Question

Apache，Php，Python

这个问题已经在这个网站上多次被问过，但对于python和php都是新手，我很难找到正确的方法来做到这一点。

目前我的客户端部分看起来像（压缩文件并发送）

  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()

服务器端看起来像

  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 ?>

当我运行我的python文件时，我得到了这个输出

上传：
类型：
尺寸：0 Kb
存储于：

这特别适用于较大的文件。

问题的第二部分，当我发送小文件时，我看到了这个

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

但那里什么都没有。正如下面的评论中提到的，我已经尝试更新php.ini.default，但即使重新启动我也没有使用phpinfo更新任何更改

Answer 1

这是我上传文件的代码

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

Answer 2

Php和Apache指的是/etc/php.ini.default，没有php.ini。我最终创建了完全相同配置的php.ini。重启后，我看到了变化。

使用python客户端将文件上传到php服务器并接收

2 个答案: