找不到输入文件 symfony [英] Could not find input file symfony

问题描述

我能够生成包.但是我在使用 propel 构建模型时遇到了问题.我尝试了 php symfony propel:build-model 来生成模型，但它抛出错误 - 无法打开输入文件:symfony.php app/console cache:clear 对我来说很好用.我是否也必须为 symfony 设置环境变量.我在window 7 32位机器上.我正在使用 symfony2.任何建议都会对我很有帮助.

I am able to generate bundle. But I face problem when building model using propel. I tried php symfony propel:build-model to generate model but it throws error - could not open input file: symfony. php app/console cache:clear works fine for me. Will I have to set environmental variable for symfony too. I am on window 7 32bit machine. I am using symfony2. Any suggestion will be very helpful to me.

推荐答案

使用 php app/console propel:build-model 而不是 php symfony propel:build-model.app/console 部分只是对 app 目录中的 console 文件的引用.该文件是 Symfony 应用程序中所有控制台命令的主要入口.

Use php app/console propel:build-model instead of php symfony propel:build-model. The app/console part is just a reference to the console file inside the app directory. That file is the main entrance for all console commands in your Symfony application.

php 部分告诉控制台使用 php 命令来执行 app/console 文件和 propel:build-模型是你执行的命令.

The php part tells the console to use the php command to execute the app/console file and propel:build-model is the command you execute.

来自评论:也不要忘记将包添加到 AppKernel::registerBundles.

From the comments: Also don't forget to add the bundle to AppKernel::registerBundles.

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