Python can'；我无法从scrapy获得输出

当我把

scrapy runspider divar.py -o data.json

在终端，我得到一个空文件。我做错什么了吗？我想从我放在start_URL中的URL中得到类别和子类别的结果，然后输入结果并打印出来，还得到一个json文件。主要是json文件

import scrapy 

class ws(scrapy.Spider):
    name = 'wsDivar'
    result = []

    start_urls =["https://divar.ir/s/tehran"]
    def parse(self, response):
        for category in response.xpath("//*/ul[@class='kt-accordion-item__header']"):
            x = {'cats' : category.xpath("//*/ul[@class='kt-accordion-item__header']/a").extract_first()}
            result.append(x)
            yield(x)
            print(result)
            

           

        next_L =response.xpath("//li[@class='next']/a/@href").extract_first()
        if next_L is not None:
            next_link = response.urljoin(next_L)
            yield scrapy.Request(url=next_link, callback=self.parse)

如果XPath工作正常<代码>产量还可以打印结果

与此相反：

scrapy runspider divar.py-o data.json

使用以下命令：

scrapy crawl wsDivar-o data.json

另外，在项目目录中运行该命令，该目录应该包括

scrapy.cfg

文件

import scrapy 

class ws(scrapy.Spider):
    name = 'wsDivar'
    start_urls =["https://divar.ir/s/tehran"]

    def parse(self, response):
        for category in response.xpath("//*/ul[@class='kt-accordion-item__header']"):
            x = {'cats' : category.xpath("//*/ul[@class='kt-accordion-item__header']/a").extract_first()}
            yield(x)

        next_L =response.xpath("//li[@class='next']/a/@href").extract_first()
        if next_L is not None:
            next_link = response.urljoin(next_L)
            yield scrapy.Request(url=next_link, callback=self.parse)