Scrapy如何增加已调度start_requests的数量？

Hey there! Let's figure out why your Scrapy spider isn't generating all 400+ requests upfront and get it scheduling all URLs efficiently. Here are the most common fixes to try:

1. 检查start_requests生成器的实现 #

First off, double-check that your start_requests method is properly yielding every URL without early termination. It’s easy to accidentally break the generator with a misplaced return or break in your loop. For example:

def start_requests(self):
    # ✅ 正确写法：遍历所有URL并逐个yield
    for url in self.url_list:
        yield scrapy.Request(url=url, callback=self.parse)
    
    # ❌ 错误写法：用return会直接终止生成器，只返回第一个请求
    # return scrapy.Request(url=self.url_list[0], callback=self.parse)

If you’ve added conditional logic inside the loop (like filtering invalid URLs), make sure it isn’t unintentionally skipping most entries in your 400-item list.

2. 调整全局并发设置 #

Scrapy’s default CONCURRENT_REQUESTS (the global maximum number of concurrent requests) is 16—exactly the number of initial requests you’re seeing. Even with per-domain delays, if your 400 URLs are from unique domains, you should be able to process multiple requests in parallel across different sites.

To let Scrapy schedule more requests upfront, increase this setting in your settings.py:

# 根据你的服务器/网络情况调整，比如设置为100
CONCURRENT_REQUESTS = 100

Leave CONCURRENT_REQUESTS_PER_DOMAIN (default 8) as-is since you’ve already set per-domain delays—this controls how many concurrent requests go to a single domain, which aligns with your rate-limiting needs.

3. 排查请求阻塞或失败 #

If the initial 15-16 requests are getting stuck (e.g., long timeouts, repeated retries, or blocked by servers), Scrapy might be waiting for them to finish before scheduling new ones. Check your logs for:

• Timeout errors (TimeoutError)
• Retry messages (look for Retrying entries)
• HTTP 403/500 responses

To fix stuck requests, adjust timeout and retry settings in settings.py:

# 缩短请求超时时间，避免长时间阻塞
DOWNLOAD_TIMEOUT = 10
# 限制重试次数，防止无限等待失败请求
RETRY_TIMES = 2

You can also add an errback to your requests to handle failures explicitly and keep the spider moving:

def start_requests(self):
    for url in self.url_list:
        yield scrapy.Request(
            url=url,
            callback=self.parse,
            errback=self.handle_error
        )

def handle_error(self, failure):
    # 记录错误并继续处理其他请求
    self.logger.error(f"Request failed: {failure.request.url} - {str(failure)}")

4. 检查中间件或扩展的干扰 #

Sometimes custom middleware or built-in extensions like RobotsMiddleware can block or delay requests. For testing, temporarily disable non-essential middleware to see if that resolves the issue:

# 在settings.py中注释掉自定义中间件
# DOWNLOADER_MIDDLEWARES = {
#    'yourproject.middlewares.YourCustomMiddleware': 543,
# }

# 临时禁用RobotsMiddleware（仅测试用）
ROBOTSTXT_OBEY = False

If disabling middleware helps, you’ll know which component is causing the problem and can adjust its logic to allow all requests through.

5. 强制调度所有请求（极端情况） #

If all else fails, you can force the spider to send all 400 requests to the scheduler at once by converting the generator to return a list of requests. Note that this loads all requests into memory immediately, which is fine for 400 entries but might not scale to thousands:

def start_requests(self):
    return [scrapy.Request(url=url, callback=self.parse) for url in self.url_list]

This bypasses incremental yielding and pushes all requests to the scheduler right away.

内容的提问来源于stack exchange，提问作者Milano