Question

用scrapy spider抓取一些数据后：

class Test_Spider(Spider):

    name = "test"
    def start_requests(self):
        for i in range(900,902,1):
            ........
            yield item

我使用sqlalchemy将数据传递给管道对象以写入sqllite表：

class SQLlitePipeline(object):

    def __init__(self):
        _engine = create_engine("sqlite:///data.db")
        _connection = _engine.connect()
        _metadata = MetaData()
        _stack_items = Table("table1", _metadata,
                             Column("id", Integer, primary_key=True),
                             Column("detail_url", Text),
        _metadata.create_all(_engine)
        self.connection = _connection
        self.stack_items = _stack_items

    def process_item(self, item, spider):
        is_valid = True

我希望能够将表名设置为变量，而不是像现在一样硬编码（＆＃34; table1＆＃34;）。怎么办呢？

Answer 1

假设您通过命令行传递此参数（例如-s table="table1"），请定义from_crawler方法。

@classmethod
def from_crawler(cls, crawler):
    # Here, you get whatever value was passed through the "table" parameter
    settings = crawler.settings
    table = settings.get('table')

    # Instantiate the pipeline with your table
    return cls(table)

def __init__(self, table):
    _engine = create_engine("sqlite:///data.db")
    _connection = _engine.connect()
    _metadata = MetaData()
    _stack_items = Table(table, _metadata,
                         Column("id", Integer, primary_key=True),
                         Column("detail_url", Text),
    _metadata.create_all(_engine)
    self.connection = _connection
    self.stack_items = _stack_items

Answer 2

class SQLlitePipeline(object):

    def __init__(self, table_name):

        _engine = create_engine("sqlite:///data.db")
        _connection = _engine.connect()
        _metadata = MetaData()
        _stack_items = Table(table_name, _metadata,
                             Column("id", Integer, primary_key=True),
                             Column("detail_url", Text),
        _metadata.create_all(_engine)
        self.connection = _connection
        self.stack_items = _stack_items

    @classmethod
    def from_crawler(cls, crawler):
        table_name = getattr(crawler.spider, 'table_name')
        return cls(table_name)

使用from_crawler，您可以使用指定的参数创建或实例化管道对象。

Answer 3

一种更简单的方法是在crawl上传递参数：

scrapy crawl -a table=table1

然后使用spider.table获取值：

class TestScrapyPipeline(object):
    def process_item(self, item, spider):
        table = spider.table

如何将参数传递给scrapy管道对象

3 个答案: