Question

我有类似下面的查询：

def connection():
    pcon = pg8000.connect(host='host', port=1234, user='user', password='password', database = 'database')
    return pcon, pcon.cursor()

pcon, pcur = connection()
query = """ SELECT * FROM db WHERE (db.foo LIKE 'string-%' OR db.foo LIKE 'bar-%')"""
db = pd.read_sql_query(query, pcon)

然而，当我尝试运行代码时，我得到了：

DatabaseError: '%'' not supported in a quoted string within the query string

我已经尝试用\和另外一个转义符号而没有运气。我怎样才能让pg8000正确地将其视为通配符？

Answer 1

“在Python中，%通常是指一个跟在字符串后面的变量。如果你想要一个文字百分号，那么你需要加倍它。%%”

- Source

LIKE 'string-%%'

否则，如果不起作用，PostgreSQL也支持underscores for pattern matching。

'abc' LIKE 'abc'    true
'abc' LIKE 'a%'     true
'abc' LIKE '_b_'    true

但是，如评论中所述，

pattern中的下划线（_）代表（匹配）任何单个字符;百分号（％）匹配零个或多个字符的任何序列

根据source code，问题似乎是%语句中LIKE之后的单引号。

if next_c == "%":
    in_param_escape = True
else:
    raise InterfaceError(
        "'%" + next_c + "' not supported in a quoted "
        "string within the query string")

因此，如果next_c == "'"代替next_c == "%"，那么您将收到错误

'%'' not supported in a quoted string within the query string

Answer 2

使用最新版本的 pg8000，% 中的 LIKE 应该不会有任何问题。例如：

>>> import pg8000.dbapi
>>>
>>> con = pg8000.dbapi.connect(user="postgres", password="cpsnow")
>>> cur = con.cursor()
>>> cur.execute("CREATE TEMPORARY TABLE book (id SERIAL, title TEXT)")
>>> for title in ("Ender's Game", "The Magus"):
...     cur.execute("INSERT INTO book (title) VALUES (%s)", [title])
>>>
>>> cur.execute("SELECT * from book WHERE title LIKE 'The %'")
>>> cur.fetchall()
([2, 'The Magus'],)

使用％wildcard和pg8000

2 个答案: