Python 字典更新序列元素 #0 的长度为 3；2 是必需的
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dictionary update sequence element #0 has length 3; 2 is required
提问by rindra
I want to add lines to the object account.bank.statement.linethrough other object But I get following error:
我想account.bank.statement.line通过其他对象向对象添加线条但出现以下错误：
"dictionary update sequence element #0 has length 3; 2 is required"
“字典更新序列元素#0 的长度为 3；需要 2”
Here is my code:
这是我的代码：
def action_account_line_create(self, cr, uid, ids):
res = False
cash_id = self.pool.get('account.bank.statement.line')
for exp in self.browse(cr, uid, ids):
company_id = exp.company_id.id
#statement_id = exp.statement_id.id
lines = []
for l in exp.line_ids:
lines.append((0, 0, {
'name': l.name,
'date': l.date,
'amount': l.amount,
'type': l.type,
'statement_id': exp.statement_id.id,
'account_id': l.account_id.id,
'account_analytic_id': l.analytic_account_id.id,
'ref': l.ref,
'note': l.note,
'company_id': l.company_id.id
}))
inv_id = cash_id.create(cr, uid, lines,context=None)
res = inv_id
return res
I changed it on that but then I ran into this error:
我改变了它，但后来我遇到了这个错误：
File "C:\Program Files (x86)\OpenERP 6.1-20121029-003136\Server\server\.\openerp\workflow\wkf_expr.py", line 68, in execute
File "C:\Program Files (x86)\OpenERP 6.1-20121029-003136\Server\server\.\openerp\workflow\wkf_expr.py", line 58, in _eval_expr
File "C:\Program Files (x86)\OpenERP 6.1-20121029-003136\Server\server\.\openerp\tools\safe_eval.py", line 241, in safe_eval
File "C:\Program Files (x86)\OpenERP 6.1-20121029-003136\Server\server\.\openerp\tools\safe_eval.py", line 108, in test_expr
File "<string>", line 0
^
SyntaxError: unexpected EOF while parsing
Code:
代码：
def action_account_line_create(self, cr, uid, ids, context=None):
res = False
cash_id = self.pool.get('account.bank.statement.line')
for exp in self.browse(cr, uid, ids):
company_id = exp.company_id.id
lines = []
for l in exp.line_ids:
res = cash_id.create ( cr, uid, {
'name': l.name,
'date': l.date,
'amount': l.amount,
'type': l.type,
'statement_id': exp.statement_id.id,
'account_id': l.account_id.id,
'account_analytic_id': l.analytic_account_id.id,
'ref': l.ref,
'note': l.note,
'company_id': l.company_id.id
}, context=None)
return res
采纳答案by MBarsi
This error raised up because you trying to update dictobject by using a wrong sequence (listor tuple) structure.
出现此错误是因为您尝试dict使用错误的序列 (list或tuple) 结构来更新对象。
cash_id.create(cr, uid, lines,context=None)trying to convert linesinto dict object:
cash_id.create(cr, uid, lines,context=None)尝试转换lines为 dict 对象：
(0, 0, {
'name': l.name,
'date': l.date,
'amount': l.amount,
'type': l.type,
'statement_id': exp.statement_id.id,
'account_id': l.account_id.id,
'account_analytic_id': l.analytic_account_id.id,
'ref': l.ref,
'note': l.note,
'company_id': l.company_id.id
})
Remove the second zero from this tuple to properly convert it into a dict object.
从此元组中删除第二个零以将其正确转换为 dict 对象。
To test it your self, try this into python shell:
要自己测试，请在 python shell 中尝试：
>>> l=[(0,0,{'h':88})]
>>> a={}
>>> a.update(l)
Traceback (most recent call last):
File "<pyshell#11>", line 1, in <module>
a.update(l)
ValueError: dictionary update sequence element #0 has length 3; 2 is required
>>> l=[(0,{'h':88})]
>>> a.update(l)
回答by Ian Michel
I was getting this error when I was updating the dictionary with the wrong syntax:
当我使用错误的语法更新字典时出现此错误：
Try with these:
试试这些：
lineItem.values.update({attribute,value})
instead of
代替
lineItem.values.update({attribute:value})
回答by AB Abhi
One of the fast ways to create a dict from equal-length tuples:
从等长元组创建 dict 的快速方法之一：
>>> t1 = (a,b,c,d)
>>> t2 = (1,2,3,4)
>>> dict(zip(t1, t2))
{'a':1, 'b':2, 'c':3, 'd':4, }
