How to read a ZIpFile using a specific char as newline separator, instead of \n?

Question

I've got this code:

from zipfile import ZipFile
...
with ZipFile(flu) as zf:
    for file in zf.namelist():
        if not file.endswith('.sql'):
            continue
        with zf.open(file,"r") as f:
            ...
        for row_b in f:
            ...

I want to use the char ';' as newline separator when opening the zip file, instead of '\n'. In this way I can run the SQL statements included in the zip file also if they are formatted in more than one line.

I find out this documentation and this documentation and it seems that I can specify the newline separator when opening a stream. I cannot understand how to do it when opening a zip file.

Answer 1

There is no option to configure the line ending character for ZipExtFile (what zip.open(...) returns).

However, you can create your own buffered reader with a simple function:

from io import BytesIO
def delimited_read(f, sep, block_size=16384):
    buf = bytearray()
    while True:
        idx = buf.find(sep)
        while idx == -1:
            block = f.read(block_size)
            if block == b'':
                break
            start = len(buf)
            buf.extend(block)
            idx = buf.find(sep, start)
        if idx == -1:
            if len(buf) > 0:
                yield bytes(buf)
            return
        else:
            yield bytes(buf[0:idx])
            buf = buf[idx+1:]

And you can simply use it as:

for line in delimited_read(f, b';'):
  print(line)