Python (json) : TypeError: expected string or buffer

Question

我想从我的数据库中读取数据，然后创建 return 一个 json 格式到我的 python 网络服务器 (web.py)。但是下面的代码没有不工作并给我这个错误：

类型错误：预期的字符串或缓冲区

在负载和负载之间切换也转储和转储（我不明白为什么）。如您所知，我对 python 和 json 格式都很陌生，如果你帮帮我（看到很多关于这个问题的 post 但仍然不知道该怎么做）

def ara_json(str):
    web.header('Content-Type','application/json; charset=utf-8', unique=True) 
    cnx = mysql.connector.connect(user='arda', password='1', database='worddb')
    cursor = cnx.cursor()
    sqlq = "SELECT * FROM names WHERE name = '%s'" %str
    cursor.execute(sqlq)
    rows = cursor.fetchall()

    result=[]
    for row in rows:
            d = dict()
            d['name'] = row[0]
            d['type'] = row[1]
                result.append(d)
            subjects = json.loads(result).read()
        return json.dump(subjects , indent=4)


class json_isimbul:
    def GET(self,isim):
    web.header('Content-Type','application/json; charset=utf-8', unique=True)    
    isim = isim.lower()
    return ara_json(isim)

---

Traceback (most recent call last):
  File "/usr/local/lib/python2.7/dist-packages/web.py-0.37-py2.7.egg/web/application.py", line 239, in process
    return self.handle()
  File "/usr/local/lib/python2.7/dist-packages/web.py-0.37-py2.7.egg/web/application.py", line 230, in handle
    return self._delegate(fn, self.fvars, args)
  File "/usr/local/lib/python2.7/dist-packages/web.py-0.37-py2.7.egg/web/application.py", line 420, in _delegate
    return handle_class(cls)
  File "/usr/local/lib/python2.7/dist-packages/web.py-0.37-py2.7.egg/web/application.py", line 396, in handle_class
    return tocall(*args)
  File "/home/arda/Downloads/arda/tmp/tornado/ps/sa2.py", line 98, in GET
    return ara_json(isim)
  File "/home/arda/Downloads/arda/tmp/tornado/ps/sa2.py", line 53, in ara_json
    subjects = json.loads(result).read()
  File "/usr/lib/python2.7/json/__init__.py", line 338, in loads
    return _default_decoder.decode(s)
  File "/usr/lib/python2.7/json/decoder.py", line 366, in decode
    obj, end = self.raw_decode(s, idx=_w(s, 0).end())
TypeError: expected string or buffer

Answer 1

json.loads() 需要一个字符串。正确的方法是首先构建数据并使用 json.dumps ，这将为您提供一个 json 对象（实际上是一个字符串）。希望这有帮助。

Answer 2

你必须改两行就可以了。

1. 删除这一行：subjects = json.loads(result).read()
2. 然后这样做：return json.dumps(result , indent=4)

希望对您有所帮助。 :)