用soaplib在Django中开发SOAP Webservice

前面写过怎么利用suds来调用webservicePython调用基于https协议的SOAP WebService,这篇讲的是如何用soaplib开发SOAP WebService(最近发现国外开源社区里把json方式的别的Web服务也叫做WebService,叫法跟Java和.Net的约定叫法不太一样,这里加上SOAP以跟json格式的WebService区分开来)。

第一步,当然是安装问题了:

下载soaplib:

这是下载地址,我选了soaplib 2.0.0-beta2下载,因为接下来要用到一个djangosnippet是基于2.0的。

在开发环境(Mac OS X 10.7.4)上安装很顺利:


    $ python setup.py install

但在测试环境(CentOS)上却碰到了一点麻烦,执行上述安装时报错:error: Setup script exited with error: command 'gcc' failed with exit status 1。网上找到有人提出的解决方案是安装libxml2, libxml2-devel, libxslt-devel,使用yum安装之。

   $yum install libxml2
   $yum install libxml2-devel
   $yum install libxslt-devel
这里有个小技巧,一般查到debian下安装这类包时使用apt-get,而CentOS使用yum,当看到以apt-get的方式安装的时叫apt-get install libxml2-dev之类的,只要把把dev换成devel,就可以使用yum在centos或者redhat下安装了,但是这次我上了个当,网上查到好几个地方的libxslt安装都是apt-get install libxslt1-dev,但是yum包没有那个“1”。。。后来我去下载lxml2.3.4源码包,看到下面那一段话

* On **Linux** (and most other well-behaved operating systems), ``easy_install`` will manage to build the source distribution as long as libxml2 and libxslt are properly installed, including development packages, i.e. header files, etc. Use your package management tool to look for packages like ``libxml2-dev`` or ``libxslt-devel`` if the build fails, and make sure they are installed.

才去掉那个“1”,安装成功。

接下来去下载一个叫SOAP web service with soaplib 2.0的东东,因为最近刚过儿童节,风紧,据说好多org域名的境外网站貌似都不问青红皂白被墙了,把代码贴出如下:


from soaplib.core import Application
from soaplib.core.server.wsgi import Application as WSGIApplication
from django.http import HttpResponse


class DjangoSoapApp(WSGIApplication):
    """
    Generic Django view for creating SOAP web services (works with soaplib 2.0)

    Based on http://djangosnippets.org/snippets/2210/
    """

    csrf_exempt = True

    def __init__(self, services, tns):
        """Create Django view for given SOAP soaplib services and tns"""

        return super(DjangoSoapApp, self).__init__(Application(services, tns))

    def __call__(self, request):
        django_response = HttpResponse()

        def start_response(status, headers):
            django_response.status_code = int(status.split(' ', 1)[0])
            for header, value in headers:
                django_response[header] = value

        response = super(DjangoSoapApp, self).__call__(request.META, start_response)
        django_response.content = '\n'.join(response)

        return django_response

djangosnippets.org是个好网站,有很多很有用的django小部件,缺点是,容易莫名其妙被墙,以及Usage太简单,很难偷懒不阅读他的源码直接使用,这个snippet的Usage就只有一行字:

Usage is the same as before: my_soap_service = DjangoSoapApp([MySOAPService], __name__)

它之前的版本也非常简单,而且那些引用在这个版本里都已经移动位置了,不得已,只得去阅读soaplib2.0的源码,把相应的位置纠正过来。

我来写个比他详细点的Usage吧:

View.py里的代码:


#这几个引用,soablib2.0的位置跟0.9+之类的版本不一样了
from soaplib.core.model.primitive import Boolean, String
from soaplib.core.service import DefinitionBase, rpc

# the class with actual web methods
class MySOAPService(DefinitionBase):
    @rpc(String, String, _returns=Boolean)
    def Test(self, f1, f2):
        return True
    @rpc(String, _returns=String)
    def HelloWorld(self, name):
        return 'Hello %s!' %name


my_soap_service = DjangoSoapApp([MySOAPService], 'laonan.net')

urls.py里的代码:


    url(r'^my-soap-service/service', 'yourproject.yourapp.views.my_soap_service'),
    url(r'^my-soap-service/service.wsdl', 'yourproject.yourapp.views.my_soap_service'),

上面这两个地址一定要配对,最好到wsdl文档里去确认下soap:address节点的location属性,最早我配的时候,第一个url少配了service那一段,还以为是soaplib2是beta版有bug,怎么老找不到,查了半天才发现是这里配错了:(

Python WebService soaplib mac centos
Views (7989) Comments (2) 2012-05-31
Dig :
CSDN上有一篇一样的,还挂了原创,http://blog.csdn.net/wangqiaoshi/article/details/9531813 有分舵?
老楠 Reply to Dig :
从来没在csdn上写过东西
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