Django group options list
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1
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Using Django, I have the following working example to make a list with group option:
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
First Method: I tried to make it dynamic using the hard way:
sub_item =''
grouped_items=''
for k in Field.objects.filter(farm=farm):
print (k.field_name )
for o in Crop.objects.filter(farming_year=farming_year, field=k ):
sub_item = sub_item + '('+ str(o.id) + ',"' +o.crop_name + '"),'
grouped_items= grouped_items + '("' + k.field_name + '", (' + sub_item + ')),'
sub_item=''
full_list_items= '[' + grouped_items + ']'
print (full_list_items)
self.fields["crop"].choices = full_list_items
print(full_list_items )
I always get the following error:
not enough values to unpack (expected 2, got 1)
Second method: To test my code I copied the print(full_list_items )'s results from the terminal and assigned it directly to the crop list as shown below, and it works again!!!
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
the question here why I cannot get the required results when I use the first method and I did when I use the second one!!
python django
asked Nov 12 at 8:55
Yousef
604
add a comment |
up vote
1
down vote
favorite
Using Django, I have the following working example to make a list with group option:
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
First Method: I tried to make it dynamic using the hard way:
sub_item =''
grouped_items=''
for k in Field.objects.filter(farm=farm):
print (k.field_name )
for o in Crop.objects.filter(farming_year=farming_year, field=k ):
sub_item = sub_item + '('+ str(o.id) + ',"' +o.crop_name + '"),'
grouped_items= grouped_items + '("' + k.field_name + '", (' + sub_item + ')),'
sub_item=''
full_list_items= '[' + grouped_items + ']'
print (full_list_items)
self.fields["crop"].choices = full_list_items
print(full_list_items )
I always get the following error:
not enough values to unpack (expected 2, got 1)
Second method: To test my code I copied the print(full_list_items )'s results from the terminal and assigned it directly to the crop list as shown below, and it works again!!!
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
the question here why I cannot get the required results when I use the first method and I did when I use the second one!!
python django
asked Nov 12 at 8:55
Yousef
604
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Using Django, I have the following working example to make a list with group option:
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
First Method: I tried to make it dynamic using the hard way:
sub_item =''
grouped_items=''
for k in Field.objects.filter(farm=farm):
print (k.field_name )
for o in Crop.objects.filter(farming_year=farming_year, field=k ):
sub_item = sub_item + '('+ str(o.id) + ',"' +o.crop_name + '"),'
grouped_items= grouped_items + '("' + k.field_name + '", (' + sub_item + ')),'
sub_item=''
full_list_items= '[' + grouped_items + ']'
print (full_list_items)
self.fields["crop"].choices = full_list_items
print(full_list_items )
I always get the following error:
not enough values to unpack (expected 2, got 1)
Second method: To test my code I copied the print(full_list_items )'s results from the terminal and assigned it directly to the crop list as shown below, and it works again!!!
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
the question here why I cannot get the required results when I use the first method and I did when I use the second one!!
python django
asked Nov 12 at 8:55
Yousef
604
Using Django, I have the following working example to make a list with group option:
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
First Method: I tried to make it dynamic using the hard way:
sub_item =''
grouped_items=''
for k in Field.objects.filter(farm=farm):
print (k.field_name )
for o in Crop.objects.filter(farming_year=farming_year, field=k ):
sub_item = sub_item + '('+ str(o.id) + ',"' +o.crop_name + '"),'
grouped_items= grouped_items + '("' + k.field_name + '", (' + sub_item + ')),'
sub_item=''
full_list_items= '[' + grouped_items + ']'
print (full_list_items)
self.fields["crop"].choices = full_list_items
print(full_list_items )
I always get the following error:
not enough values to unpack (expected 2, got 1)
Second method: To test my code I copied the print(full_list_items )'s results from the terminal and assigned it directly to the crop list as shown below, and it works again!!!
self.fields["crop"].choices = [( "First", ( ('1','aaa'), ('2','bbb'), ) ), ( "Second", ( ('3','ccc'), ) ), ]
the question here why I cannot get the required results when I use the first method and I did when I use the second one!!
python django
python django
asked Nov 12 at 8:55
Yousef
604
asked Nov 12 at 8:55
Yousef
604
asked Nov 12 at 8:55
Yousef
604
asked Nov 12 at 8:55
Yousef
604
asked Nov 12 at 8:55
Yousef
604
604
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You're building up a single string, not a list of tuples. I don't know why you are doing that. Instead of all this string concatenation, you should keep things as Python data structures:
grouped_items =
for k in Field.objects.filter(farm=farm):
sub_item =
for o in Crop.objects.filter(farming_year=farming_year, field=k):
sub_item.append((str(o.id), o.crop_name))
grouped_items.append((k.field_name, sub_item))
self.fields["crop"].choices = grouped_items
Note that the inner loop could be rewritten as a list comprehension, and simplified further by using the reverse relation:
sub_item = [(str(o.id), o.crop_name) for o in k.crop_set.filter(farming_year = farming_year)]
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
I used a list of tuples before and it gave me the same error, for sure I did mistake. Your answer works for me, Thanks
– Yousef
Nov 12 at 9:18
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You're building up a single string, not a list of tuples. I don't know why you are doing that. Instead of all this string concatenation, you should keep things as Python data structures:
grouped_items =
for k in Field.objects.filter(farm=farm):
sub_item =
for o in Crop.objects.filter(farming_year=farming_year, field=k):
sub_item.append((str(o.id), o.crop_name))
grouped_items.append((k.field_name, sub_item))
self.fields["crop"].choices = grouped_items
Note that the inner loop could be rewritten as a list comprehension, and simplified further by using the reverse relation:
sub_item = [(str(o.id), o.crop_name) for o in k.crop_set.filter(farming_year = farming_year)]
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
I used a list of tuples before and it gave me the same error, for sure I did mistake. Your answer works for me, Thanks
– Yousef
Nov 12 at 9:18
add a comment |
up vote
1
down vote
accepted
You're building up a single string, not a list of tuples. I don't know why you are doing that. Instead of all this string concatenation, you should keep things as Python data structures:
grouped_items =
for k in Field.objects.filter(farm=farm):
sub_item =
for o in Crop.objects.filter(farming_year=farming_year, field=k):
sub_item.append((str(o.id), o.crop_name))
grouped_items.append((k.field_name, sub_item))
self.fields["crop"].choices = grouped_items
Note that the inner loop could be rewritten as a list comprehension, and simplified further by using the reverse relation:
sub_item = [(str(o.id), o.crop_name) for o in k.crop_set.filter(farming_year = farming_year)]
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
I used a list of tuples before and it gave me the same error, for sure I did mistake. Your answer works for me, Thanks
– Yousef
Nov 12 at 9:18
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You're building up a single string, not a list of tuples. I don't know why you are doing that. Instead of all this string concatenation, you should keep things as Python data structures:
grouped_items =
for k in Field.objects.filter(farm=farm):
sub_item =
for o in Crop.objects.filter(farming_year=farming_year, field=k):
sub_item.append((str(o.id), o.crop_name))
grouped_items.append((k.field_name, sub_item))
self.fields["crop"].choices = grouped_items
Note that the inner loop could be rewritten as a list comprehension, and simplified further by using the reverse relation:
sub_item = [(str(o.id), o.crop_name) for o in k.crop_set.filter(farming_year = farming_year)]
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
You're building up a single string, not a list of tuples. I don't know why you are doing that. Instead of all this string concatenation, you should keep things as Python data structures:
grouped_items =
for k in Field.objects.filter(farm=farm):
sub_item =
for o in Crop.objects.filter(farming_year=farming_year, field=k):
sub_item.append((str(o.id), o.crop_name))
grouped_items.append((k.field_name, sub_item))
self.fields["crop"].choices = grouped_items
Note that the inner loop could be rewritten as a list comprehension, and simplified further by using the reverse relation:
sub_item = [(str(o.id), o.crop_name) for o in k.crop_set.filter(farming_year = farming_year)]
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
answered Nov 12 at 8:59
Daniel Roseman
442k41573629
442k41573629
I used a list of tuples before and it gave me the same error, for sure I did mistake. Your answer works for me, Thanks
– Yousef
Nov 12 at 9:18
add a comment |
I used a list of tuples before and it gave me the same error, for sure I did mistake. Your answer works for me, Thanks
– Yousef
Nov 12 at 9:18
I used a list of tuples before and it gave me the same error, for sure I did mistake. Your answer works for me, Thanks
– Yousef
Nov 12 at 9:18
I used a list of tuples before and it gave me the same error, for sure I did mistake. Your answer works for me, Thanks
– Yousef
Nov 12 at 9:18
add a comment |
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