Pythonda cgi orqali zip fayllarni (yoki boshqa ikkilik fayllarni) qanday joylashtirish mumkin?

Men Python va CGI bilan kichik veb-saytni kodlayapman, u erda foydalanuvchilar zip fayllarni yuklashlari va boshqa foydalanuvchilar tomonidan yuklangan fayllarni yuklab olishlari mumkin. Hozirda men zip-ni to'g'ri yuklay olaman, lekin foydalanuvchiga fayllarni to'g'ri yuborishda muammolarga duch keldim. Mening birinchi yondashuvim:

file = open('../../data/code/' + filename + '.zip','rb')

print("Content-type: application/octet-stream")
print("Content-Disposition: filename=%s.zip" %(filename))
print(file.read())

file.close()

Ammo tez orada men faylni ikkilik sifatida yuborishim kerakligini angladim, shuning uchun men harakat qildim:

print("Content-type: application/octet-stream")
print("Content-Disposition: filename=%s.zip" %(filename))
print('Content-transfer-encoding: base64\r')
print( base64.b64encode(file.read()).decode(encoding='UTF-8') )

Va uning turli xil variantlari. Bu shunchaki ishlamaydi; Apache "skriptdan noto'g'ri shakllangan sarlavha" xatosini ko'taradi, shuning uchun faylni boshqa yo'l bilan kodlashim kerak deb o'ylayman.


python-3.x python binary zip cgi
person dvilela    schedule 20.08.2013    source manba

Javoblar (3)


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Sarlavhalardan keyin boʻsh qatorni chop etishingiz kerak va Content-disposition sarlavhasida (attachment) turi yoʻq:

print("Content-type: application/octet-stream")
print("Content-Disposition: attachment; filename=%s.zip" %(filename))
print()

Olingan faylni yuklashning yanada samarali usulidan ham foydalanishni xohlashingiz mumkin; ma'lumotlarni sys.stdout.buffer ga nusxalash uchun shutil.copyfileobj() dan foydalaning:

from shutil import copyfileobj
import sys

print("Content-type: application/octet-stream")
print("Content-Disposition: attachment; filename=%s.zip" %(filename))
print()

with open('../../data/code/' + filename + '.zip','rb') as zipfile:
    copyfileobj(zipfile, sys.stdout.buffer)

Hech qanday holatda ikkilik ma'lumotlar uchun print() dan foydalanmaslik kerak; faqat b'...' bayt literal sintaksisini olasiz. sys.stdout.buffer ob'ekti asosiy ikkilik kiritish-chiqarish buferidir, ikkilik ma'lumotlarni to'g'ridan-to'g'ri unga nusxalash.

person Martijn Pieters    schedule 20.08.2013
comment
Men Python 2.7 dan foydalanmoqdaman, bu farq qilishi mumkin, lekin bu kodni ishlashi uchun men oxirgi qatorni o'zgartirishim kerak edi copyfileobj(zipfile, sys.stdout). - person Daniel Griscom; 19.07.2015
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@DanielGriscom: ha, bu butunlay to'g'ri; Mening javobim Python 3 uchun yozilgan bo'lib, unda io kutubxonasi barcha kiritish-chiqarish bilan shug'ullanadi va sys.stdout.buffer asosiy io.BufferedIOBase tatbiq, (kodlash) io.TextIOBase; Python 2 da sys.stdout fayl obyekti bevosita kodlangan baytlarni qabul qiladi. - person Martijn Pieters; 20.07.2015
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Python 2 da ishlagan holda, men uchinchi qatordagi print() ni olib tashlashim kerak edi, shuningdek, yuqorida @DanielGriscom taklif qilganidek, oxirgi qatorni copyfileobj(zipfile, sys.stdout) ga o'zgartirishim kerak edi. - person eeScott; 19.01.2020
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@eeScott ha, chunki bu savol maxsus python-3.x bilan belgilangan. Python 2 da sys.stdout oddiy Python 2 fayl obyekti, io.TextIOWrapper() obyekti emas. - person Martijn Pieters; 19.01.2020
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Rozi bo'ldim, men buni keyingi odam uchun hujjatlashtirmoqchi edim, chunki bu savolga python 2 ga moslashtirilgan odamlar unchalik ko'p emas... - person eeScott; 19.01.2020

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Sarlavha noto'g'ri tuzilgan, chunki ba'zi sabablarga ko'ra Python faylni yuborganidan keyin uni yuboradi.

Nima qilishingiz kerak bo'lsa, sarlavhadan keyin darhol stdout-ni o'chiring:

sys.stdout.flush()

Keyin fayl nusxasini qo'ying

person Alecz    schedule 16.06.2014
comment
Bu yaxshi javob ekanligiga ishonchim komil emas, lekin biroz ko'proq tushuntirish yaxshi bo'lishi mumkin. - person Aaron Hall; 16.06.2014

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Bu men uchun ishlagan narsa, men Apache2 ni ishga tushiryapman va ushbu skriptni cgi orqali yuklayapman. Python 3 mening tilim.

Birinchi qatorni python 3 bin yo'li bilan almashtirishingiz kerak bo'lishi mumkin.

#!/usr/bin/python3
import cgitb
import cgi
from zipfile import ZipFile
import sys

# Files to include in archive
source_file = ["/tmp/file1.txt", "/tmp/file2.txt"]

# Name and Path to our zip file.
zip_name = "zipfiles.zip"
zip_path = "/tmp/{}".format(zip_name)

with ZipFile( zip_path,'w' ) as zipFile:
    for f in source_file:
        zipFile.write(f);

# Closing File.
zipFile.close()

# Setting Proper Header.
print ( 'Content-Type:application/octet-stream; name="{}"'.format(zip_name) );
print ( 'Content-Disposition:attachment; filename="{}"\r\n'.format(zip_name) );

# Flushing Out stdout.
sys.stdout.flush()

bstdout = open(sys.stdout.fileno(), 'wb', closefd=False)
file = open(zip_path,'rb')
bstdout.write(file.read())
bstdout.flush()
person Community    schedule 21.11.2016