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PassFab PowerPoint Password Recovery 8.3.0 Cracked Fixed

PassFab PowerPoint Password Recovery 8.3.0 Cracked Fixed


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PassFab PowerPoint Password Recovery 8.3.0 Cracked

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Distinct with join

I have the following query. I want to select distinct entries that has a higher score than the best score of the entry. So I need to find the best score of the entry together with the account that has that score.
SELECT DISTINCT a
FROM entry e
INNER JOIN
(
SELECT max(s.score) score, e1.account
FROM entry e1
INNER JOIN score s ON e1.score = s.score
GROUP BY e1.account
) scores ON e.account = scores.account

This query is working but only if there is one single entry on account.
If there are multiple entries that have the same score and I would like to use distinct, how do I add it in there?

A:

I believe you just have to move your subquery into the SELECT clause:
SELECT DISTINCT a
FROM entry e
INNER JOIN
(
SELECT MAX(s.score) score, e1.account
FROM entry e1
INNER JOIN score s ON e1.score = s.score
GROUP BY e1.account
) scores ON e.account = scores.account

This should work in Oracle.

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Iterating in python, what does a=’|’.join([‘abc’,’def’,’xyz’,’pqr’]) print?

In the below snippet, what does the a=’|’.join([‘abc’,’def’,’xyz’,’pqr’]) print?
How is a=’|’.join([‘abc’,’def’,’xyz’,’pqr’]) evaluated?
a = ‘|’.join([‘abc’,’def’,’xyz’,’pqr’])
for i in a:
print(i, end=’ ‘)

A:

In Python, strings are immutable. Meaning, a new string object is not created. Whenever you change a value inside a string, it is actually changing the existing object.
This is the concept of immutable in python. For more understanding, you may read more on it.
In your case, a is a new string object, which is created. Now, a value of the string object ‘|’.join([‘abc’,’def’,’xyz’,’pqr’]) is assigned to a. Now, you are iterating over a.
for i in a:
print(i, end=’ ‘)

The above loop will iterate over the string object a. Each iteration, will print the character in the string object a.

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