You should be able to query it as a ForeignKey
Jobs.objects.filter(names__is_a_problem=True).distinct()
(distinct to prevent repeated jobs with more than one worker problem)
On Wed, Dec 23, 2015 at 9:31 AM, Jonty Needham <jontyneedham@gmail.com> wrote:
--is_a_problem = Trueclass Workers(modesl.Model):I have a model of the form:class Jobs(models.Model):
names = models.ManyToManyField(Workers)And I want the set of jobs that are done by workers who are a problem.Something likeJobs.objects.filter(names__is_contained_in=Workers.objects.filter(is_a_problem=True))Obviously is_contained_in is not a valid lookup, so I would like to know how to do this please.ThanksJonty
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