上傳文件到PHP服務器使用Python客戶端和接收

Question

了Apache，PHP，Python的

這個問題已經被問在多個場合這個網站上，但是被很新的Python和PHP，這是我很難找出一個正確的方法來做到這一點。

目前 我的客戶端的部分看起來像（拉鍊文件，並將其發送）

1 #!/usr/bin/python 
    2 
    3 import os 
    4 import zipfile 
    5 import sys 
    6 import hashlib 
    7 from poster.encode import multipart_encode 
    8 from poster.streaminghttp import register_openers 
    9 import urllib2 
10 
11 def zip(src, dst): 
12  zf = zipfile.ZipFile("%s.zip" % (dst), "w", zipfile.ZIP_DEFLATED) 
13  abs_src = os.path.abspath(src) 
14  for dirname, subdirs, files in os.walk(src): 
15   for filename in files: 
16    absname = os.path.abspath(os.path.join(dirname, filename)) 
17    arcname = absname[len(abs_src) + 1:] 
18    print 'zipping %s as %s' % (os.path.join(dirname, filename), arcname) 
19    zf.write(absname, arcname) 
20  zf.close() 
21 
22 # zip the file using source to the destination.. can do some error checks here 
23 zip(sys.argv[1], sys.argv[2]) 
24 
25 # create md5 
26 md5 = hashlib.md5(open(sys.argv[2]+".zip", 'rb').read()).hexdigest() 
27 
28 # Register the streaming http handlers with urllib2 
29 register_openers() 
30 
31 filename=sys.argv[2]+".zip" 
32 
33 # headers contains the necessary Content-Type and Content-Length 
34 # datagen is a generator object that yields the encoded parameters 
35 datagen, headers = multipart_encode({ 
36  'type'  :  'zip', 
37  'name'  :  "hello.zip", 
38  'file'  :  open(filename) 
39 }) 
40 
41 # make a call 
42 request = urllib2.Request("http://localhost/upload.php", datagen, headers) 
43 
44 
45 # Actually do the request, and get the response 
46 print urllib2.urlopen(request).read() 

服務器側面看上去就像是

1 <?php 
    2 if ($_FILES["file"]["error"] > 0) { 
    3   echo "Error: " . $_FILES["file"]["error"] . "<br />"; 
    4  } else { 
    5   echo "Upload: " . $_FILES["file"]["name"] . "<br />"; 
    6   echo "Type: " . $_FILES["file"]["type"] . "<br />"; 
    7   echo "Size: " . ($_FILES["file"]["size"]/1024) . " Kb<br />"; 
    8   echo "Stored in: " . $_FILES["file"]["tmp_name"]; 
    9  } 
15 
16 
17 ?> 

當我跑我的Python文件，我得到這個輸出

上傳：
類型：
大小：0 KB
在存儲：

這種情況特別適用於較大的文件。問題的

第二部分，當我發送的小文件，我看到這個

Upload: upload.php<br />Type: application/x-httpd-php<br />Size: 0.4775390625 Kb<br />Stored in: /private/var/tmp/phpHapPaO 
Beautiful-iMac:~ agauravdeep$ open /private/var/tmp/ 
Beautiful-iMac:~ agauravdeep$ cd /private/var/tmp/phpHapPaO 
-bash: cd: /private/var/tmp/phpHapPaO: No such file or directory 
Beautiful-iMac:~ agauravdeep$ vi /private/var/tmp/phpHapPaO 
You have new mail in /var/mail/agauravdeep 

但什麼也沒有。正如在下面的評論中提到，我試圖更新php.ini.default但我沒有得到任何改變與更新的phpinfo即使重啓

Answer 1

PHP和代碼Apache指的是/etc/php.ini.default，並沒有php.ini。我最終創建了完全相同配置的php.ini。重新啓動後，我看到了更改。

Answer 2

這是我上傳的文件

import poster, urllib2, gzip, StringIO, sys 

def getUrllib2(upload = False, redirect = False): 
    if upload: 
     handlers = poster.streaminghttp.get_handlers() 
    else: 
     handlers = [] 
    err = HTTPError() 
    handlers.append(err) 
    handlers.append(XmlParseTest.getCookie()) 

    try: 
     opener = urllib2.build_opener(*handlers) 
    except Exception, e: 
     print err.getErrorMsg() 
     raise e 
    return opener,err 

def sendMultipartPost(url, params, files): 
    #params like this: {'id': '20010', 'type':'zip'} 
    #files like thie : {'uploadfile':'/tmp/testfile.zip'} 
    posterParams = [] 
    for key in params: 
     value = params[key] 
     try:     
      posterParams.append(poster.encode.MultipartParam(key, value)) 
     except Exception, e: 
      print e, key, value  
      raise e 

    for key in files: 
     value = files[key] 
     try: 
      value = value.encode(sys.getfilesystemencoding()) 
      posterParams.append(poster.encode.MultipartParam.from_file(key, value)) 
     except Exception, e: 
      print e, key, value  
      raise e 

    try:  
     datagen, headers = poster.encode.multipart_encode(posterParams) 
    except Exception, e: 
     print e, key, value  
     raise e 

    if headers is None: 
     headers = {}    

    try: 
     request = urllib2.Request(url, datagen, headers) 
     request.add_header('Accept-encoding', 'gzip') 
     request.add_header("Accept", "*/*")  
#    print request 
#    print request.get_data() 
     opener,err = getUrllib2(True, False) 
     response = opener.open(request) 
    except Exception, e: 
     print e, url, files 
     print err.getErrorMsg() 
     raise e 
    data = response.read() 
    '''if response is too large, use this 
    data = response.read(16*1024) 
    length = len(data) 
    _data = None 
    while length: 
     if _data: data += _data 
     _data = response.read(16*1024) 
     length = len(_data) 
    '''  
    if 'gzip' == response.headers.get('content-encoding', ''): 
     compressedstream = StringIO.StringIO(data) 
     gzipper = gzip.GzipFile(fileobj=compressedstream) 
     data =gzipper.read()    

    return data