edit or delete child or detail table on SQLFORM.grid with left join

[黄祥]

is it possible to edit or delete child or detail table on SQLFORM.grid with left join?

i've tried before it seems the edit is for the first table (parent or header), the child or detail table cannot be edited or deleted with left join in SQLFORM.grid.

e.g.

def report_production_result():

table_header = db.production_result_header

table_detail = db.production_result_detail

orderby = ~table_header.id

left = table_detail.on(table_header.id == table_detail.production_result_no)

fields = [

 # header

 table_header.id, table_header.production_result_no, table_header.production_result_date, 

 table_header.shift, table_header.origin, table_header.status, 

 # detail

 table_detail.stock, table_detail.quantity]

grid = SQLFORM.grid(table_header, fields = fields, left = left, orderby = orderby)

return locals()

thanks and best regards,

stifan

[Massimo Di Pierro]

Unfortunately not out of the box. You need a custom button that does a ajax query. 

[Fabiano Almeida]

Thanks Massimo, I will try this.

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[黄祥]

could anyone give an example to custom button in grid that does an ajax query?

another question is oot :

what is the best practice to define table for transaction that separated for header and detail? 

when use header and detail it good for normalisation but not good in operational, in case want to edit or delete it, when using 1 table for header and detail, the data is for header is repeated in the rows.

thanks and best regards,

stifan

[Massimo Di Pierro]

For example:

db.define_table('thing',Field('name'),Field('checked','boolean',readable=False))

def index():

   grid = SQLFORM.grid(db.thing, links=[lambda row: A('check',_onclick="jQuery.post('%s')" % URL('check',args=row.id))])

   return dict(grid=grid)

def check():

  db(db.thing.id==request.args(0)).update(checked=True)

[黄祥]

a, i c, thank you so much for your example massimo, that's why my old code not work because i'm using _href

e.g.

def index():

grid = SQLFORM.grid(db.thing, links=[lambda row: A('check',_onclick="jQuery.post('%s')" % URL('check',args=row.id))])

#grid = SQLFORM.grid(db.thing, links=[lambda row: A('check',_href=URL('check',args=row.id) ) ] ) # not work use _onclick = jQuery.post()

return dict(grid=grid)

[Massimo Di Pierro]

The A helper also has a A(..., callback=URL()) that posts to the callback. But I think _onclick is more explicit.