symfony、var-folder 作为符号链接由于错误的供应商包含而无法在 apache 上工作

问题描述 投票:0回答:1

我有一个空的symfony,通过composer.json安装,包含apache-pack和api-platform(参见下面的文件)。

环境

我们需要提供 var 文件夹作为符号链接。

  • 应用程序文件夹:/app
  • Var-文件夹:/contents/var

应用程序文件夹结构导致:

[...] 
templates 
var -> /contents/var 
vendor

问题

Symfony 在 var/cache 中创建类,其中包括如下所示的供应商类:

include_once \dirname(__DIR__, 4).'/vendor/symfony/kernel/ErrorHandler.php';

问题是, __ DIR __ 返回符号链接的源路径,而不是目标应用程序路径,因此找不到该类:

警告:include_once(/contents/vendor/symfony/kernel/ErrorHandler.php):无法打开流:没有这样的文件或目录

所以它说供应商在 /contents/vendor 而不是 /应用程序/供应商

如果 DIR 返回应用程序路径,一切都会好起来的: /app/vendor/symfony/kernel/ErrorHandler.php

问题

有人知道如何解决这个问题吗?由于我们无法更改 gitlab-setup,因此无法更改环境。

composer.json

{
    "type": "project",
    "license": "proprietary",
    "minimum-stability": "stable",
    "prefer-stable": true,
    "require": {
        "php": ">=8.1",
        "ext-ctype": "*",
        "ext-iconv": "*",
        "api-platform/core": "^3.2",
        "doctrine/doctrine-bundle": "^2.10",
        "doctrine/doctrine-migrations-bundle": "^3.2",
        "doctrine/orm": "^2.16",
        "nelmio/cors-bundle": "^2.3",
        "phpdocumentor/reflection-docblock": "^5.3",
        "phpstan/phpdoc-parser": "^1.24",
        "symfony/apache-pack": "^1.0",
        "symfony/asset": "6.3.*",
        "symfony/console": "6.3.*",
        "symfony/dotenv": "6.3.*",
        "symfony/expression-language": "6.3.*",
        "symfony/flex": "^2",
        "symfony/framework-bundle": "6.3.*",
        "symfony/property-access": "6.3.*",
        "symfony/property-info": "6.3.*",
        "symfony/runtime": "6.3.*",
        "symfony/security-bundle": "6.3.*",
        "symfony/serializer": "6.3.*",
        "symfony/twig-bundle": "6.3.*",
        "symfony/validator": "6.3.*",
        "symfony/yaml": "6.3.*"
    },
    "require-dev": {
        "symfony/maker-bundle": "^1.51"
    },
    "config": {
        "allow-plugins": {
            "php-http/discovery": true,
            "symfony/flex": true,
            "symfony/runtime": true
        },
        "sort-packages": true
    },
    "autoload": {
        "psr-4": {
            "App\\": "src/"
        }
    },
    "autoload-dev": {
        "psr-4": {
            "App\\Tests\\": "tests/"
        }
    },
    "replace": {
        "symfony/polyfill-ctype": "*",
        "symfony/polyfill-iconv": "*",
        "symfony/polyfill-php72": "*",
        "symfony/polyfill-php73": "*",
        "symfony/polyfill-php74": "*",
        "symfony/polyfill-php80": "*",
        "symfony/polyfill-php81": "*"
    },
    "scripts": {
        "auto-scripts": {
            "cache:clear": "symfony-cmd",
            "assets:install %PUBLIC_DIR%": "symfony-cmd"
        },
        "post-install-cmd": [
            "@auto-scripts"
        ],
        "post-update-cmd": [
            "@auto-scripts"
        ]
    },
    "conflict": {
        "symfony/symfony": "*"
    },
    "extra": {
        "symfony": {
            "allow-contrib": "true",
            "require": "6.3.*"
        }
    }
}
php symfony persistence environment symlink
1个回答
0
投票

好吧,在这种情况下,它有点太神奇了。 设置 server-var APP_CACHE_DIR 可以解决问题,但是:

如果应用程序根缓存是例如 /var/www/api/var/cache

APP_CACHE_DIR = /var/contents/api/var/cache -> 错误

不起作用,因为最后三个文件夹“api/var/cache”是相同的。 如果您将缓存重命名为 appcache 或其他名称,它就可以工作。

APP_CACHE_DIR = /var/contents/api/var/appcache ->成功

© www.soinside.com 2019 - 2024. All rights reserved.