Django ORM拆分名称并搜索

Question

I have a model named Suburb. with one field called name.

name of the suburb can be mulitple words say "East west country". I want all the records whose part words starts with given string.

class Suburb(models.Model):
    name = models.CharField(_('suburb name'), blank=False, max_length=200)

search_string = "we"

# Give me all records whose part words starts with "search_string"

Suburb.objects.filter(...) # with ignore case ??? 

Example 
----------
1) "East west one"
2) "We east two"
3) "North south three"

result should be 1) and 2) 

I have a pythonic solution. but its performance is quite bad.

DB = MYSQL

Thanks in advance

Answer 1

Best Django solution without engaging Full Text Search or search engine would be to use icontains

So in your case

Suburb.objects.filter(name__icontains=search_string)

Edit : for start of each word use iregex:

For mysql [[:<:]] signifies start of word

So something like this would be solution

Suburb.objects.filter.(name__iregex=r'[[:<:]]' + re.escape(search_string)

Also note that following solution doesn't scale as there is wildcard regex

Answer 2

You can use a combination of istartswith and icontains with the Q object

search_string = 'we'
Sample.objects.filter(Q(name__istartswith=search_string) | Q(name__icontains=' ' + search_string))

With this solution I am assuming that your words are always delimited by a space character.

Using the iregex filter as another option which could be used but I have not created an example for.

Answer 3

Search by part word:

search_string = "we"
Suburb.objects.filter(name__icontains = search_string)
    Result:
    1) "East west one"
    2) "We east two"

Search from starting word:

search_string = "we"
Suburb.objects.filter(name__startswith = search_string)
    Result:
    2) "We east two"

Search exact work:

search_string = "we"
Suburb.objects.filter(name__iexact = search_string)
Result: no result
search_string = "We east two"
Result:
2) "We east two"