecatedPHP扩展uuid-php.x86_64过时了吗?它与PHP 5“不兼容”。有哪些替代方案?
只是倾向于强调这不是一个复制.
$foo = NULL;
uuid_create($foo);将产生:
Type: Run-time warnings (non-fatal error).
Message: uuid_create(): uuid_create: parameter wasn't passed by reference发布于 2012-06-02 09:00:44
PHP没有uuid_create方法,在文档中也没有提到它,所以如果它来自一个扩展,它很可能不是正式的,而且可能已经过时了。函数期望out参数而不是返回值这一事实已经是一个非常明显的迹象,表明该函数相当糟糕。
但是,编写PHP代码来生成uuid4非常容易,因为它对所有字段都使用随机值,也就是说,您不需要访问特定于系统的东西,比如MAC地址:
function uuid4() {
return sprintf('%04x%04x-%04x-%04x-%04x-%04x%04x%04x',
// 32 bits for "time_low"
mt_rand(0, 0xffff), mt_rand(0, 0xffff),
// 16 bits for "time_mid"
mt_rand(0, 0xffff),
// 16 bits for "time_hi_and_version",
// four most significant bits holds version number 4
mt_rand(0, 0x0fff) | 0x4000,
// 16 bits, 8 bits for "clk_seq_hi_res",
// 8 bits for "clk_seq_low",
// two most significant bits holds zero and one for variant DCE1.1
mt_rand(0, 0x3fff) | 0x8000,
// 48 bits for "node"
mt_rand(0, 0xffff), mt_rand(0, 0xffff), mt_rand(0, 0xffff)
);
}https://stackoverflow.com/questions/10860978
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