如何从 sqlite 查询中获取 dict？

Question

db = sqlite.connect("test.sqlite")
res = db.execute("select * from table")

通过迭代，我得到与行对应的列表。

for row in res:
    print row

我可以获取列的名称

col_name_list = [tuple[0] for tuple in res.description]

但是是否有一些函数或设置可以获取字典而不是列表？

{'col1': 'value', 'col2': 'value'}

还是我必须自己做？

Answer 1

您可以使用 row_factory，如文档中的示例所示：

import sqlite3

def dict_factory(cursor, row):
    d = {}
    for idx, col in enumerate(cursor.description):
        d[col[0]] = row[idx]
    return d

con = sqlite3.connect(":memory:")
con.row_factory = dict_factory
cur = con.cursor()
cur.execute("select 1 as a")
print cur.fetchone()["a"]

或者遵循文档中此示例之后给出的建议：

如果返回元组还不够
并且您想要基于名称的访问
列，您应该考虑设置
row_factory 到高度优化的
sqlite3.行类型。 Row 提供了两者
基于索引且不区分大小写
基于名称的列访问
几乎没有内存开销。它将
可能比你自己的更好
基于自定义词典的方法或
甚至是基于 db_row 的解决方案。

这是第二个解决方案的代码：

con = sqlite3.connect(…)
con.row_factory = sqlite3.Row   #   add this row
cursor = con.cursor()

Answer 2

我想我回答了这个问题，尽管 Adam Schmideg 和 Alex Martelli 的答案中都部分提到了这个问题。为了让和我有同样疑问的人能够轻松找到答案。

conn = sqlite3.connect(":memory:")

#This is the important part, here we are setting row_factory property of
#connection object to sqlite3.Row(sqlite3.Row is an implementation of
#row_factory)
conn.row_factory = sqlite3.Row
c = conn.cursor()
c.execute('select * from stocks')

result = c.fetchall()
#returns a list of dictionaries, each item in list(each dictionary)
#represents a row of the table

Answer 3

连接到 SQLite 后：
con = sqlite3.connect(.....) 只需运行即可：

con.row_factory = sqlite3.Row

瞧！

Answer 4

即使使用 sqlite3.Row 类，您仍然不能使用以下形式的字符串格式：

print "%(id)i - %(name)s: %(value)s" % row

为了解决这个问题，我使用一个辅助函数来获取行并转换为字典。我只在字典对象优于 Row 对象时才使用它（例如，对于字符串格式化之类的事情，其中​​ Row 对象本身也不支持字典 API）。但其他时候都使用 Row 对象。

def dict_from_row(row):
    return dict(zip(row.keys(), row))       

Answer 5

正如 @gandalf 的回答所提到的，必须使用 conn.row_factory = sqlite3.Row，但结果不是直接的字典。必须在最后一个循环中向 dict 添加额外的“转换”：

import sqlite3
conn = sqlite3.connect(":memory:")
conn.execute('create table t (a text, b text, c text)')
conn.execute('insert into t values ("aaa", "bbb", "ccc")')
conn.execute('insert into t values ("AAA", "BBB", "CCC")')
conn.row_factory = sqlite3.Row
c = conn.cursor()
c.execute('select * from t')
for r in c.fetchall():
    print(dict(r))

# {'a': 'aaa', 'b': 'bbb', 'c': 'ccc'}
# {'a': 'AAA', 'b': 'BBB', 'c': 'CCC'}

Answer 6

来自 PEP 249：

Question: 

   How can I construct a dictionary out of the tuples returned by
   .fetch*():

Answer:

   There are several existing tools available which provide
   helpers for this task. Most of them use the approach of using
   the column names defined in the cursor attribute .description
   as basis for the keys in the row dictionary.

   Note that the reason for not extending the DB API specification
   to also support dictionary return values for the .fetch*()
   methods is that this approach has several drawbacks:

   * Some databases don't support case-sensitive column names or
     auto-convert them to all lowercase or all uppercase
     characters.

   * Columns in the result set which are generated by the query
     (e.g.  using SQL functions) don't map to table column names
     and databases usually generate names for these columns in a
     very database specific way.

   As a result, accessing the columns through dictionary keys
   varies between databases and makes writing portable code
   impossible.

所以是的，你自己做吧。

Answer 7

较短的版本：

db.row_factory = lambda c, r: dict([(col[0], r[idx]) for idx, col in enumerate(c.description)])

Answer 8

我的测试中最快：

conn.row_factory = lambda c, r: dict(zip([col[0] for col in c.description], r))
c = conn.cursor()

%timeit c.execute('SELECT * FROM table').fetchall()
19.8 µs ± 1.05 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)

vs：

conn.row_factory = lambda c, r: dict([(col[0], r[idx]) for idx, col in enumerate(c.description)])
c = conn.cursor()

%timeit c.execute('SELECT * FROM table').fetchall()
19.4 µs ± 75.6 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

你决定:)

Answer 9

与前面提到的解决方案类似，但最紧凑：

db.row_factory = lambda C, R: { c[0]: R[i] for i, c in enumerate(C.description) }

Answer 10

获取查询结果

output_obj = con.execute(query)
results = output_obj.fetchall()

选项 1）带 Zip 的显式循环

for row in results:
    col_names = [tup[0] for tup in output_obj.description]
    row_values = [i for i in row]
    row_as_dict = dict(zip(col_names,row_values))

选项 2）带 Dict Comp 的更快循环

for row in results:
    row_as_dict = {output_obj.description[i][0]:row[i] for i in range(len(row))}

Answer 11

我认为你走在正确的道路上。让我们保持这个非常简单并完成您想要做的事情：

import sqlite3
db = sqlite3.connect("test.sqlite3")
cur = db.cursor()
res = cur.execute("select * from table").fetchall()
data = dict(zip([c[0] for c in cur.description], res[0]))

print(data)

缺点是 .fetchall()，如果您的表非常多，它会严重消耗您的内存大的。但对于仅处理几千行文本和数字列的琐碎应用程序，这种简单的方法就足够了。

对于严肃的事情，您应该研究行工厂，正如许多其他答案中所建议的那样。

Answer 12

或者您可以将 sqlite3.Rows 转换为字典，如下所示。这将给出一个字典，其中每行都有一个列表。

    def from_sqlite_Row_to_dict(list_with_rows):
    ''' Turn a list with sqlite3.Row objects into a dictionary'''
    d ={} # the dictionary to be filled with the row data and to be returned

    for i, row in enumerate(list_with_rows): # iterate throw the sqlite3.Row objects            
        l = [] # for each Row use a separate list
        for col in range(0, len(row)): # copy over the row date (ie. column data) to a list
            l.append(row[col])
        d[i] = l # add the list to the dictionary   
    return d

Answer 13

一个通用的替代方案，仅使用三行

def select_column_and_value(db, sql, parameters=()):
    execute = db.execute(sql, parameters)
    fetch = execute.fetchone()
    return {k[0]: v for k, v in list(zip(execute.description, fetch))}

con = sqlite3.connect('/mydatabase.db')
c = con.cursor()
print(select_column_and_value(c, 'SELECT * FROM things WHERE id=?', (id,)))

，但如果您的查询不返回任何内容，将导致错误。在这种情况下...

def select_column_and_value(self, sql, parameters=()):
    execute = self.execute(sql, parameters)
    fetch = execute.fetchone()

    if fetch is None:
        return {k[0]: None for k in execute.description}

    return {k[0]: v for k, v in list(zip(execute.description, fetch))}

或者

def select_column_and_value(self, sql, parameters=()):
    execute = self.execute(sql, parameters)
    fetch = execute.fetchone()

    if fetch is None:
        return {}

    return {k[0]: v for k, v in list(zip(execute.description, fetch))}

Answer 14

import sqlite3

db = sqlite3.connect('mydatabase.db')
cursor = db.execute('SELECT * FROM students ORDER BY CREATE_AT')
studentList = cursor.fetchall()

columnNames = list(map(lambda x: x[0], cursor.description)) #students table column names list
studentsAssoc = {} #Assoc format is dictionary similarly


#THIS IS ASSOC PROCESS
for lineNumber, student in enumerate(studentList):
    studentsAssoc[lineNumber] = {}

    for columnNumber, value in enumerate(student):
        studentsAssoc[lineNumber][columnNames[columnNumber]] = value


print(studentsAssoc)

结果肯定是真的，但我不知道最好的。

Answer 15

python 中的字典提供对其元素的任意访问。
因此，任何带有“名称”的字典虽然一方面可能提供信息（又名字段名称是什么），但会“取消排序”字段，这可能是不需要的。

最好的方法是将名称放在单独的列表中，然后根据需要自行将它们与结果组合起来。

try:
         mycursor = self.memconn.cursor()
         mycursor.execute('''SELECT * FROM maintbl;''')
         #first get the names, because they will be lost after retrieval of rows
         names = list(map(lambda x: x[0], mycursor.description))
         manyrows = mycursor.fetchall()

         return manyrows, names

另请记住，在所有方法中，名称都是您在查询中提供的名称，而不是数据库中的名称。例外是 SELECT * FROM

如果您唯一关心的是使用字典获取结果，那么一定要使用 conn.row_factory = sqlite3.Row （已在另一个答案中说明） ）。

Answer 16

def getUsers(self,assoc=False):
        result = self.cursor.execute("SELECT * FROM users").fetchall()
        result_len = len(result)
        if(result_len == False): return
        if(assoc != True):
            return result
        else:
            result_formated = []
            columns = [column[0] for column in self.cursor.description]
            for row in result:
                row_dict = {}
                i = 0
                # print(result_len)
                while(i <= result_len):
                    row_dict[columns[i]] = row[i]
                    i += 1
                result_formated.append(row_dict)
            return result_formated

我将把我的错误代码留在这里