Php 在预期输出目录下的目录中推进构建模型
我在根目录中的spreep.yml中有以下设置:Php 在预期输出目录下的目录中推进构建模型,php,propel,Php,Propel,我在根目录中的spreep.yml中有以下设置: propel: paths: # The directory where Propel expects to find your `schema.xml` file. schemaDir: ./gateway/Propel/propelschema # The directory where Propel should output generated object model classes. phpDir
propel:
paths:
# The directory where Propel expects to find your `schema.xml` file.
schemaDir: ./gateway/Propel/propelschema
# The directory where Propel should output generated object model classes.
phpDir: ./gateway/Propel/Model
# The directory where Propel should output the compiled runtime configuration.
phpConfDir: ./gateway/Propel
# The directory where Propel should output the generated migrations.
migrationDir: ./gateway/Propel/Migrations
# The directory where Propel should output the generated DDL (or data insert statements, etc.)
sqlDir: ./gateway
database:
connections:
...
我在gateway/prople/propelschema/schema.xml
中有以下内容:
<database name="default" defaultIdMethod="native" namespace="Propel\Model" defaultPhpNamingMethod="underscore">
当我运行./spreep model:build时,我希望模型是在/gateway/spreep/model
中构建的,但是它们是在/gateway/spreep/model/spreep/model
中构建的,但是在文件中具有正确的命名空间
我不知道这里发生了什么,所以任何事情都会很好
如果需要,可以提供更多详细信息。在schema.xml中,使用\spreep\Model而不是spreep\Model。这是一个具有多个PRs的已知错误
"autoload": {
"psr-4": {
"App\\": "app/",
"Gateway\\": "gateway/",
"Propel\\Model\\": "gateway/Propel/Model"
},
"classmap": [
"database/"
]
}