CODEWITHSUNDEEP
pythondjango
no_freedom
no_freedom

Reputation: 1961

Generate slug field in existing table

I have table with data. Is it possible slug field automatically generated on existing table? Or is there any other alternative? Thanks Here is my table

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Upvotes: 4

Views: 3369

Answers (3)

Tom
Tom

Reputation: 22841

I have a convenience model I use in all projects for these kinds of things. I think it's a good example of how to do what you want.

from django.template.defaultfilters import slugify


class NameAndSlug(models.Model):
    '''
    convenience model for things that are just names
    '''
    name = models.CharField(max_length=200, unique=True)
    slug = models.SlugField()

    def __unicode__(self):
        return self.name

    def save(self, *args, **kwargs):
        """
        Slugify name if it doesn't exist. IMPORTANT: doesn't check to see
        if slug is a dupe!
        """
        if not self.slug:
            self.slug = slugify(self.name)
        super(NameAndSlug, self).save(*args, **kwargs)

    class Meta:
        abstract = True

Upvotes: 4

Alasdair
Alasdair

Reputation: 308999

Using the slugify template filter, you can write a script, or loop through the objects in the shell.

>>> from django.template.defaultfilters import slugify
>>> for obj in MyModel.objects.all():
...     obj.slug = slugify(obj.title)
...     obj.save()

Upvotes: 18

Andrew Kozak
Andrew Kozak

Reputation: 1660

You can do this in MySQL like so (substitute the name of your table for "tableName"):

UPDATE `tableName` SET `slug`=LOWER(REPLACE( `title` , ' ' , '-' ));

This changes titles like "This Is A Title" to slugs like "this-is-a-title".

EDIT To handle parentheses and remove double spaces use:

UPDATE `tableName` SET `slug`=LOWER( REPLACE( REPLACE( REPLACE( REPLACE( `title` , '(' , '' ) , ')' , '' ) , '  ' , ' ' ) , ' ' , '-' ) );

Upvotes: 3

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