我是PhoneGap/Cordova-CLI的新手(和在线构建).
我在互联网上搜索并研究了如何使用Awesome Phonegap开始开发应用程序.他们说它很容易设置......!
我正在使用Winows 7 Ultimate 64位
我安装了
我根据需要设置了我的Java_home,ANT_Home和PATH变量.
我的ANT_HOME路径是:C:\ phonegap\ant \(即%ANT_HOME%/ bin/..)
然后我去命令提示(我使用ConEmu 64位)并输入以下命令
C:\phonegap>phonegap create phonegapone come.myapps.phonegapone phonegapone
[phonegap] created project at C:\phonegap\phonegapone
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项目创建!然后尝试了以下命令
C:\phonegap\phonegapone>phonegap build android
[phonegap] detecting Android SDK environment...
[phonegap] using the local environment
[phonegap] adding the Android platform...
[error] An error occured during creation of android sub-project. Creating Cordova project for the Android platform:
Run Code Online (Sandbox Code Playgroud)我不明白为什么我会收到这个错误
[error]在创建android子项目期间发生错误.为Android平台创建Cordova项目:
请帮帮我...
欢迎任何链接到适当的文档或维基.
PS:
最初我已经开始使用Phonegap …
我遇到了一种情况,我的scrapy代码在从命令行使用时工作正常,但是当我在部署(scrapy-deploy)和使用scrapyd api进行调度后使用相同的蜘蛛时,它会在"scrapy.extensions.feedexport.FeedExporter"中抛出错误"上课.
1."open_spider"信号错误
2016-05-14 12:09:38 [scrapy] INFO: Spider opened
2016-05-14 12:09:38 [scrapy] ERROR: Error caught on signal handler: <bound method ?.open_spider of <scrapy.extensions.feedexport.FeedExporter object at 0x7fafb1ce4a90>>
Traceback (most recent call last):
File "/home/jonsnow/venv/scrapy1/lib/python2.7/site-packages/twisted/internet/defer.py", line 150, in maybeDeferred
result = f(*args, **kw)
File "/home/jonsnow/venv/scrapy1/lib/python2.7/site-packages/pydispatch/robustapply.py", line 55, in robustApply
return receiver(*arguments, **named)
File "/home/jonsnow/venv/scrapy1/lib/python2.7/site-packages/scrapy/extensions/feedexport.py", line 185, in open_spider
uri = self.urifmt % self._get_uri_params(spider)
TypeError: float argument required, not dict
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2."item_scraped"信号错误:
2016-05-14 12:09:49 [scrapy] DEBUG: Scraped from <200 https://someurl.>
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