我有一個文件，其中包含多個存儲過程，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', $$ DECLARE a INT:= 0; b INT:= 0;BEGIN END; $$;我需要把它分成每個以---- Name:此外，各個文件應以架構名稱命名。在這種情況下，文件名應為 abc.sql、bcd.sql、cde.sql..到目前為止我試過這個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)
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TA貢獻1829條經驗 獲得超13個贊
嘗試這個：
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()
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