小编Jos*_*eda的帖子

定义我的django表单时出现一个奇怪的错误.我收到错误:

__init__() got multiple values for keyword argument 'choices'

这种情况发生在TestForm和SpeciesForm上(引用如下); 基本上两种形式都带有'choices'关键字参数.永远不会显式调用init(),甚至还没有在视图中实例化表单.有一个ModelForm和一个普通Form.

from django import forms as f
from orders.models import *

class TestForm(f.Form):
    species = f.ChoiceField('Species', choices=Specimen.SPECIES)
    tests = f.MultipleChoiceField('Test', choices=Test.TESTS, widget=f.CheckboxSelectMultiple())
    dna_extraction = f.CharField('DNA extraction', help_text='If sending pre-extracted DNA, we require at least 900 ng')

class SpeciesForm(f.ModelForm):
    TYPE_CHOICES = (
        ('blood', 'Blood'),
        ('dna', 'Extracted DNA'),
    )
    dam_provided = f.BooleanField('DAM', help_text='Is dam for this specimen included in sample shipment?')
    sample_type = f.ChoiceField('Type of sample', choices=TYPE_CHOICES)
    dna_concentration = f.CharField('DNA …

尝试使用Zend_Db_Select. 任何指针？

SELECT 
  compounds.id as compounds_id,
  reactions.id as reactions_id, 
  reaction_compound.number as reaction_compound_number  
FROM compounds, reactions, reaction_compound 
WHERE  
  compounds.id IN (68,74,112) 
  AND compounds.id = reaction_compound.compound  
  AND reactions.id = reaction_compound.reaction;

具体来说，我遇到的一些问题是在 Zend 中执行多个表连接。我不确定如何使用他们的查询构建器跨多个表进行连接。

任何帮助表示赞赏！

J

嘿伙计们,对阿帕奇有疑问.我有一个已经重新设计的网站,但是我希望捕获人们可能已经收藏或从搜索引擎到新域名下的旧网站的所有"旧"链接.如何让apache只重定向404找不到旧网站？

TIA,

Ĵ