Python 使用pycurl从服务器获取响应

程序如上所述，但我无法从服务器获得正确的响应。 错误是： )

com.sun.grizzly.SelectionKeyContextTask.call（SelectionKeyContextTask.java:59） com.sun.grizzly.ContextTask.run（ContextTask.java:71） com.sun.grizzly.util.AbstractThreadPool$Worker.doWork（AbstractThreadPool.java:532） com.sun.grizzly.util.AbstractThreadPool$Worker.run（AbstractThreadPool.java:513） run（Thread.java:701） . . .

这只是错误的快照

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如何修复它？

您可以尝试此代码

def curlDBPedia(DB_url):
    data = json.dumps({"text":"President Obama called Wednesday on Congress to extend a tax break for students included in last year's economic stimulus package, arguing that the policy provides more generous assistance.",
    "confidence": "0.2", "support": "20"
    })
    c = pycurl.Curl()
    c.setopt(pycurl.URL, DB_url)
    c.setopt(pycurl.POST, 1)
    c.setopt(pycurl.POSTFIELDS, data)
    c.perform()
curlDBPedia("http://spotlight.dbpedia.org/rest/annotate")

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我有点困惑；看起来您已经粘贴了Python代码，但没有粘贴java stacktrace。你能提供更多的上下文吗？更新：在将该URL发布到浏览器（我意识到它将执行GET请求）后，我看到了类似的输出。输出不是程序的堆栈跟踪，而是服务器的输出。所以请求中的某些内容需要有所不同。异常的开头是什么？这是重要的部分，而不是结尾。看起来web服务器有一些bug。

def curlDBPedia(DB_url):
        data = json.dumps({"text":"President Obama called Wednesday on Congress to extend a tax break for students included in last year's economic stimulus package, arguing that the policy provides more generous assistance.",
        "confidence": "0.2", "support": "20"
        })
        buffer = StringIO()
        c = pycurl.Curl()
        c.setopt(pycurl.URL, DB_url)
        c.setopt(pycurl.POST, 1)
        c.setopt(pycurl.POSTFIELDS, data)
        c.setopt(c.WRITEFUNCTION, buffer.write)
        c.perform()
        c.close()
        body = buffer.getvalue()#here we got the response data
    curlDBPedia("http://spotlight.dbpedia.org/rest/annotate")