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Ticket Resolution Summary Owner Reporter
#168 invalid Appengine query models xi martinhoarantes@…

Reported by martinhoarantes@…, 4 years ago.


Hi, I´m using Google Appengine, language Python. I´d like to know how to query a model.

I´m using this code, but the output is miserable:

class GerarYAML(webapp.RequestHandler):
    def post(self):
        user = users.get_current_user() 
        if user: 
            formacaos = models.Formacao().all()
            experiencias = models.Experiencia().all()
            investigacaos = models.Investigacao().all() 
            distribuicaos = models.Distribuicao().all()
            docentes = models.Docente.all()
            docentes.filter('user =', users.get_current_user())
            for docente in docentes:
                for formacao in formacaos:
                    for experiencia in experiencias:
                        for investigacao in investigacaos:
                            for distribuicao in distribuicaos:
                                formacao.docente = docente.key()
                                'Nome: %s' % (docente.nome)
                                'Apelido: %s' % (docente.apelido)
                                'Unidade: %s' % (docente.unidade)
                                'Categoria: %s' % (docente.categoria)
                                'Regime: %s' % (docente.regime)
                                'Formacao: %s' % (formacao.formAno)
                                'Grau de Formacao: %s' % (formacao.formGrau)
                                'Area de Formacao: %s' % (formacao.formArea)
                                'Instituicao Academica: %s' % (formacao.formInstituicao)
                                'Classificacao Final: %s' % (formacao.formClassificacao)
                                yaml.dump(docente, sys.stdout)
                                yaml.dump(formacao, sys.stdout)  
                                yaml.dump(experiencia, sys.stdout)
                                yaml.dump(investigacao, sys.stdout)
                                yaml.dump(distribuicao, sys.stdout)  
                                self.response.headers['Content-Type'] = 'application/yaml'
                                self.response.headers['Content-Disposition'] = 'filename = my.yaml'

Can someone point out, how can i solve this issue?

I´d like to download a file automatically when user submit a button with .yaml format and more or less formatted.

#167 worksforme Please help saving dict to yaml xi anonymous

Reported by anonymous, 4 years ago.

Hi, is there any way of saving dict to a file in yaml format? I'm using:
import yaml
y = """
         - subject1:
            desc: msg1
         - subject2:
            desc: msg2

settings['root'][0]['subject1']['desc'] = "new msg"
print settings

f = open("temp", "w")
but failing miserably, contents of the temp file:
{'root': [{'subject1': {'desc': 'new msg'}}, {'subject2': {'desc': 'msg2'}}]}

expected result:
         - subject1:
            desc: new msg
         - subject2:
            desc: msg2

Is there anything like Any help greatly appreciated!

#166 wontfix PyYAML should use libyaml if libyaml available xi sgwong

Reported by sgwong, 4 years ago.


Currently PyYAML only use python implementation although the libyaml available. The following simple changes on should do: (I'm not sure whats the use of __with_libyaml__)

__version__ = '3.09'

    from cyaml import *
    __with_libyaml__ = True
    Loader = CLoader
    Dumper = CDumper
except ImportError:
    __with_libyaml__ = False 
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