Python - 每次出现特定字符串时将文件拆分为多个文件

Question

我有一个文件，其中包含多个存储过程，如下所示。---- Name: abc(character varying); Type: FUNCTION; Schema: schema_name; Owner: ---CREATE FUNCTION schema_name.abc(schema_name character varying) RETURNS opaque    LANGUAGE plpgsql    AS '54:-1', $$    DECLARE a  INT:= 0; b  INT:= 0;BEGIN           END;$$;---- Name: bcd(); Type: FUNCTION; Schema: schema_name; Owner: ---CREATE FUNCTION schema_name.bcd() RETURNS opaque    LANGUAGE plpgsql    AS '-1', $$         DECLARE a  INT:= 0; b  INT:= 0;BEGIN       END;    $$;---- Name: cde(); Type: FUNCTION; Schema: schema_name; Owner: ---CREATE FUNCTION schema_name.cde() RETURNS opaque    LANGUAGE plpgsql    AS '-1', $$         DECLARE a  INT:= 0; b  INT:= 0;BEGIN       END;    $$;我需要把它分成每个以---- Name:此外，各个文件应以架构名称命名。在这种情况下，文件名应为 abc.sql、bcd.sql、cde.sql..到目前为止我试过这个token = 'CREATE FUNCTION'chunks = []current_chunk = []for line in open('procedures.txt'):   if line.startswith(token) and current_chunk:       chunks.append(current_chunk[:])      current_chunk = []   current_chunk.append(line)   print (current_chunk)chunks.append(current_chunk) 

Answer 1

尝试这个：

import os

path = r"D:\test"

token = '-- Name: '

chunks = {}

current_chunk = []

with open (os.path.join(path, "test.txt"), "r") as myfile:

    data=myfile.readlines()

for line in data:

    if line.startswith(token):

        start = line.find("token")+len(token)

        end = line.find("(")

        schema_name = line[start:end].strip()

        

        current_chunk = []

        current_chunk.append(line)

        chunks[schema_name] = current_chunk

    else:

        current_chunk.append(line)

print (chunks)

for name, storage in chunks.items():

    print(name)

    with open(os.path.join(path, name + '.sql'), 'w') as file:

        file.write(" ".join(storage))

        file.close()