1
我在我的APIController.php
上有错误Only variables should be passed by reference
,我已经阅读了关于这个错误的所有问题,但没有解决它。异常'只有变量应该通过引用传递
我的代码:
$ip = $_SERVER['REMOTE_ADDR'];
if (array_key_exists('HTTP_X_FORWARDED_FOR', $_SERVER)) {
$ip = explode(',', $_SERVER['HTTP_X_FORWARDED_FOR']);
$ip = array_pop(end($ip));
}
error.log中:
[2016-12-06 15:43:00] production.ERROR: exception 'ErrorException' with message 'Only variables should be passed by reference' in /var/www/app/Http/Controllers/Api/ApiController.php:33
Stack trace:
#0 /var/www/app/Http/Controllers/Api/ApiController.php(33): Illuminate\Foundation\Bootstrap\HandleExceptions->handleError(2048, 'Only variables ...', '/var/www/app/Ht...', 33, Array)
谢谢。
错误'ErrorException'与消息'array_pop()期望参数1是数组,字符串给定'在/var/www/app/Http/Controllers/Api/ApiController.php:34 – Pixel
是的错误是非常正确的,你应该在array_pop()中传递数组'' –
如何修复我的代码以获取用户IP? – Pixel