Warning: file_get_contents(/data/phpspider/zhask/data//catemap/0/laravel/11.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Php 如何在使用include时将字符串转换为laravel刀片中的参数_Php_Laravel_Laravel Blade - Fatal编程技术网

Php 如何在使用include时将字符串转换为laravel刀片中的参数

Php 如何在使用include时将字符串转换为laravel刀片中的参数,php,laravel,laravel-blade,Php,Laravel,Laravel Blade,我正在尝试转换刀片服务器上的字符串,该字符串是在include上发送的参数: @include('pages.groups.adventureCard', ['title' => 'GREEN WEEKEND','primaryPrice' => '225,00 €']) 我知道我可以在我的adventureCard.blade.php上使用@lang($title),但这不允许我在使用插件(kkomelin/laravel Translateable string exporte

我正在尝试转换刀片服务器上的字符串,该字符串是在include上发送的参数:

@include('pages.groups.adventureCard', ['title' => 'GREEN WEEKEND','primaryPrice' => '225,00 €'])
我知道我可以在我的
adventureCard.blade.php
上使用
@lang($title)
,但这不允许我在使用插件
(kkomelin/laravel Translateable string exporter)
时跟踪所有未翻译的字符串,该插件获取
@lang('***')
之间的内容,但忽略变量

我试过
trans()
@lang()
\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

如果视图未自动刷新,请确保在更改它们时清除视图缓存

php artisan view:clear
希望这有帮助

\uuuu()
应该可以工作

如果视图未自动刷新,请确保在更改它们时清除视图缓存

php artisan view:clear

希望这有帮助

\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu__()应该可以工作,请确保清除您的视图,然后查看哇谢谢,我只需要清除缓存,它就可以工作了!