Trying to push through a prop to the .like filtering option in supabase function. Due to .like needing % in value, im not sure how to push the prop through. Please see below.
const fetchData = async () => {
const { data, error } = await supabase
.from('sneaker')
.select()
.eq('brand', props.model?.brand)
.like('name', '%Yeezy%')
filteredList1.value = data
console.log(data)
}
I would like to have props.model?.name instead of '%Yeezy%'. I have tried adding in the prop like stated but does not work. Much appreciated.
CodePudding user response:
Have you tried concatenating the string?
.like('name', `%${props.model?.name}%`)