我有一个很长的疑问。我想用Python将它分成几行。在JavaScript中实现这一点的一种方法是使用几个句子并用+运算符将它们连接起来(我知道,也许这不是最有效的方法,但我并不真正关心这个阶段的性能,只是代码的可读性)。例子:

var long_string = 'some text not important. just garbage to' +
                      'illustrate my example';

我尝试在Python中做类似的事情,但没有成功,所以我使用了\来拆分长字符串。然而,我不确定这是否是唯一/最好/最蟒蛇的做法。看起来很尴尬。实际代码:

query = 'SELECT action.descr as "action", '\
    'role.id as role_id,'\
    'role.descr as role'\
    'FROM '\
    'public.role_action_def,'\
    'public.role,'\
    'public.record_def, '\
    'public.action'\
    'WHERE role.id = role_action_def.role_id AND'\
    'record_def.id = role_action_def.def_id AND'\
    'action.id = role_action_def.action_id AND'\
    'role_action_def.account_id = ' + account_id + ' AND'\
    'record_def.account_id=' + account_id + ' AND'\
    'def_id=' + def_id

当前回答

通常,我对多行注释/string使用list和join。

lines = list()
lines.append('SELECT action.enter code here descr as "action", ')
lines.append('role.id as role_id,')
lines.append('role.descr as role')
lines.append('FROM ')
lines.append('public.role_action_def,')
lines.append('public.role,')
lines.append('public.record_def, ')
lines.append('public.action')
query = " ".join(lines)

您可以使用任何字符串连接所有这些列表元素,如“\n”(换行符)或“,”(逗号)或“”(空格)。

其他回答

添加@Levon的答案。。。。

1.创建如下多行字符串:

paragraph = """this is a very
        long string if I had the
        energy to type more and more ..."""

print(paragraph)

输出:

'this is a very\n        long string if I had the\n        energy to type more and more ...'

此字符串将包含换行符和空格。因此,请移除它们。

2.使用正则表达式删除多余的空格

paragraph = re.sub('\s+', ' ', paragraph)
print(paragraph)

输出:

'this is a very long string if I had the energy to type more and more ...'

我认为当代码(例如,变量)缩进并且输出字符串应该是单行(没有换行)时,另一个选项更可读:

def some_method():

    long_string = """
A presumptuous long string
which looks a bit nicer
in a text editor when
written over multiple lines
""".strip('\n').replace('\n', ' ')

    return long_string

结合以下观点:

Levon或Jesse、Faheel和ddrscott

根据我的格式建议,您可以将查询写成:

query = ('SELECT'
             ' action.descr as "action"'
             ',role.id as role_id'
             ',role.descr as role'
         ' FROM'
             ' public.role_action_def'
             ',public.role'
             ',public.record_def'
             ',public.action'
         ' WHERE'
             ' role.id = role_action_def.role_id'
             ' AND'
             ' record_def.id = role_action_def.def_id'
             ' AND'
             ' action.id = role_action_def.action_id'
             ' AND'
             ' role_action_def.account_id = ?' # account_id
             ' AND'
             ' record_def.account_id = ?'      # account_id
             ' AND'
             ' def_id = ?'                     # def_id
         )

 vars = (account_id, account_id, def_id)     # A tuple of the query variables
 cursor.execute(query, vars)                 # Using Python's sqlite3 module

或类似:

vars = []
query = ('SELECT'
             ' action.descr as "action"'
             ',role.id as role_id'
             ',role.descr as role'
         ' FROM'
             ' public.role_action_def'
             ',public.role'
             ',public.record_def'
             ',public.action'
         ' WHERE'
             ' role.id = role_action_def.role_id'
             ' AND'
             ' record_def.id = role_action_def.def_id'
             ' AND'
             ' action.id = role_action_def.action_id'
             ' AND'
             ' role_action_def.account_id = '
                 vars.append(account_id) or '?'
             ' AND'
             ' record_def.account_id = '
                 vars.append(account_id) or '?'
             ' AND'
             ' def_id = '
                 vars.append(def_id) or '?'
         )

 cursor.execute(query, tuple(vars))  # Using Python's sqlite3 module

与“IN”和“vars.extend(options)”或“n_options(len(option))”一起使用可能很有趣,其中:

def n_options(count):
    return '(' + ','.join(count*'?') + ')'

或者从darkcaline那里得到的提示是,您可能仍然会在使用前导空格和分隔符以及命名占位符时出错:

SPACE_SEP = ' '
COMMA_SEP = ', '
AND_SEP   = ' AND '

query = SPACE_SEP.join((
    'SELECT',
        COMMA_SEP.join((
        'action.descr as "action"',
        'role.id as role_id',
        'role.descr as role',
        )),
    'FROM',
        COMMA_SEP.join((
        'public.role_action_def',
        'public.role',
        'public.record_def',
        'public.action',
        )),
    'WHERE',
        AND_SEP.join((
        'role.id = role_action_def.role_id',
        'record_def.id = role_action_def.def_id',
        'action.id = role_action_def.action_id',
        'role_action_def.account_id = :account_id',
        'record_def.account_id = :account_id',
        'def_id = :def_id',
        )),
    ))

vars = {'account_id':account_id,'def_id':def_id}  # A dictionary of the query variables
cursor.execute(query, vars)                       # Using Python's sqlite3 module

请参阅Cursor.execute-function的文档。

“这是[最Python]的方式!”-。。。

我发现,当构建长字符串时,您通常会做一些类似于构建SQL查询的事情,在这种情况下,这是最好的:

query = ' '.join((  # Note double parentheses. join() takes an iterable
    "SELECT foo",
    "FROM bar",
    "WHERE baz",
))

Levon的建议很好,但可能容易出错:

query = (
    "SELECT foo"
    "FROM bar"
    "WHERE baz"
)

query == "SELECT fooFROM barWHERE baz"  # Probably not what you want

我使用递归函数来构建复杂的SQL查询。此技术通常可用于构建大型字符串,同时保持代码可读性。

# Utility function to recursively resolve SQL statements.
# CAUTION: Use this function carefully, Pass correct SQL parameters {},
# TODO: This should never happen but check for infinite loops
def resolveSQL(sql_seed, sqlparams):
    sql = sql_seed % (sqlparams)
    if sql == sql_seed:
        return ' '.join([x.strip() for x in sql.split()])
    else:
        return resolveSQL(sql, sqlparams)

P.S.:看看很棒的python-sqlparse库,如果需要,可以打印SQL查询。