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# MySQL - Find the lowest unique value in table

By : user2176038
Date : October 14 2020, 02:22 PM
hope this fix your issue Assume your table has the bid value stored in the field name bet, then below is a worked solution:
code :
``````SELECT * FROM YourTableName
WHERE bet = (SELECT MIN(bet) FROM YourTableName)
GROUP BY bet
HAVING COUNT(*) = 1;
``````

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## Find FIRST lowest value in table AND date on which that lowest value occurs

By : Ray Liu
Date : March 29 2020, 07:55 AM
Hope this helps If I have a table with the following schema: , Simple and fast solution:
code :
``````SELECT weight, entry_date
FROM   diet_watch
WHERE  entry_date BETWEEN '2001-01-02' AND '2001-01-06'
ORDER  BY 1, 2
LIMIT  1;
``````
``````SELECT weight, entry_date
FROM  (
SELECT weight, entry_date
,row_number() OVER (ORDER BY weight, entry_date) AS rn
FROM   diet_watch
WHERE  entry_date between '2001-01-02' and '2001-01-06'
) AS x
WHERE rn = 1;
``````
``````SELECT weight, entry_date
FROM   diet_watch
WHERE  entry_date BETWEEN '2001-01-02' AND '2001-01-06'
ORDER  BY weight, entry_date
FETCH  FIRST 1 ROWS ONLY;
``````

## Find the index of n unique rows with the lowest values

By : Bertalan Verebelyi
Date : March 29 2020, 07:55 AM
it fixes the issue First, get rid of the repeated values by making them Inf so that they won't be mistaken for being the lowest values:
code :
``````A1 = tril(A);
A1(A1==0) = Inf;
``````
``````[~,idx] = sort(A1(:));
[r,c] = ind2sub(size(A), idx(1:n));
``````
``````[~,idx] = sort(A(:));
[r,c] = ind2sub(size(A), idx);
rows = unique(r,'stable');
result = rows(1:n)
``````

## Find lowest value between 4 columns values from MySQL table

By : Kireeti Vallabhu
Date : March 29 2020, 07:55 AM

## Find the index of the lowest unique value in an array

By : Sonali Sharma
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , You could use LINQ and group the elements to get the count (uniqueness). Then find the lowest value.

## Django ORM: How to find all instances with unique names which has the lowest value?

By : Fbuk Knows Toomuch
Date : March 29 2020, 07:55 AM
help you fix your problem fetch the stores and then fetch the products from them, choose the first one after arranging them in ascending order
code :
``````stores = Store.objects.all()
for st in stores:
product = Product.objects.fiter(store=st).order_by('-price')[:1]
print(product)
``````